Einstein Photoelectric Equation | NEET, JEE Main & JEE Advanced Questions
Kumar Physics Classes
Phone / WhatsApp: +91 9958461445 kumarsirphysics@gmail.com kumarphysicsclasses.com
Dual Nature of Radiation and Matter

Einstein Photoelectric Equation

A full-width premium Physics resource for CBSE, NEET, JEE Main, JEE Advanced, IB, IGCSE and A-Level students, with concepts, formulas, graphs, exam tips, solved numericals and original practice questions.

hν = Φ + Kmax photon energy = work function + maximum kinetic energy
Call Kumar Sir WhatsApp Kumar Sir Visit Website

NEET and JEE Photoelectric Effect Practice Starts With One Energy Balance

The photoelectric effect becomes simple when students stop memorising scattered facts and read every problem as an energy transaction between one photon and one surface electron.

What the Experiment Shows

When light of sufficiently high frequency falls on a clean metal surface, electrons are emitted almost immediately. The emission is not controlled by total brightness alone; it depends on the energy carried by each photon.

Why It Matters

The experiment gives direct evidence for particle-like behaviour of light. It explains threshold frequency, stopping potential, maximum kinetic energy and the failure of classical wave theory.

Exam View

Most NEET and JEE questions test the difference between intensity and frequency, the graph slope or intercept, or a short calculation using E = hc/λ and Kmax = hν - Φ.

Einstein's Photon Theory Explained Clearly

Einstein treated light as a stream of photons. Each photon has energy hν, and a surface electron absorbs energy from one photon in the ordinary photoelectric process.

Classical expectation Energy spread over wavefront Weak light should accumulate Intensity should change Kmax Photon explanation hνhνhνhν Energy arrives in packets Emission needs hν ≥ Φ Kmax depends on frequency
Photon theory explains instantaneous emission, threshold frequency and the frequency dependence of kinetic energy.

Core Idea

A photon either has enough energy to eject an electron or it does not. If hν is less than the work function, increasing the number of photons cannot rescue emission in the basic one-photon model.

Exam tip: In conceptual questions, translate intensity as "number of photons per second" and frequency as "energy per photon". This one sentence solves many NEET questions.

Einstein Photoelectric Equation and Meaning of Each Term

The equation is not just a formula. It is the complete energy accounting of the fastest emitted photoelectron.

Main Equation

hν = Φ + Kmax

Photon energy hν first pays the work function Φ. The remaining energy appears as maximum kinetic energy of the emitted electron.

Useful Forms

Kmax = hν - Φ Kmax = eV0 = ½mvmax2

Use eV units whenever possible: if Kmax = 2.3 eV, then V0 = 2.3 V.

Photon Energy

E = hν = hc/λ. Higher frequency means higher photon energy; shorter wavelength also means higher photon energy.

Maximum Kinetic Energy

Kmax belongs to the fastest emitted electrons. Not every emitted electron has this value because electrons may lose energy inside the metal before escaping.

Exam tip: Start every numerical by writing E = Φ + Kmax. Then decide whether E is given as hν, hc/λ or directly in eV.

Work Function, Threshold Frequency and Threshold Wavelength

These three quantities describe the minimum condition for photoelectric emission from a particular metal.

Work Function

Φ

The minimum energy required to liberate an electron from the metal surface. It depends on the metal and surface condition.

Threshold Frequency

Φ = hν0

At ν0, electrons just escape with Kmax = 0.

Threshold Wavelength

Φ = hc/λ0

Emission occurs for λ ≤ λ0, not for wavelengths longer than threshold.

Exam tip: Students often reverse the wavelength condition. Higher frequency means shorter wavelength, so photoemission needs ν ≥ ν0 or λ ≤ λ0.
Common mistake: Writing "large intensity causes emission below threshold". In the ordinary model, if hν < Φ, increasing intensity only increases the number of insufficient-energy photons.

Stopping Potential and Photoelectric Current

Stopping potential measures the energy of the fastest electrons. Saturation current measures how many electrons are being collected per second.

Emitter Collector Photoelectrons Reverse voltage stops fastest electrons
At stopping potential, the retarding electric field prevents even the fastest photoelectrons from reaching the collector.

