Photoelectric Effect Complete Notes | Kumar Physics Classes
e⁻Dual Nature · Chapter 01

Photoelectric Effect

Master photon emission, threshold frequency, stopping potential and every important photoelectric graph.

CBSENEETJEE MainJEE AdvancedIB PhysicsIGCSEA-Level
hν = Φ + Kmax
1. Photoelectric Effect
Hertz: UV assists spark dischargeHallwachs: charged zinc loses electrons
Early experiments revealed light-induced electron emission from metal surfaces.

Definition

Photoelectric effect is the emission of electrons from a material surface when electromagnetic radiation of sufficient frequency falls on it.

Energy Transfer

One photon transfers its energy hν to one electron. Part overcomes the work function and the remainder becomes kinetic energy.

Instantaneous Emission

Energy is delivered in a single photon-electron interaction, so no classical accumulation time is required.

Historical Experiments

Hertz observed UV-assisted spark discharge; Hallwachs observed loss of negative charge from illuminated zinc.

2. Photoelectric Emission

Work Function

Φ = hν₀

Minimum surface escape energy; depends on material and surface condition.

Threshold Frequency

ν₀ = Φ/h

No emission occurs for ν < ν₀, regardless of intensity.

Threshold Wavelength

λ₀ = c/ν₀ = hc/Φ

Emission requires λ ≤ λ₀.

Electron Escape

Surface electrons need less energy to escape than deeply bound electrons; emitted electrons therefore have a range of kinetic energies.

3. Experimental Setup
Cathode CAnode AMonochromatic lightPhotoelectronsμAVariable batteryReverse polarity → retarding potential
Evacuated photoelectric tube with illuminated cathode, collecting anode, variable potential and microammeter.

The variable battery can accelerate electrons toward the anode or reverse the field to retard them. The microammeter measures collected photoelectric current in the evacuated tube.

4. Experimental Observations

Threshold

Every cathode material has a minimum frequency ν₀.

No Time Lag

Emission begins essentially as soon as suitable light arrives.

Intensity

Higher intensity increases current by increasing photon flux.

Frequency

Higher frequency increases Kmax and stopping potential.

Energy Range

Electrons emerge with energies from nearly zero up to Kmax.

Saturation

At sufficiently positive anode potential, all emitted electrons are collected.

5. Threshold Frequency

Definition

The least frequency capable of liberating electrons from a given surface.

ν₀ = Φ/h

Threshold Wavelength

λ₀ = hc/Φ

For λ > λ₀, emission is impossible.

Example

For Φ = 2.0 eV, λ₀ = 1240/2.0 = 620 nm.

Concept Trap

Increasing intensity below ν₀ only adds more low-energy photons; it cannot cause emission.

6. Stopping Potential

Einstein Equation

hν = Φ + KmaxKmax = hν − Φ

Stopping Relation

Kmax = eV₀hν = Φ + eV₀

Physical Meaning

V₀ is the reverse potential just sufficient to prevent even the fastest photoelectrons from reaching the anode.

Dependence

V₀ rises linearly with frequency and is independent of intensity.

7. Saturation Current
Light intensitySaturation current
Saturation current is directly proportional to incident intensity above threshold frequency.

Saturation current is reached when every emitted photoelectron is collected. It increases with photon flux and therefore with intensity, but its ideal value is not controlled by stopping potential.

8. Effect of Intensity
Applied potentialPhotoelectric currentI1I2I3−V₀
Increasing intensity raises saturation current; stopping potential remains unchanged.

At fixed ν > ν₀, greater intensity means more photons per second, more emitted electrons and larger saturation current. Photon energy, Kmax and V₀ stay unchanged.

9. Effect of Frequency
Applied potentialPhotoelectric current−V1−V2−V3Same saturation current
Increasing frequency makes stopping potential more negative while saturation current stays fixed for constant intensity.

At fixed intensity, increasing frequency increases photon energy and makes the stopping potential larger in magnitude. In the idealised comparison, saturation current remains the same.

10. All Important Graphs
Applied potentialPhotoelectric currentI1I2I3−V₀
Increasing intensity raises saturation current; stopping potential remains unchanged.
Applied potentialPhotoelectric current−V1−V2−V3Same saturation current
Increasing frequency makes stopping potential more negative while saturation current stays fixed for constant intensity.
ν₀Slope = hFrequency νKmaxY-intercept = −Φ
Kmax = hν − Φ: slope h and threshold-frequency x-intercept ν₀.
ν₀Slope = h/eFrequency νV₀
V₀ = (h/e)ν − Φ/e; the intercept on the frequency axis is ν₀.
Light intensitySaturation current
Saturation current is directly proportional to incident intensity above threshold frequency.
11. Laws of Photoelectric Emission

Law 1

For each metal, emission occurs only when ν ≥ ν₀.

Law 2

Photoelectric current is proportional to incident intensity above threshold.

Law 3

Kmax and stopping potential depend on frequency, not intensity.

Law 4

Emission begins without measurable time lag.

Law 5

At fixed frequency and intensity, current reaches saturation at sufficient positive potential.