Stopping Relation

eV0 = Kmax V0 = (h/e)ν - Φ/e
Exam tip: A negative collector potential is used experimentally, but stopping potential is usually reported as a positive magnitude.

Stopping Potential Graph and Kmax Versus Frequency Graph

Graph questions are high-scoring because the same equation gives slope, intercept and threshold information.

ν0 ν Kmax slope = h no emission below threshold
Kmax versus frequency: slope is h and the frequency-axis intercept is threshold frequency.
ν0 ν V0 slope = h/e intercept gives work function
V0 versus frequency: slope is h/e and the vertical intercept is -Φ/e.
Exam tip: If the graph is Kmax versus ν, slope is h. If it is V0 versus ν, slope is h/e. Do not mix the two.
Common mistake: Reading the x-intercept as work function. The x-intercept is threshold frequency. Work function is h times that intercept.

How Kumar Sir Explains This Concept

Kumar Sir connects the formula with a simple payment picture, so students can handle conceptual, numerical and graph-based problems without panic.

Step 1: Photon Budget

Every photon brings a fixed budget hν. The electron cannot use intensity as extra personal energy; it receives energy from one photon.

Step 2: Surface Exit Cost

The work function Φ is the minimum exit cost. If the photon cannot pay this cost, emission does not occur.

Step 3: Remaining Energy

Whatever remains after paying Φ becomes Kmax. Stopping potential is simply the voltage needed to take this remaining energy away.

NEET Photoelectric Effect Questions

Twenty fresh NEET-level MCQs on photon energy, threshold frequency, threshold wavelength, work function, stopping potential, intensity, frequency and graphs.

NEET 1Light of frequency 8.0 × 1014 Hz falls on a metal of threshold frequency 5.0 × 1014 Hz. What mainly decides the maximum kinetic energy of emitted electrons?
  1. Intensity of the light
  2. Difference between incident frequency and threshold frequency
  3. Area of the metal plate
  4. Time for which light is incident
Correct answer: B
Explanation: From Kmax = h(ν - ν0), maximum kinetic energy depends on how much the incident frequency exceeds threshold frequency.
NEET 2A photon has wavelength 400 nm. Using hc = 1240 eV nm, its energy is nearest to:
  1. 1.55 eV
  2. 2.48 eV
  3. 3.10 eV
  4. 4.96 eV
Correct answer: C
Explanation: E = hc/λ = 1240/400 = 3.10 eV.
NEET 3If the work function of a metal is 2.2 eV, the threshold wavelength is approximately:
  1. 282 nm
  2. 364 nm
  3. 564 nm
  4. 720 nm
Correct answer: C
Explanation: λ0 = 1240/Φ = 1240/2.2 = 563.6 nm.
NEET 4For a given metal, increasing intensity at fixed frequency above threshold will increase:
  1. Stopping potential only
  2. Maximum kinetic energy only
  3. Photoelectric current
  4. Threshold frequency
Correct answer: C
Explanation: Higher intensity means more photons per second, so more electrons are emitted per second. Kmax and stopping potential remain unchanged.
NEET 5No photoelectrons are emitted when red light falls on a certain metal, but emission occurs with violet light. The best explanation is:
  1. Red light has too much intensity
  2. Red light has frequency below threshold
  3. Violet light has smaller speed in air
  4. Violet light has no photons
Correct answer: B
Explanation: Emission requires hν ≥ Φ. Violet light has higher frequency and higher photon energy than red light.
NEET 6The stopping potential for photoelectrons is 1.5 V. Their maximum kinetic energy is:
  1. 1.5 eV
  2. 0.75 eV
  3. 3.0 eV
  4. 1.5 J
Correct answer: A
Explanation: Kmax = eV0. When V0 is in volts, the numerical energy in eV is equal to V0.
NEET 7On a graph of stopping potential V0 versus frequency ν, the slope is:
  1. h
  2. h/e
  3. e/h
  4. Φ/h
Correct answer: B
Explanation: Since V0 = (h/e)ν - Φ/e, the slope is h/e.
NEET 8A metal has threshold frequency ν0. At incident frequency exactly equal to ν0, emitted electrons have:
  1. Maximum kinetic energy equal to hν0
  2. Maximum kinetic energy zero
  3. Infinite speed
  4. Stopping potential equal to Φ
Correct answer: B
Explanation: At threshold, hν0 = Φ, so no energy is left as kinetic energy.
NEET 9A photon of energy 5.0 eV ejects an electron from a metal of work function 3.2 eV. Kmax is:
  1. 1.8 eV
  2. 3.2 eV
  3. 5.0 eV
  4. 8.2 eV
Correct answer: A
Explanation: Kmax = E - Φ = 5.0 - 3.2 = 1.8 eV.
NEET 10Which statement is correct for photoelectric effect in the ordinary one-photon process?
  1. Intensity controls photon energy
  2. Frequency controls photon energy
  3. Threshold frequency depends on light intensity
  4. Stopping potential is independent of frequency
Correct answer: B
Explanation: Photon energy is E = hν, so frequency decides the energy of each photon.
NEET 11If wavelength of incident light is decreased while intensity is kept same and emission already occurs, Kmax:
  1. Decreases
  2. Increases
  3. Becomes zero
  4. Depends only on metal area
Correct answer: B
Explanation: Decreasing wavelength increases frequency and photon energy, so Kmax increases.
NEET 12The work function of a metal is the minimum energy required to:
  1. Stop all photons
  2. Remove an electron from the metal surface
  3. Double the intensity of light
  4. Increase the speed of light
Correct answer: B
Explanation: Work function is the minimum surface binding energy that must be supplied to liberate an electron.
NEET 13For two metals A and B, ΦA < ΦB. Their threshold frequencies satisfy:
  1. ν0A < ν0B
  2. ν0A > ν0B
  3. ν0A = ν0B
  4. No relation is possible
Correct answer: A
Explanation: Φ = hν0, so the metal with smaller work function has smaller threshold frequency.
NEET 14A photon energy is less than the work function of a metal. What happens if intensity is increased greatly?
  1. Electrons are emitted with low speed
  2. Electrons are emitted after a long delay
  3. No emission in the ordinary photoelectric effect
  4. Threshold frequency becomes zero
Correct answer: C
Explanation: If each photon lacks enough energy, increasing photon number does not help in the usual one-photon explanation.
NEET 15Which graph is a straight line for photoelectric emission from a metal?
  1. Kmax versus frequency above threshold
  2. Kmax versus wavelength
  3. Photon energy versus wavelength
  4. Stopping potential versus intensity at fixed frequency
Correct answer: A
Explanation: Kmax = hν - Φ is linear in frequency above threshold.
NEET 16If the threshold wavelength of a metal is 620 nm, light of which wavelength can eject electrons?
  1. 700 nm
  2. 650 nm
  3. 6200 nm
  4. 500 nm
Correct answer: D
Explanation: Emission needs λ ≤ λ0. Among the choices, 500 nm has higher photon energy.
NEET 17Saturation current in a photoelectric experiment is mainly proportional to:
  1. Number of photons incident per second
  2. Threshold frequency of the metal only
  3. Stopping potential only
  4. Planck constant only
Correct answer: A
Explanation: More incident photons per second can release more electrons per second, raising saturation current.
NEET 18If V0 is the stopping potential, the fastest electron is stopped because the retarding field does work equal to:
  1. hν0
  2. eV0
  3. Φ/e
  4. hcλ
Correct answer: B
Explanation: The retarding electric work eV0 equals Kmax.
NEET 19A metal has Φ = 4.0 eV and incident photon energy is 3.5 eV. The photoelectric current is:
  1. Large
  2. Small but non-zero
  3. Zero
  4. Independent of metal
Correct answer: C
Explanation: Photon energy is below work function, so electrons are not emitted in the ordinary photoelectric process.
NEET 20Which quantity is represented by the intercept on the frequency axis in a Kmax versus ν graph?
  1. Threshold frequency
  2. Stopping potential
  3. Saturation current
  4. Intensity
Correct answer: A
Explanation: At Kmax = 0, ν = ν0, the threshold frequency.

JEE Main Photoelectric Effect Questions

Twenty original mixed conceptual and numerical questions with step-by-step answers for JEE Main practice.