50 Conceptual Questions
C1. What is photoelectric emission?
Answer: Electron emission caused by incident electromagnetic radiation.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C2. Why is a clean metal surface preferred?
Answer: It avoids extra work-function changes from contamination.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C3. What is work function?
Answer: Minimum energy needed to remove a surface electron.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C4. What is threshold frequency?
Answer: Minimum incident frequency capable of emission.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C5. What is threshold wavelength?
Answer: Maximum incident wavelength capable of emission.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C6. Why is emission instantaneous?
Answer: A photon transfers its energy in a single interaction.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C7. Why does intensity affect current?
Answer: More photons per second release more electrons per second.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C8. Why does intensity not affect Kmax?
Answer: Each photon still has the same energy hν.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C9. Why does frequency affect Kmax?
Answer: Photon energy increases with ν.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C10. What is stopping potential?
Answer: Minimum reverse potential that stops the fastest electrons.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C11. What is saturation current?
Answer: Electron emission caused by incident electromagnetic radiation.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C12. Can sub-threshold light emit electrons at high intensity?
Answer: It avoids extra work-function changes from contamination.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C13. Why is ultraviolet effective for many metals?
Answer: Minimum energy needed to remove a surface electron.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C14. What happens when frequency equals threshold frequency?
Answer: Minimum incident frequency capable of emission.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C15. What happens above threshold frequency?
Answer: Maximum incident wavelength capable of emission.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C16. What is the meaning of Kmax?
Answer: A photon transfers its energy in a single interaction.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C17. Why do emitted electrons have a range of energies?
Answer: More photons per second release more electrons per second.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C18. What does the Kmax–ν slope represent?
Answer: Each photon still has the same energy hν.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C19. What does the V₀–ν slope represent?
Answer: Photon energy increases with ν.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C20. How is work function obtained graphically?
Answer: Minimum reverse potential that stops the fastest electrons.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C21. Why is current zero at stopping potential?
Answer: Electron emission caused by incident electromagnetic radiation.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C22. Does stopping potential depend on intensity?
Answer: It avoids extra work-function changes from contamination.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C23. Does saturation current depend on frequency?
Answer: Minimum energy needed to remove a surface electron.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C24. What is photon momentum?
Answer: Minimum incident frequency capable of emission.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C25. How does photon energy depend on wavelength?
Answer: Maximum incident wavelength capable of emission.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C26. What did Hertz observe?
Answer: A photon transfers its energy in a single interaction.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C27. What did Hallwachs observe?
Answer: More photons per second release more electrons per second.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C28. Why did classical wave theory fail?
Answer: Each photon still has the same energy hν.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C29. What is Einstein's key assumption?
Answer: Photon energy increases with ν.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C30. Where is photon energy spent?
Answer: Minimum reverse potential that stops the fastest electrons.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C31. What is retarding potential?
Answer: Electron emission caused by incident electromagnetic radiation.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C32. What is accelerating potential?
Answer: It avoids extra work-function changes from contamination.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C33. Why does current saturate?
Answer: Minimum energy needed to remove a surface electron.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C34. What happens if anode area increases?
Answer: Minimum incident frequency capable of emission.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C35. How does surface oxidation affect emission?
Answer: Maximum incident wavelength capable of emission.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C36. Which metals have low work function?
Answer: A photon transfers its energy in a single interaction.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C37. Why are alkali metals photosensitive?
Answer: More photons per second release more electrons per second.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C38. Does frequency change at a medium boundary?
Answer: Each photon still has the same energy hν.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C39. How is threshold wavelength related to work function?
Answer: Photon energy increases with ν.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C40. What is one electron volt?
Answer: Minimum reverse potential that stops the fastest electrons.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C41. How is Planck constant measured by this experiment?
Answer: Electron emission caused by incident electromagnetic radiation.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C42. What is the role of the microammeter?
Answer: It avoids extra work-function changes from contamination.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C43. Why is the tube evacuated?
Answer: Minimum energy needed to remove a surface electron.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C44. What is dark current?
Answer: Minimum incident frequency capable of emission.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C45. How does cathode material alter graphs?
Answer: Maximum incident wavelength capable of emission.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C46. Can every photon eject an electron?
Answer: A photon transfers its energy in a single interaction.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C47. Why is emission a one-photon process at ordinary intensity?
Answer: More photons per second release more electrons per second.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C48. What proves particle nature?
Answer: Each photon still has the same energy hν.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C49. Why is photoelectric current not ordinary conduction current?
Answer: Photon energy increases with ν.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
C50. What are the main laws of photoelectric emission?
Answer: Minimum reverse potential that stops the fastest electrons.
Explanation: Use Einstein's photon equation hν = Φ + Kmax and distinguish photon number from photon energy.
12. 55 Solved Numericals
NEET

1. Numerical

Light frequency is 5.0×10¹⁴ Hz and work function is 2 eV. Find Kmax.

Answer: 0.07 eV

Solution: Kmax = hν − Φ = 0.07 eV.

NEET

2. Numerical

A metal has work function 2.0 eV. Find threshold frequency.

Answer: 4.835e+14 Hz

Solution: ν₀ = Φ/h after converting eV to joules.

NEET

3. Numerical

Find threshold wavelength for work function 3.0 eV.

Answer: 413.3 nm

Solution: λ₀ = hc/Φ; using hc = 1240 eV·nm gives 413.3 nm.

NEET

4. Numerical

Maximum kinetic energy is 2.5 eV. Find stopping potential.

Answer: 2.5 V

Solution: eV₀ = Kmax, so the numerical value in volts equals the energy in eV.

NEET

5. Numerical

Find photon energy for wavelength 380 nm.

Answer: 3.26 eV

Solution: E = hc/λ = 1240/380 = 3.26 eV.

NEET

6. Numerical

A metal has ν₀ = 4.0×10¹⁴ Hz and stopping potential 1.75 V. Estimate incident frequency.

Answer: 8.23×10¹⁴ Hz

Solution: hν = hν₀ + eV₀, so ν = ν₀ + eV₀/h.

NEET

7. Numerical

Light frequency is 5.6×10¹⁴ Hz and work function is 2 eV. Find Kmax.

Answer: 0.32 eV

Solution: Kmax = hν − Φ = 0.32 eV.

NEET

8. Numerical

A metal has work function 2.4 eV. Find threshold frequency.

Answer: 5.803e+14 Hz

Solution: ν₀ = Φ/h after converting eV to joules.

NEET

9. Numerical

Find threshold wavelength for work function 3.0 eV.

Answer: 413.3 nm

Solution: λ₀ = hc/Φ; using hc = 1240 eV·nm gives 413.3 nm.

NEET

10. Numerical

Maximum kinetic energy is 3.0 eV. Find stopping potential.

Answer: 3.0 V

Solution: eV₀ = Kmax, so the numerical value in volts equals the energy in eV.

NEET

11. Numerical

Find photon energy for wavelength 300 nm.

Answer: 4.13 eV

Solution: E = hc/λ = 1240/300 = 4.13 eV.

NEET

12. Numerical

A metal has ν₀ = 4.0×10¹⁴ Hz and stopping potential 1.75 V. Estimate incident frequency.

Answer: 8.23×10¹⁴ Hz

Solution: hν = hν₀ + eV₀, so ν = ν₀ + eV₀/h.

NEET

13. Numerical

Light frequency is 6.2×10¹⁴ Hz and work function is 2 eV. Find Kmax.

Answer: 0.56 eV

Solution: Kmax = hν − Φ = 0.56 eV.