JEE Main 1A metal has work function 2.0 eV. Light of wavelength 310 nm falls on it. Find Kmax in eV. Take hc = 1240 eV nm.
  1. Photon energy E = 1240/310 = 4.0 eV.
  2. Kmax = E - Φ = 4.0 - 2.0 = 2.0 eV.
Final answer: 2.0 eV
JEE Main 2The threshold frequency of a metal is 6.0 × 1014 Hz. If incident frequency is 9.0 × 1014 Hz, calculate Kmax. Use h = 6.63 × 10-34 J s.
  1. Kmax = h(ν - ν0).
  2. Δν = 3.0 × 1014 Hz.
  3. Kmax = 6.63 × 10-34 × 3.0 × 1014 = 1.989 × 10-19 J.
Final answer: 1.99 × 10-19 J
JEE Main 3A stopping potential of 2.4 V is measured. What is the maximum kinetic energy in joule? Use e = 1.6 × 10-19 C.
  1. Kmax = eV0.
  2. Kmax = 1.6 × 10-19 × 2.4 = 3.84 × 10-19 J.
Final answer: 3.84 × 10-19 J
JEE Main 4For a metal, λ0 = 500 nm. Will light of wavelength 650 nm produce photoemission?
  1. Emission requires wavelength less than or equal to threshold wavelength.
  2. 650 nm is greater than 500 nm, so photon energy is lower than work function.
Final answer: No emission
JEE Main 5A V0 versus ν graph has slope 4.14 × 10-15 V s. Estimate Planck constant.
  1. For V0 versus ν, slope = h/e.
  2. h = e × slope = 1.6 × 10-19 × 4.14 × 10-15.
  3. h = 6.624 × 10-34 J s.
Final answer: 6.62 × 10-34 J s
JEE Main 6The work function is 3.1 eV. What is the threshold frequency? Use h = 4.14 × 10-15 eV s.
  1. ν0 = Φ/h.
  2. ν0 = 3.1/(4.14 × 10-15) = 7.49 × 1014 Hz.
Final answer: 7.5 × 1014 Hz
JEE Main 7Two lights of the same frequency but different intensities fall on the same metal. Compare their stopping potentials.
  1. Stopping potential is determined by Kmax.
  2. Kmax depends on frequency, not intensity.
  3. Same frequency gives same stopping potential if frequency is above threshold.
Final answer: Same stopping potential
JEE Main 8The maximum kinetic energy is 1.2 eV when photon energy is 4.5 eV. Find the work function.
  1. Einstein equation: E = Φ + Kmax.
  2. Φ = 4.5 - 1.2 = 3.3 eV.
Final answer: 3.3 eV
JEE Main 9A metal emits photoelectrons for 450 nm radiation but not for 540 nm radiation. Which interval contains its threshold wavelength?
  1. Emission occurs when λ < λ0 and fails when λ > λ0.
  2. Since 450 nm works, λ0 is at least 450 nm.
  3. Since 540 nm fails, λ0 is less than 540 nm.
Final answer: 450 nm ≤ λ0 < 540 nm
JEE Main 10Light of frequency 1.0 × 1015 Hz falls on a metal of work function 2.5 eV. Find stopping potential. Use h = 4.14 × 10-15 eV s.
  1. Photon energy E = hν = 4.14 eV.
  2. Kmax = 4.14 - 2.5 = 1.64 eV.
  3. V0 in volts has the same numerical value as Kmax in eV.
Final answer: 1.64 V
JEE Main 11If incident frequency is doubled, does Kmax double?
  1. Kmax = hν - Φ.
  2. After doubling frequency, K'max = 2hν - Φ.
  3. This is not equal to 2(hν - Φ) unless Φ = 0, which is not true for a real metal.
Final answer: No, it increases linearly but does not generally double
JEE Main 12A graph of Kmax versus frequency cuts the energy axis at -2.8 eV. Find the work function.
  1. Kmax = hν - Φ.
  2. At ν = 0, the intercept is -Φ.
  3. If intercept is -2.8 eV, Φ = 2.8 eV.
Final answer: 2.8 eV
JEE Main 13Find the de Broglie wavelength of the fastest emitted electron if Kmax = 1.6 × 10-19 J. Use m = 9.1 × 10-31 kg and h = 6.63 × 10-34 J s.
  1. Momentum p = √(2mK).
  2. p = √(2 × 9.1 × 10-31 × 1.6 × 10-19) = 5.40 × 10-25 kg m/s.
  3. λ = h/p = 6.63 × 10-34 / 5.40 × 10-25 = 1.23 × 10-9 m.
Final answer: 1.23 nm
JEE Main 14A metal plate is illuminated by photons of energy 6 eV. If stopping potential is 2 V, find threshold wavelength.
  1. Kmax = eV0 = 2 eV.
  2. Φ = E - Kmax = 6 - 2 = 4 eV.
  3. λ0 = 1240/4 = 310 nm.
Final answer: 310 nm
JEE Main 15If the number of photons incident per second is doubled without changing frequency, what happens to saturation current and stopping potential?
  1. Doubling photon rate can double emitted electron rate, so saturation current increases.
  2. Frequency is unchanged, so Kmax and stopping potential are unchanged.
Final answer: Saturation current increases; stopping potential remains same
JEE Main 16A photoelectric graph gives V0 = 0 at ν = 5 × 1014 Hz and V0 = 2.0 V at ν = 9.8 × 1014 Hz. Find h/e from the data.
  1. Slope h/e = ΔV0/Δν.
  2. h/e = 2.0 / (4.8 × 1014) = 4.17 × 10-15 V s.
Final answer: 4.17 × 10-15 V s
JEE Main 17For a metal, Φ = 2.4 eV. Which is larger: Kmax for 300 nm light or 400 nm light, and by how much?
  1. E300 = 1240/300 = 4.13 eV.
  2. E400 = 1240/400 = 3.10 eV.
  3. K values differ by E300 - E400 = 1.03 eV because Φ cancels.
Final answer: 300 nm gives larger Kmax by about 1.03 eV
JEE Main 18A photoelectron leaves with maximum speed 8.0 × 105 m/s. Calculate stopping potential. Use m = 9.1 × 10-31 kg, e = 1.6 × 10-19 C.
  1. Kmax = 1/2 mv2 = 0.5 × 9.1 × 10-31 × (8.0 × 105)2.
  2. Kmax = 2.91 × 10-19 J.
  3. V0 = K/e = 2.91/1.6 = 1.82 V.
Final answer: 1.82 V
JEE Main 19A metal has threshold wavelength 400 nm. What is its work function in eV?
  1. Φ = hc/λ0.
  2. Using hc = 1240 eV nm, Φ = 1240/400 = 3.10 eV.
Final answer: 3.10 eV
JEE Main 20The photocurrent becomes zero at a collector potential of -1.8 V. What does the negative sign physically indicate?
  1. The collector is made negative relative to the emitter.
  2. This retarding potential repels electrons.
  3. When its magnitude is 1.8 V, even the fastest emitted electrons cannot reach the collector.
Final answer: The applied field is retarding; stopping potential magnitude is 1.8 V

JEE Advanced Photoelectric Effect Questions

Ten higher-standard original questions using multi-step numericals, graph analysis, multiple-correct logic, assertion-reason and case-based reasoning.

JEE Advanced 1A metal surface is illuminated first by 300 nm light and then by 450 nm light. The stopping potentials are 2.1 V and 0.72 V respectively. Find Planck constant and work function using the two observations.

AdvancedMulti-step numerical

  1. For each wavelength, eV0 = hc/λ - Φ.
  2. Subtract the two equations: e(V1 - V2) = hc(1/λ1 - 1/λ2).
  3. Using e = 1.6 × 10-19 C, c = 3 × 108 m/s, λ1 = 300 nm, λ2 = 450 nm, and ΔV = 1.38 V gives h ≈ 6.62 × 10-34 J s.
  4. Photon energy for 300 nm is about 4.13 eV, so Φ = 4.13 - 2.1 = 2.03 eV.
Final answer: h ≈ 6.62 × 10-34 J s; Φ ≈ 2.03 eV
Key concept tested: Using two stopping-potential observations to eliminate work function
JEE Advanced 2Light of fixed frequency above threshold is incident on a metal. Intensity is increased without changing frequency. Which statements are correct? A: Saturation current increases. B: Stopping potential increases. C: Number of emitted electrons per second can increase. D: Maximum kinetic energy remains unchanged.