NEET

14. Numerical

A metal has work function 2.0 eV. Find threshold frequency.

Answer: 4.835e+14 Hz

Solution: ν₀ = Φ/h after converting eV to joules.

NEET

15. Numerical

Find threshold wavelength for work function 3.0 eV.

Answer: 413.3 nm

Solution: λ₀ = hc/Φ; using hc = 1240 eV·nm gives 413.3 nm.

NEET

16. Numerical

Maximum kinetic energy is 1.0 eV. Find stopping potential.

Answer: 1.0 V

Solution: eV₀ = Kmax, so the numerical value in volts equals the energy in eV.

NEET

17. Numerical

Find photon energy for wavelength 320 nm.

Answer: 3.88 eV

Solution: E = hc/λ = 1240/320 = 3.88 eV.

NEET

18. Numerical

A metal has ν₀ = 4.0×10¹⁴ Hz and stopping potential 1.75 V. Estimate incident frequency.

Answer: 8.23×10¹⁴ Hz

Solution: hν = hν₀ + eV₀, so ν = ν₀ + eV₀/h.

NEET

19. Numerical

Light frequency is 6.8×10¹⁴ Hz and work function is 2 eV. Find Kmax.

Answer: 0.81 eV

Solution: Kmax = hν − Φ = 0.81 eV.

NEET

20. Numerical

A metal has work function 2.4 eV. Find threshold frequency.

Answer: 5.803e+14 Hz

Solution: ν₀ = Φ/h after converting eV to joules.

JEE Main

21. Numerical

Find threshold wavelength for work function 3.0 eV.

Answer: 413.3 nm

Solution: λ₀ = hc/Φ; using hc = 1240 eV·nm gives 413.3 nm.

JEE Main

22. Numerical

Maximum kinetic energy is 1.5 eV. Find stopping potential.

Answer: 1.5 V

Solution: eV₀ = Kmax, so the numerical value in volts equals the energy in eV.

JEE Main

23. Numerical

Find photon energy for wavelength 340 nm.

Answer: 3.65 eV

Solution: E = hc/λ = 1240/340 = 3.65 eV.

JEE Main

24. Numerical

A metal has ν₀ = 4.0×10¹⁴ Hz and stopping potential 1.75 V. Estimate incident frequency.

Answer: 8.23×10¹⁴ Hz

Solution: hν = hν₀ + eV₀, so ν = ν₀ + eV₀/h.

JEE Main

25. Numerical

Light frequency is 7.4×10¹⁴ Hz and work function is 2 eV. Find Kmax.

Answer: 1.06 eV

Solution: Kmax = hν − Φ = 1.06 eV.

JEE Main

26. Numerical

A metal has work function 2.0 eV. Find threshold frequency.

Answer: 4.835e+14 Hz

Solution: ν₀ = Φ/h after converting eV to joules.

JEE Main

27. Numerical

Find threshold wavelength for work function 3.0 eV.

Answer: 413.3 nm

Solution: λ₀ = hc/Φ; using hc = 1240 eV·nm gives 413.3 nm.

JEE Main

28. Numerical

Maximum kinetic energy is 2.0 eV. Find stopping potential.

Answer: 2.0 V

Solution: eV₀ = Kmax, so the numerical value in volts equals the energy in eV.

JEE Main

29. Numerical

Find photon energy for wavelength 360 nm.

Answer: 3.44 eV

Solution: E = hc/λ = 1240/360 = 3.44 eV.

JEE Main

30. Numerical

A metal has ν₀ = 4.0×10¹⁴ Hz and stopping potential 1.75 V. Estimate incident frequency.

Answer: 8.23×10¹⁴ Hz

Solution: hν = hν₀ + eV₀, so ν = ν₀ + eV₀/h.

JEE Main

31. Numerical

Light frequency is 8.0×10¹⁴ Hz and work function is 2 eV. Find Kmax.

Answer: 1.31 eV

Solution: Kmax = hν − Φ = 1.31 eV.

JEE Main

32. Numerical

A metal has work function 2.4 eV. Find threshold frequency.

Answer: 5.803e+14 Hz

Solution: ν₀ = Φ/h after converting eV to joules.

JEE Main

33. Numerical

Find threshold wavelength for work function 3.0 eV.

Answer: 413.3 nm

Solution: λ₀ = hc/Φ; using hc = 1240 eV·nm gives 413.3 nm.

JEE Main

34. Numerical

Maximum kinetic energy is 2.5 eV. Find stopping potential.

Answer: 2.5 V

Solution: eV₀ = Kmax, so the numerical value in volts equals the energy in eV.

JEE Main

35. Numerical

Find photon energy for wavelength 380 nm.

Answer: 3.26 eV

Solution: E = hc/λ = 1240/380 = 3.26 eV.

JEE Main

36. Numerical

A metal has ν₀ = 4.0×10¹⁴ Hz and stopping potential 1.75 V. Estimate incident frequency.

Answer: 8.23×10¹⁴ Hz

Solution: hν = hν₀ + eV₀, so ν = ν₀ + eV₀/h.

JEE Main

37. Numerical

Light frequency is 8.6×10¹⁴ Hz and work function is 2 eV. Find Kmax.

Answer: 1.56 eV

Solution: Kmax = hν − Φ = 1.56 eV.

JEE Main

38. Numerical

A metal has work function 2.0 eV. Find threshold frequency.

Answer: 4.835e+14 Hz

Solution: ν₀ = Φ/h after converting eV to joules.

JEE Main

39. Numerical

Find threshold wavelength for work function 3.0 eV.

Answer: 413.3 nm

Solution: λ₀ = hc/Φ; using hc = 1240 eV·nm gives 413.3 nm.

JEE Main

40. Numerical

Maximum kinetic energy is 3.0 eV. Find stopping potential.

Answer: 3.0 V

Solution: eV₀ = Kmax, so the numerical value in volts equals the energy in eV.

JEE Advanced

41. Numerical

Find photon energy for wavelength 300 nm.

Answer: 4.13 eV

Solution: E = hc/λ = 1240/300 = 4.13 eV.

JEE Advanced

42. Numerical

A metal has ν₀ = 4.0×10¹⁴ Hz and stopping potential 1.75 V. Estimate incident frequency.

Answer: 8.23×10¹⁴ Hz

Solution: hν = hν₀ + eV₀, so ν = ν₀ + eV₀/h.