Moderate-AdvancedMultiple-correct style

  1. Intensity at fixed frequency increases photon number per second.
  2. More photons can eject more electrons per second, so saturation current can increase.
  3. Energy of each photon is unchanged; therefore Kmax and V0 remain unchanged.
Final answer: A, C and D
Key concept tested: Separating photon number from photon energy
JEE Advanced 3For a metal, a Kmax versus ν graph is a straight line passing through (ν = 6 × 1014 Hz, K = 0) and (ν = 10 × 1014 Hz, K = 1.66 eV). Estimate h in eV s and the work function.

AdvancedGraph-based

  1. Slope = ΔK/Δν = 1.66/(4 × 1014) eV s.
  2. h = 4.15 × 10-15 eV s.
  3. Φ = hν0 = 4.15 × 10-15 × 6 × 1014 = 2.49 eV.
Final answer: h ≈ 4.15 × 10-15 eV s; Φ ≈ 2.49 eV
Key concept tested: Slope and threshold intercept in Kmax graphs
JEE Advanced 4Assertion: Below threshold frequency, increasing intensity cannot cause photoemission in the ordinary photoelectric effect. Reason: Increasing intensity at fixed frequency increases energy of each photon.

AdvancedAssertion-reason style

  1. The assertion is true because each electron absorbs energy from a photon and needs at least Φ.
  2. The reason is false. Intensity increases the number of photons, not energy per photon.
Final answer: Assertion is true; Reason is false
Key concept tested: Threshold condition and meaning of intensity
JEE Advanced 5A metal is illuminated by monochromatic light. When the frequency is ν, stopping potential is V. When frequency is increased by 30 percent, stopping potential becomes 2V. Express the work function in terms of hν.

AdvancedParagraph/case-based

  1. First condition: eV = hν - Φ.
  2. Second condition: 2eV = 1.3hν - Φ.
  3. Subtract first from second: eV = 0.3hν.
  4. Substitute in the first: 0.3hν = hν - Φ, so Φ = 0.7hν.
Final answer: Φ = 0.7hν
Key concept tested: Algebraic relation between frequency change and stopping potential
JEE Advanced 6For photoelectric emission from a clean metal surface, which are correct? A: V0 versus ν is linear. B: Kmax versus intensity is linear at fixed frequency. C: The ν-axis intercept of Kmax graph is ν0. D: The slope of V0 versus ν is h/e.

AdvancedMultiple-correct style

  1. A follows from V0 = (h/e)ν - Φ/e.
  2. B is false because Kmax does not depend on intensity at fixed frequency.
  3. C is true because Kmax becomes zero at threshold.
  4. D is true from the coefficient of ν in the stopping-potential equation.
Final answer: A, C and D
Key concept tested: Graph interpretation in photoelectric effect
JEE Advanced 7The fastest photoelectrons from a surface have de Broglie wavelength 1.0 nm. Find the stopping potential. Use h = 6.63 × 10-34 J s, m = 9.1 × 10-31 kg, e = 1.6 × 10-19 C.

AdvancedMulti-step numerical

  1. For the fastest electron, p = h/λ = 6.63 × 10-25 kg m/s.
  2. K = p2/(2m) = (6.63 × 10-25)2/(2 × 9.1 × 10-31) = 2.42 × 10-19 J.
  3. V0 = K/e = 2.42/1.6 = 1.51 V.
Final answer: 1.51 V
Key concept tested: Connecting photoelectric Kmax with de Broglie wavelength
JEE Advanced 8Two metals have V0 versus ν graphs with equal slopes but different frequency-axis intercepts. What can be concluded about h, work functions and threshold frequencies?

AdvancedGraph-based

  1. Equal slope means both graphs have slope h/e, so Planck constant is universal.
  2. Different frequency-axis intercepts mean different threshold frequencies.
  3. Since Φ = hν0, different threshold frequencies mean different work functions.
Final answer: Same h; different ν0; different work functions
Key concept tested: Comparing metals using stopping-potential graphs
JEE Advanced 9A photon of energy 7 eV ejects an electron from a metal. The stopping potential is 4 V. What is the threshold wavelength in nm? Use hc = 1240 eV nm.