JEE Advanced

43. Numerical

Light frequency is 9.2×10¹⁴ Hz and work function is 2 eV. Find Kmax.

Answer: 1.81 eV

Solution: Kmax = hν − Φ = 1.81 eV.

JEE Advanced

44. Numerical

A metal has work function 2.4 eV. Find threshold frequency.

Answer: 5.803e+14 Hz

Solution: ν₀ = Φ/h after converting eV to joules.

JEE Advanced

45. Numerical

Find threshold wavelength for work function 3.0 eV.

Answer: 413.3 nm

Solution: λ₀ = hc/Φ; using hc = 1240 eV·nm gives 413.3 nm.

JEE Advanced

46. Numerical

Maximum kinetic energy is 1.0 eV. Find stopping potential.

Answer: 1.0 V

Solution: eV₀ = Kmax, so the numerical value in volts equals the energy in eV.

JEE Advanced

47. Numerical

Find photon energy for wavelength 320 nm.

Answer: 3.88 eV

Solution: E = hc/λ = 1240/320 = 3.88 eV.

JEE Advanced

48. Numerical

A metal has ν₀ = 4.0×10¹⁴ Hz and stopping potential 1.75 V. Estimate incident frequency.

Answer: 8.23×10¹⁴ Hz

Solution: hν = hν₀ + eV₀, so ν = ν₀ + eV₀/h.

JEE Advanced

49. Numerical

Light frequency is 9.8×10¹⁴ Hz and work function is 2 eV. Find Kmax.

Answer: 2.05 eV

Solution: Kmax = hν − Φ = 2.05 eV.

JEE Advanced

50. Numerical

A metal has work function 2.0 eV. Find threshold frequency.

Answer: 4.835e+14 Hz

Solution: ν₀ = Φ/h after converting eV to joules.

JEE Advanced

51. Numerical

Find threshold wavelength for work function 3.0 eV.

Answer: 413.3 nm

Solution: λ₀ = hc/Φ; using hc = 1240 eV·nm gives 413.3 nm.

JEE Advanced

52. Numerical

Maximum kinetic energy is 1.5 eV. Find stopping potential.

Answer: 1.5 V

Solution: eV₀ = Kmax, so the numerical value in volts equals the energy in eV.

JEE Advanced

53. Numerical

Find photon energy for wavelength 340 nm.

Answer: 3.65 eV

Solution: E = hc/λ = 1240/340 = 3.65 eV.

JEE Advanced

54. Numerical

A metal has ν₀ = 4.0×10¹⁴ Hz and stopping potential 1.75 V. Estimate incident frequency.

Answer: 8.23×10¹⁴ Hz

Solution: hν = hν₀ + eV₀, so ν = ν₀ + eV₀/h.

JEE Advanced

55. Numerical

Light frequency is 10.4×10¹⁴ Hz and work function is 2 eV. Find Kmax.

Answer: 2.30 eV

Solution: Kmax = hν − Φ = 2.30 eV.

13. Exam-Style Question Banks

These are original examination-style questions, not verbatim copyrighted past papers.

CBSE · 20 Questions

Q1. CBSE style: select the correct photoelectric result.

A. Kmax = hν − Φ

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q2. CBSE style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. V₀ = Kmax/e

Answer: D. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q3. CBSE style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. ν₀ = Φ/h

D. frequency depends on intensity

Answer: C. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q4. CBSE style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. saturation current ∝ intensity

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q5. CBSE style: select the correct photoelectric result.

A. Kmax is independent of intensity

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.
Q6. CBSE style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. Kmax = hν − Φ

Answer: D. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q7. CBSE style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. V₀ = Kmax/e

D. frequency depends on intensity

Answer: C. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q8. CBSE style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. ν₀ = Φ/h

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q9. CBSE style: select the correct photoelectric result.

A. saturation current ∝ intensity

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q10. CBSE style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. Kmax is independent of intensity

Answer: D. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.
Q11. CBSE style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. Kmax = hν − Φ

D. frequency depends on intensity

Answer: C. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q12. CBSE style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. V₀ = Kmax/e

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q13. CBSE style: select the correct photoelectric result.

A. ν₀ = Φ/h

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q14. CBSE style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. saturation current ∝ intensity

Answer: D. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q15. CBSE style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. Kmax is independent of intensity

D. frequency depends on intensity

Answer: C. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.
Q16. CBSE style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. Kmax = hν − Φ

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q17. CBSE style: select the correct photoelectric result.

A. V₀ = Kmax/e

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q18. CBSE style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. ν₀ = Φ/h

Answer: D. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q19. CBSE style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. saturation current ∝ intensity

D. frequency depends on intensity

Answer: C. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q20. CBSE style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. Kmax is independent of intensity

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.

NEET · 20 Questions

Q1. NEET style: select the correct photoelectric result.

A. Kmax = hν − Φ

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q2. NEET style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. V₀ = Kmax/e

Answer: D. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q3. NEET style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. ν₀ = Φ/h

D. frequency depends on intensity

Answer: C. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q4. NEET style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. saturation current ∝ intensity

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q5. NEET style: select the correct photoelectric result.

A. Kmax is independent of intensity

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.
Q6. NEET style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. Kmax = hν − Φ

Answer: D. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q7. NEET style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. V₀ = Kmax/e

D. frequency depends on intensity

Answer: C. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q8. NEET style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. ν₀ = Φ/h

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q9. NEET style: select the correct photoelectric result.

A. saturation current ∝ intensity

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q10. NEET style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. Kmax is independent of intensity

Answer: D. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.
Q11. NEET style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. Kmax = hν − Φ

D. frequency depends on intensity

Answer: C. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q12. NEET style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. V₀ = Kmax/e

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q13. NEET style: select the correct photoelectric result.

A. ν₀ = Φ/h

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q14. NEET style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. saturation current ∝ intensity

Answer: D. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q15. NEET style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. Kmax is independent of intensity

D. frequency depends on intensity

Answer: C. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.
Q16. NEET style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. Kmax = hν − Φ

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q17. NEET style: select the correct photoelectric result.

A. V₀ = Kmax/e

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q18. NEET style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. ν₀ = Φ/h

Answer: D. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q19. NEET style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. saturation current ∝ intensity

D. frequency depends on intensity

Answer: C. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q20. NEET style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. Kmax is independent of intensity

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.

JEE Main · 20 Questions

Q1. JEE Main style: select the correct photoelectric result.

A. Kmax = hν − Φ

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q2. JEE Main style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. V₀ = Kmax/e

Answer: D. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q3. JEE Main style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. ν₀ = Φ/h

D. frequency depends on intensity

Answer: C. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q4. JEE Main style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. saturation current ∝ intensity

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q5. JEE Main style: select the correct photoelectric result.

A. Kmax is independent of intensity

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.
Q6. JEE Main style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. Kmax = hν − Φ

Answer: D. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q7. JEE Main style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. V₀ = Kmax/e

D. frequency depends on intensity

Answer: C. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q8. JEE Main style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. ν₀ = Φ/h

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q9. JEE Main style: select the correct photoelectric result.

A. saturation current ∝ intensity

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q10. JEE Main style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. Kmax is independent of intensity

Answer: D. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.
Q11. JEE Main style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. Kmax = hν − Φ

D. frequency depends on intensity

Answer: C. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q12. JEE Main style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. V₀ = Kmax/e

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q13. JEE Main style: select the correct photoelectric result.

A. ν₀ = Φ/h

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q14. JEE Main style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. saturation current ∝ intensity

Answer: D. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q15. JEE Main style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. Kmax is independent of intensity

D. frequency depends on intensity

Answer: C. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.
Q16. JEE Main style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. Kmax = hν − Φ

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q17. JEE Main style: select the correct photoelectric result.

A. V₀ = Kmax/e

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q18. JEE Main style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. ν₀ = Φ/h

Answer: D. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q19. JEE Main style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. saturation current ∝ intensity

D. frequency depends on intensity

Answer: C. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q20. JEE Main style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. Kmax is independent of intensity

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.

JEE Advanced · 20 Questions

Q1. JEE Advanced style: select the correct photoelectric result.

A. Kmax = hν − Φ

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q2. JEE Advanced style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. V₀ = Kmax/e

Answer: D. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q3. JEE Advanced style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. ν₀ = Φ/h

D. frequency depends on intensity

Answer: C. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q4. JEE Advanced style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. saturation current ∝ intensity

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q5. JEE Advanced style: select the correct photoelectric result.

A. Kmax is independent of intensity

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.
Q6. JEE Advanced style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. Kmax = hν − Φ

Answer: D. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q7. JEE Advanced style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. V₀ = Kmax/e

D. frequency depends on intensity

Answer: C. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q8. JEE Advanced style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. ν₀ = Φ/h

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q9. JEE Advanced style: select the correct photoelectric result.

A. saturation current ∝ intensity

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q10. JEE Advanced style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. Kmax is independent of intensity

Answer: D. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.
Q11. JEE Advanced style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. Kmax = hν − Φ

D. frequency depends on intensity

Answer: C. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q12. JEE Advanced style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. V₀ = Kmax/e

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q13. JEE Advanced style: select the correct photoelectric result.

A. ν₀ = Φ/h

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q14. JEE Advanced style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. saturation current ∝ intensity

Answer: D. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q15. JEE Advanced style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. Kmax is independent of intensity

D. frequency depends on intensity

Answer: C. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.
Q16. JEE Advanced style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. Kmax = hν − Φ

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q17. JEE Advanced style: select the correct photoelectric result.

A. V₀ = Kmax/e

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q18. JEE Advanced style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. ν₀ = Φ/h

Answer: D. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q19. JEE Advanced style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. saturation current ∝ intensity

D. frequency depends on intensity

Answer: C. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q20. JEE Advanced style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. Kmax is independent of intensity

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.

IB Physics · 20 Questions

Q1. IB Physics style: select the correct photoelectric result.

A. Kmax = hν − Φ

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q2. IB Physics style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. V₀ = Kmax/e

Answer: D. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q3. IB Physics style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. ν₀ = Φ/h

D. frequency depends on intensity

Answer: C. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q4. IB Physics style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. saturation current ∝ intensity

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q5. IB Physics style: select the correct photoelectric result.

A. Kmax is independent of intensity

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.
Q6. IB Physics style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. Kmax = hν − Φ

Answer: D. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q7. IB Physics style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. V₀ = Kmax/e

D. frequency depends on intensity

Answer: C. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q8. IB Physics style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. ν₀ = Φ/h

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q9. IB Physics style: select the correct photoelectric result.

A. saturation current ∝ intensity

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q10. IB Physics style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. Kmax is independent of intensity

Answer: D. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.
Q11. IB Physics style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. Kmax = hν − Φ

D. frequency depends on intensity

Answer: C. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q12. IB Physics style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. V₀ = Kmax/e

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q13. IB Physics style: select the correct photoelectric result.

A. ν₀ = Φ/h

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q14. IB Physics style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. saturation current ∝ intensity

Answer: D. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q15. IB Physics style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. Kmax is independent of intensity

D. frequency depends on intensity

Answer: C. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.
Q16. IB Physics style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. Kmax = hν − Φ

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q17. IB Physics style: select the correct photoelectric result.

A. V₀ = Kmax/e

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q18. IB Physics style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. ν₀ = Φ/h

Answer: D. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q19. IB Physics style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. saturation current ∝ intensity

D. frequency depends on intensity

Answer: C. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q20. IB Physics style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. Kmax is independent of intensity

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.

IGCSE · 20 Questions

Q1. IGCSE style: select the correct photoelectric result.

A. Kmax = hν − Φ

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q2. IGCSE style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. V₀ = Kmax/e

Answer: D. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q3. IGCSE style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. ν₀ = Φ/h

D. frequency depends on intensity

Answer: C. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q4. IGCSE style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. saturation current ∝ intensity

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q5. IGCSE style: select the correct photoelectric result.

A. Kmax is independent of intensity

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.
Q6. IGCSE style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. Kmax = hν − Φ

Answer: D. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q7. IGCSE style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. V₀ = Kmax/e

D. frequency depends on intensity

Answer: C. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q8. IGCSE style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. ν₀ = Φ/h

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q9. IGCSE style: select the correct photoelectric result.

A. saturation current ∝ intensity

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q10. IGCSE style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. Kmax is independent of intensity

Answer: D. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.
Q11. IGCSE style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. Kmax = hν − Φ

D. frequency depends on intensity

Answer: C. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q12. IGCSE style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. V₀ = Kmax/e

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q13. IGCSE style: select the correct photoelectric result.