Moderate-AdvancedInteger/numerical style

  1. Kmax = eV0 = 4 eV.
  2. Φ = 7 - 4 = 3 eV.
  3. λ0 = 1240/3 = 413.3 nm.
Final answer: Approximately 413 nm
Key concept tested: Finding threshold wavelength from photon energy and stopping potential
JEE Advanced 10A metal of work function 2.5 eV is illuminated simultaneously by two monochromatic beams: 700 nm and 350 nm. Which beam contributes to Kmax, and what is the stopping potential?

AdvancedCase with limiting condition

  1. Photon energy for 700 nm is 1240/700 = 1.77 eV, below work function, so it cannot eject electrons.
  2. Photon energy for 350 nm is 1240/350 = 3.54 eV.
  3. Kmax = 3.54 - 2.5 = 1.04 eV.
  4. V0 = 1.04 V.
Final answer: 350 nm beam; stopping potential about 1.04 V
Key concept tested: Dominant photon energy when two wavelengths are incident

Case-Based Questions on Einstein Photoelectric Equation

Three compact case sets for experiment setup, stopping potential graph and Kmax versus frequency graph.

Case 1: Photoelectric Experiment Setup

A clean metal emitter and a collector are placed inside an evacuated tube. Monochromatic light of adjustable frequency falls on the emitter. The collector potential can be made positive or negative, and the current is measured with a microammeter.

1. Why is the tube evacuated?

To prevent emitted electrons from losing energy by collisions with gas molecules.

2. What happens when collector potential is made sufficiently negative?

Photocurrent becomes zero at the stopping potential.

3. Which change increases saturation current at fixed frequency above threshold?

Increasing intensity, because photon rate increases.

4. Which change increases stopping potential?

Increasing frequency of incident light.

Case 2: Stopping Potential Graph

For a metal, stopping potential is measured for different frequencies. The graph of V0 versus ν is a straight line that cuts the frequency axis at ν0.

1. What is the slope of this graph?

h/e.

2. What does the frequency-axis intercept represent?

Threshold frequency of the metal.

3. How is work function obtained from the intercept?

Φ = hν0.

4. If a second metal has a larger intercept, what is larger?

Its work function is larger.

Case 3: Kmax Versus Frequency Graph

A student plots maximum kinetic energy of photoelectrons against incident light frequency and obtains a straight line above threshold.

1. Why is the graph linear?

Because Kmax = hν - Φ.

2. What is the slope?

Planck constant h.

3. What is the energy-axis intercept?

-Φ.

4. Why is the graph not drawn for frequencies below ν0?

No photoelectrons are emitted in the ordinary one-photon effect.

Assertion-Reason Questions for NEET and JEE

Ten original assertion-reason questions focused on the logic behind threshold frequency, intensity, stopping potential and graphs.

AR 1Assertion: Increasing frequency above threshold increases maximum kinetic energy.

Reason: Photon energy is directly proportional to frequency.

Answer: Both A and R are true, and R correctly explains A.
AR 2Assertion: At threshold frequency, photoelectric current is large.

Reason: At threshold frequency, emitted electrons have zero maximum kinetic energy.

Answer: A is false, R is true.
AR 3Assertion: Stopping potential is independent of intensity for fixed frequency.

Reason: Stopping potential is determined by maximum kinetic energy.

Answer: Both A and R are true, and R correctly explains A.
AR 4Assertion: A metal with larger work function has larger threshold frequency.

Reason: Work function is related to threshold frequency by Φ = hν0.

Answer: Both A and R are true, and R correctly explains A.
AR 5Assertion: For light below threshold frequency, photoelectric emission can occur by waiting longer.

Reason: Energy from many weak photons always accumulates in one electron in the basic Einstein model.

Answer: Both A and R are false for the ordinary one-photon photoelectric effect.
AR 6Assertion: Kmax versus frequency graph is a straight line above threshold.

Reason: Einstein's equation can be written as Kmax = hν - Φ.

Answer: Both A and R are true, and R correctly explains A.
AR 7Assertion: Threshold wavelength is inversely proportional to work function.