A. ν₀ = Φ/h

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q14. IGCSE style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. saturation current ∝ intensity

Answer: D. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q15. IGCSE style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. Kmax is independent of intensity

D. frequency depends on intensity

Answer: C. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.
Q16. IGCSE style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. Kmax = hν − Φ

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q17. IGCSE style: select the correct photoelectric result.

A. V₀ = Kmax/e

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q18. IGCSE style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. ν₀ = Φ/h

Answer: D. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q19. IGCSE style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. saturation current ∝ intensity

D. frequency depends on intensity

Answer: C. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q20. IGCSE style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. Kmax is independent of intensity

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.

A-Level · 20 Questions

Q1. A-Level style: select the correct photoelectric result.

A. Kmax = hν − Φ

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q2. A-Level style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. V₀ = Kmax/e

Answer: D. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q3. A-Level style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. ν₀ = Φ/h

D. frequency depends on intensity

Answer: C. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q4. A-Level style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. saturation current ∝ intensity

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q5. A-Level style: select the correct photoelectric result.

A. Kmax is independent of intensity

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.
Q6. A-Level style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. Kmax = hν − Φ

Answer: D. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q7. A-Level style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. V₀ = Kmax/e

D. frequency depends on intensity

Answer: C. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q8. A-Level style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. ν₀ = Φ/h

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q9. A-Level style: select the correct photoelectric result.

A. saturation current ∝ intensity

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q10. A-Level style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. Kmax is independent of intensity

Answer: D. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.
Q11. A-Level style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. Kmax = hν − Φ

D. frequency depends on intensity

Answer: C. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q12. A-Level style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. V₀ = Kmax/e

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q13. A-Level style: select the correct photoelectric result.

A. ν₀ = Φ/h

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q14. A-Level style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. saturation current ∝ intensity

Answer: D. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q15. A-Level style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. Kmax is independent of intensity

D. frequency depends on intensity

Answer: C. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.
Q16. A-Level style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. Kmax = hν − Φ

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. Kmax = hν − Φ
Explanation: The correct relation is Kmax = hν − Φ.
Q17. A-Level style: select the correct photoelectric result.

A. V₀ = Kmax/e

B. frequency depends on intensity

C. sub-threshold light always emits

D. stopping potential is proportional to intensity

Answer: A. V₀ = Kmax/e
Explanation: The correct relation is V₀ = Kmax/e.
Q18. A-Level style: select the correct photoelectric result.

A. frequency depends on intensity

B. sub-threshold light always emits

C. stopping potential is proportional to intensity

D. ν₀ = Φ/h

Answer: D. ν₀ = Φ/h
Explanation: The correct relation is ν₀ = Φ/h.
Q19. A-Level style: select the correct photoelectric result.

A. sub-threshold light always emits

B. stopping potential is proportional to intensity

C. saturation current ∝ intensity

D. frequency depends on intensity

Answer: C. saturation current ∝ intensity
Explanation: The correct relation is saturation current ∝ intensity.
Q20. A-Level style: select the correct photoelectric result.

A. stopping potential is proportional to intensity

B. Kmax is independent of intensity

C. frequency depends on intensity

D. sub-threshold light always emits

Answer: B. Kmax is independent of intensity
Explanation: The correct relation is Kmax is independent of intensity.
15 Case Study Questions
CS1. Case 1: A laboratory changes light intensity in a photoelectric tube. Predict the current, Kmax and stopping-potential response.
Answer: Apply photon-number and photon-energy rules separately.
Explanation: Intensity primarily controls emitted-electron rate and saturation current. Frequency controls photon energy, Kmax and stopping potential. Cathode material controls Φ and ν₀; applied voltage controls collection.
CS2. Case 2: A laboratory changes light frequency in a photoelectric tube. Predict the current, Kmax and stopping-potential response.
Answer: Apply photon-number and photon-energy rules separately.
Explanation: Intensity primarily controls emitted-electron rate and saturation current. Frequency controls photon energy, Kmax and stopping potential. Cathode material controls Φ and ν₀; applied voltage controls collection.
CS3. Case 3: A laboratory changes cathode material in a photoelectric tube. Predict the current, Kmax and stopping-potential response.
Answer: Apply photon-number and photon-energy rules separately.
Explanation: Intensity primarily controls emitted-electron rate and saturation current. Frequency controls photon energy, Kmax and stopping potential. Cathode material controls Φ and ν₀; applied voltage controls collection.
CS4. Case 4: A laboratory changes applied potential in a photoelectric tube. Predict the current, Kmax and stopping-potential response.
Answer: Apply photon-number and photon-energy rules separately.
Explanation: Intensity primarily controls emitted-electron rate and saturation current. Frequency controls photon energy, Kmax and stopping potential. Cathode material controls Φ and ν₀; applied voltage controls collection.
CS5. Case 5: A laboratory changes surface condition in a photoelectric tube. Predict the current, Kmax and stopping-potential response.
Answer: Apply photon-number and photon-energy rules separately.
Explanation: Intensity primarily controls emitted-electron rate and saturation current. Frequency controls photon energy, Kmax and stopping potential. Cathode material controls Φ and ν₀; applied voltage controls collection.
CS6. Case 6: A laboratory changes light intensity in a photoelectric tube. Predict the current, Kmax and stopping-potential response.
Answer: Apply photon-number and photon-energy rules separately.
Explanation: Intensity primarily controls emitted-electron rate and saturation current. Frequency controls photon energy, Kmax and stopping potential. Cathode material controls Φ and ν₀; applied voltage controls collection.
CS7. Case 7: A laboratory changes light frequency in a photoelectric tube. Predict the current, Kmax and stopping-potential response.
Answer: Apply photon-number and photon-energy rules separately.
Explanation: Intensity primarily controls emitted-electron rate and saturation current. Frequency controls photon energy, Kmax and stopping potential. Cathode material controls Φ and ν₀; applied voltage controls collection.
CS8. Case 8: A laboratory changes cathode material in a photoelectric tube. Predict the current, Kmax and stopping-potential response.
Answer: Apply photon-number and photon-energy rules separately.
Explanation: Intensity primarily controls emitted-electron rate and saturation current. Frequency controls photon energy, Kmax and stopping potential. Cathode material controls Φ and ν₀; applied voltage controls collection.
CS9. Case 9: A laboratory changes applied potential in a photoelectric tube. Predict the current, Kmax and stopping-potential response.
Answer: Apply photon-number and photon-energy rules separately.
Explanation: Intensity primarily controls emitted-electron rate and saturation current. Frequency controls photon energy, Kmax and stopping potential. Cathode material controls Φ and ν₀; applied voltage controls collection.
CS10. Case 10: A laboratory changes surface condition in a photoelectric tube. Predict the current, Kmax and stopping-potential response.
Answer: Apply photon-number and photon-energy rules separately.
Explanation: Intensity primarily controls emitted-electron rate and saturation current. Frequency controls photon energy, Kmax and stopping potential. Cathode material controls Φ and ν₀; applied voltage controls collection.
CS11. Case 11: A laboratory changes light intensity in a photoelectric tube. Predict the current, Kmax and stopping-potential response.
Answer: Apply photon-number and photon-energy rules separately.
Explanation: Intensity primarily controls emitted-electron rate and saturation current. Frequency controls photon energy, Kmax and stopping potential. Cathode material controls Φ and ν₀; applied voltage controls collection.
CS12. Case 12: A laboratory changes light frequency in a photoelectric tube. Predict the current, Kmax and stopping-potential response.
Answer: Apply photon-number and photon-energy rules separately.
Explanation: Intensity primarily controls emitted-electron rate and saturation current. Frequency controls photon energy, Kmax and stopping potential. Cathode material controls Φ and ν₀; applied voltage controls collection.
CS13. Case 13: A laboratory changes cathode material in a photoelectric tube. Predict the current, Kmax and stopping-potential response.
Answer: Apply photon-number and photon-energy rules separately.
Explanation: Intensity primarily controls emitted-electron rate and saturation current. Frequency controls photon energy, Kmax and stopping potential. Cathode material controls Φ and ν₀; applied voltage controls collection.
CS14. Case 14: A laboratory changes applied potential in a photoelectric tube. Predict the current, Kmax and stopping-potential response.
Answer: Apply photon-number and photon-energy rules separately.
Explanation: Intensity primarily controls emitted-electron rate and saturation current. Frequency controls photon energy, Kmax and stopping potential. Cathode material controls Φ and ν₀; applied voltage controls collection.
CS15. Case 15: A laboratory changes surface condition in a photoelectric tube. Predict the current, Kmax and stopping-potential response.
Answer: Apply photon-number and photon-energy rules separately.
Explanation: Intensity primarily controls emitted-electron rate and saturation current. Frequency controls photon energy, Kmax and stopping potential. Cathode material controls Φ and ν₀; applied voltage controls collection.
25 Assertion-Reason Questions
AR1. Assertion: What is photoelectric emission? Reason: Photoelectric behaviour follows quantised photon energy and Einstein's equation.

A. Both true; reason explains assertion

B. Both true; reason does not explain assertion

C. Assertion true; reason false

D. Assertion false; reason true

Answer: A
Explanation: The photoelectric observation is explained by one-photon energy transfer and the energy-balance equation.
AR2. Assertion: Why is a clean metal surface preferred? Reason: Photoelectric behaviour follows quantised photon energy and Einstein's equation.

A. Both true; reason explains assertion

B. Both true; reason does not explain assertion

C. Assertion true; reason false

D. Assertion false; reason true

Answer: A
Explanation: The photoelectric observation is explained by one-photon energy transfer and the energy-balance equation.
AR3. Assertion: What is work function? Reason: Photoelectric behaviour follows quantised photon energy and Einstein's equation.

A. Both true; reason explains assertion

B. Both true; reason does not explain assertion

C. Assertion true; reason false

D. Assertion false; reason true

Answer: A
Explanation: The photoelectric observation is explained by one-photon energy transfer and the energy-balance equation.
AR4. Assertion: What is threshold frequency? Reason: Photoelectric behaviour follows quantised photon energy and Einstein's equation.

A. Both true; reason explains assertion

B. Both true; reason does not explain assertion

C. Assertion true; reason false

D. Assertion false; reason true

Answer: A
Explanation: The photoelectric observation is explained by one-photon energy transfer and the energy-balance equation.
AR5. Assertion: What is threshold wavelength? Reason: Photoelectric behaviour follows quantised photon energy and Einstein's equation.

A. Both true; reason explains assertion

B. Both true; reason does not explain assertion

C. Assertion true; reason false

D. Assertion false; reason true

Answer: A
Explanation: The photoelectric observation is explained by one-photon energy transfer and the energy-balance equation.
AR6. Assertion: Why is emission instantaneous? Reason: Photoelectric behaviour follows quantised photon energy and Einstein's equation.

A. Both true; reason explains assertion

B. Both true; reason does not explain assertion

C. Assertion true; reason false

D. Assertion false; reason true

Answer: A
Explanation: The photoelectric observation is explained by one-photon energy transfer and the energy-balance equation.
AR7. Assertion: Why does intensity affect current? Reason: Photoelectric behaviour follows quantised photon energy and Einstein's equation.

A. Both true; reason explains assertion

B. Both true; reason does not explain assertion

C. Assertion true; reason false

D. Assertion false; reason true

Answer: A
Explanation: The photoelectric observation is explained by one-photon energy transfer and the energy-balance equation.
AR8. Assertion: Why does intensity not affect Kmax? Reason: Photoelectric behaviour follows quantised photon energy and Einstein's equation.

A. Both true; reason explains assertion

B. Both true; reason does not explain assertion

C. Assertion true; reason false

D. Assertion false; reason true

Answer: A
Explanation: The photoelectric observation is explained by one-photon energy transfer and the energy-balance equation.
AR9. Assertion: Why does frequency affect Kmax? Reason: Photoelectric behaviour follows quantised photon energy and Einstein's equation.

A. Both true; reason explains assertion

B. Both true; reason does not explain assertion

C. Assertion true; reason false

D. Assertion false; reason true

Answer: A
Explanation: The photoelectric observation is explained by one-photon energy transfer and the energy-balance equation.
AR10. Assertion: What is stopping potential? Reason: Photoelectric behaviour follows quantised photon energy and Einstein's equation.