Reason: Φ = hc/λ0.

Answer: Both A and R are true, and R correctly explains A.
AR 8Assertion: Photoelectric current becomes zero at stopping potential.

Reason: The retarding potential then prevents even the fastest electrons from reaching the collector.

Answer: Both A and R are true, and R correctly explains A.
AR 9Assertion: Increasing intensity at fixed frequency changes the slope of V0 versus ν graph.

Reason: The slope is h/e.

Answer: A is false, R is true.
AR 10Assertion: If wavelength is decreased, photon energy increases.

Reason: Photon energy is E = hc/λ.

Answer: Both A and R are true, and R correctly explains A.

Common Mistakes in Einstein Photoelectric Equation Numericals

These errors appear frequently in CBSE, NEET, JEE Main and JEE Advanced preparation.

1. Treating intensity as photon energy. Intensity changes photon count, while frequency changes energy per photon.
2. Forgetting the wavelength condition. Emission needs wavelength shorter than or equal to threshold wavelength.
3. Mixing graph slopes. Kmax versus ν has slope h, while V0 versus ν has slope h/e.
4. Using nm directly in SI formulas. Convert nm to m, or use hc = 1240 eV nm consistently.
5. Calling negative Kmax a physical answer. A negative value means no photoelectric emission.
6. Ignoring surface material. Work function and threshold frequency are metal-dependent.

Final Revision Table

Use this table for quick revision before tests and practice sessions.

Quantity Formula Meaning Exam Note
Photon energy E = hν = hc/λ Energy of one photon Frequency up, energy up; wavelength down, energy up
Work function Φ = hν0 = hc/λ0 Minimum energy to remove surface electron Depends on metal
Einstein equation hν = Φ + Kmax Energy balance for fastest emitted electron Most important formula
Stopping potential eV0 = Kmax Reverse voltage that stops fastest electrons K in eV has same number as V0 in volts
Kmax graph Kmax = hν - Φ Straight line above threshold Slope h, x-intercept ν0
V0 graph V0 = (h/e)ν - Φ/e Stopping potential versus frequency Slope h/e

Einstein Photoelectric Equation FAQs

Short answers to the most searched student questions.

FAQ 1What is Einstein's photoelectric equation?
Einstein's photoelectric equation is h ν = Φ + Kmax. It states that photon energy is used first to overcome the metal work function and the remaining energy becomes the maximum kinetic energy of the emitted electron.
FAQ 2What is work function?
Work function is the minimum energy needed to remove an electron from the surface of a metal. It is equal to h ν0 or hc divided by threshold wavelength.
FAQ 3What is stopping potential?
Stopping potential is the minimum reverse potential required to stop even the fastest emitted photoelectrons from reaching the collector. It satisfies eV0 = Kmax.
FAQ 4Why does intensity not change maximum kinetic energy?
At fixed frequency, intensity changes the νmber of photons falling per second, not the energy of each photon. Since Kmax depends on photon energy, it remains unchanged.
FAQ 5How is Planck's constant found from graph?
From a Kmax versus frequency graph, the slope is h. From a stopping potential versus frequency graph, the slope is h/e, so h is found by multiplying the slope by electronic charge.
FAQ 6How should students prepare photoelectric effect for NEET and JEE?
Students should master the energy equation, threshold frequency, threshold wavelength, stopping potential, intensity-frequency distinction, and graph slopes and intercepts, then practise mixed conceptual and νmerical questions.

Need One-to-One Help With Photoelectric Effect?

Kumar Sir has more than 30 years of teaching experience and guides students for CBSE, NEET, JEE Main, JEE Advanced, IB, IGCSE and A-Level Physics.

Kumar Physics Classes Phone / WhatsApp: +91 9958461445 Email: kumarsirphysics@gmail.com Website: https://kumarphysicsclasses.com
Call Kumar Sir WhatsApp Kumar Sir Visit Website

Kumar Physics Classes

Einstein Photoelectric Equation, Photoelectric Effect NEET Questions, JEE Main numericals and JEE Advanced graph practice by Kumar Sir.

Call Kumar Sir WhatsApp Kumar Sir
Scroll to Top