A. Both true; reason explains assertion

B. Both true; reason does not explain assertion

C. Assertion true; reason false

D. Assertion false; reason true

Answer: A
Explanation: The photoelectric observation is explained by one-photon energy transfer and the energy-balance equation.
AR11. Assertion: What is saturation current? Reason: Photoelectric behaviour follows quantised photon energy and Einstein's equation.

A. Both true; reason explains assertion

B. Both true; reason does not explain assertion

C. Assertion true; reason false

D. Assertion false; reason true

Answer: A
Explanation: The photoelectric observation is explained by one-photon energy transfer and the energy-balance equation.
AR12. Assertion: Can sub-threshold light emit electrons at high intensity? Reason: Photoelectric behaviour follows quantised photon energy and Einstein's equation.

A. Both true; reason explains assertion

B. Both true; reason does not explain assertion

C. Assertion true; reason false

D. Assertion false; reason true

Answer: A
Explanation: The photoelectric observation is explained by one-photon energy transfer and the energy-balance equation.
AR13. Assertion: Why is ultraviolet effective for many metals? Reason: Photoelectric behaviour follows quantised photon energy and Einstein's equation.

A. Both true; reason explains assertion

B. Both true; reason does not explain assertion

C. Assertion true; reason false

D. Assertion false; reason true

Answer: A
Explanation: The photoelectric observation is explained by one-photon energy transfer and the energy-balance equation.
AR14. Assertion: What happens when frequency equals threshold frequency? Reason: Photoelectric behaviour follows quantised photon energy and Einstein's equation.

A. Both true; reason explains assertion

B. Both true; reason does not explain assertion

C. Assertion true; reason false

D. Assertion false; reason true

Answer: A
Explanation: The photoelectric observation is explained by one-photon energy transfer and the energy-balance equation.
AR15. Assertion: What happens above threshold frequency? Reason: Photoelectric behaviour follows quantised photon energy and Einstein's equation.

A. Both true; reason explains assertion

B. Both true; reason does not explain assertion

C. Assertion true; reason false

D. Assertion false; reason true

Answer: A
Explanation: The photoelectric observation is explained by one-photon energy transfer and the energy-balance equation.
AR16. Assertion: What is the meaning of Kmax? Reason: Photoelectric behaviour follows quantised photon energy and Einstein's equation.

A. Both true; reason explains assertion

B. Both true; reason does not explain assertion

C. Assertion true; reason false

D. Assertion false; reason true

Answer: A
Explanation: The photoelectric observation is explained by one-photon energy transfer and the energy-balance equation.
AR17. Assertion: Why do emitted electrons have a range of energies? Reason: Photoelectric behaviour follows quantised photon energy and Einstein's equation.

A. Both true; reason explains assertion

B. Both true; reason does not explain assertion

C. Assertion true; reason false

D. Assertion false; reason true

Answer: A
Explanation: The photoelectric observation is explained by one-photon energy transfer and the energy-balance equation.
AR18. Assertion: What does the Kmax–ν slope represent? Reason: Photoelectric behaviour follows quantised photon energy and Einstein's equation.

A. Both true; reason explains assertion

B. Both true; reason does not explain assertion

C. Assertion true; reason false

D. Assertion false; reason true

Answer: A
Explanation: The photoelectric observation is explained by one-photon energy transfer and the energy-balance equation.
AR19. Assertion: What does the V₀–ν slope represent? Reason: Photoelectric behaviour follows quantised photon energy and Einstein's equation.

A. Both true; reason explains assertion

B. Both true; reason does not explain assertion

C. Assertion true; reason false

D. Assertion false; reason true

Answer: A
Explanation: The photoelectric observation is explained by one-photon energy transfer and the energy-balance equation.
AR20. Assertion: How is work function obtained graphically? Reason: Photoelectric behaviour follows quantised photon energy and Einstein's equation.

A. Both true; reason explains assertion

B. Both true; reason does not explain assertion

C. Assertion true; reason false

D. Assertion false; reason true

Answer: A
Explanation: The photoelectric observation is explained by one-photon energy transfer and the energy-balance equation.
AR21. Assertion: Why is current zero at stopping potential? Reason: Photoelectric behaviour follows quantised photon energy and Einstein's equation.

A. Both true; reason explains assertion

B. Both true; reason does not explain assertion

C. Assertion true; reason false

D. Assertion false; reason true

Answer: A
Explanation: The photoelectric observation is explained by one-photon energy transfer and the energy-balance equation.
AR22. Assertion: Does stopping potential depend on intensity? Reason: Photoelectric behaviour follows quantised photon energy and Einstein's equation.

A. Both true; reason explains assertion

B. Both true; reason does not explain assertion

C. Assertion true; reason false

D. Assertion false; reason true

Answer: A
Explanation: The photoelectric observation is explained by one-photon energy transfer and the energy-balance equation.
AR23. Assertion: Does saturation current depend on frequency? Reason: Photoelectric behaviour follows quantised photon energy and Einstein's equation.

A. Both true; reason explains assertion

B. Both true; reason does not explain assertion

C. Assertion true; reason false

D. Assertion false; reason true

Answer: A
Explanation: The photoelectric observation is explained by one-photon energy transfer and the energy-balance equation.
AR24. Assertion: What is photon momentum? Reason: Photoelectric behaviour follows quantised photon energy and Einstein's equation.

A. Both true; reason explains assertion

B. Both true; reason does not explain assertion

C. Assertion true; reason false

D. Assertion false; reason true

Answer: A
Explanation: The photoelectric observation is explained by one-photon energy transfer and the energy-balance equation.
AR25. Assertion: How does photon energy depend on wavelength? Reason: Photoelectric behaviour follows quantised photon energy and Einstein's equation.

A. Both true; reason explains assertion

B. Both true; reason does not explain assertion

C. Assertion true; reason false

D. Assertion false; reason true

Answer: A
Explanation: The photoelectric observation is explained by one-photon energy transfer and the energy-balance equation.

One-Page Quick Revision

Core Formulae

Φ = hν₀Kmax = hν − Φ = eV₀

Graph Rules

Intensity changes saturation current. Frequency changes stopping potential and Kmax.

Common Mistakes

High intensity cannot overcome a sub-threshold frequency. V₀ does not depend on intensity.

Exam Tips

Use hc = 1240 eV·nm. Convert work function carefully. Read graph slope and intercepts.

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