Master photon energy, work function, maximum kinetic energy, stopping potential and every important graph.
Classical wave theory treated light energy as continuously distributed across a wavefront. It therefore suggested that stronger intensity should give each electron more energy and that weak radiation might require an observable accumulation time. Experiments contradicted both predictions: emission was immediate, maximum kinetic energy followed frequency, and no emission occurred below a material-specific threshold frequency.
The ultraviolet catastrophe showed that classical equipartition could not describe black-body spectra. Planck resolved the crisis by proposing that oscillators exchange energy in discrete units proportional to frequency. This introduced the constant h and prepared the conceptual path for photons.
Planck proposed energy values E = nhν, where n is an integer. Energy is emitted or absorbed in packets hν rather than arbitrary continuous amounts. Higher-frequency radiation therefore has larger quantum energy.
E = nhνEinstein treated each light quantum as a localized photon carrying energy hν and momentum h/λ. In ordinary photoemission, a photon transfers its energy to one electron in a single interaction.
E = hν = hc/λFor radiation of frequency ν, Planck’s quantum hypothesis assigns each photon the energy
\(E=h\nu\)The proportionality constant h = 6.626 × 10⁻³⁴ J s.
For electromagnetic radiation in vacuum, c = νλ. Substituting ν = c/λ into E = hν gives
\(E=\dfrac{hc}{\lambda}\)Photon energy is directly proportional to frequency. If frequency doubles, energy per photon doubles. This is why ultraviolet photons can eject electrons from surfaces for which red photons fail.
Photon energy is inversely proportional to wavelength. Short wavelength means high photon energy. A useful numerical relation is E(eV) = 1240/λ(nm).
\(E(\mathrm{eV})=\dfrac{1240}{\lambda(\mathrm{nm})}\)Example: a 400 nm photon has energy 1240/400 = 3.10 eV. A 200 nm photon has twice this energy, 6.20 eV.
The work function is the minimum energy required to remove an electron from a particular material surface. It depends on material, crystal face and surface condition.
\(\Phi=h\nu_0\)The least incident frequency capable of photoemission. At threshold, the fastest electron emerges with zero kinetic energy.
\(\nu_0=\dfrac{\Phi}{h}\)The greatest wavelength capable of photoemission. Since c = ν₀λ₀,
\(\lambda_0=\dfrac{c}{\nu_0}=\dfrac{hc}{\Phi}\)Photoemission requires hν ≥ Φ, equivalently ν ≥ ν₀ or λ ≤ λ₀. High intensity cannot compensate for a photon frequency below threshold.
At the threshold, the electron just escapes and Kmax = 0.
Apply energy conservation: hν₀ = Φ + 0.
Rearrange for threshold frequency.
Use c = ν₀λ₀ to obtain threshold wavelength.
Conceptual example: If Φ = 2.0 eV, then λ₀ = 1240/2.0 = 620 nm. Radiation of 700 nm cannot eject electrons, while 500 nm radiation can.
Einstein applied conservation of energy to a single photon absorbed by a single electron. The photon disappears in the interaction. The minimum energy Φ is spent in escaping the surface, and the remainder is available as electron kinetic energy.
m is electron mass and vmax is the greatest speed among emitted electrons.
Electrical work eV₀ equals the kinetic energy of the fastest electron.
hν is fixed by frequency. Increasing intensity at the same ν supplies more photons but does not change this energy per photon.
The equation is an energy ledger. It explains why frequency determines electron energy and why a minimum frequency is required.
Kmax increases linearly with frequency above threshold.
Electron speed varies as the square root of excess photon energy.
At fixed metal surface, an increase Δν produces ΔKmax = hΔν. The slope is universal and equal to h.
Intensity changes photon flux and electron count, not hν. Therefore it changes current but not the upper endpoint of the kinetic-energy distribution.
NEET/JEE key distinction: frequency controls the energy of individual photons; intensity controls how many photons arrive per second.
A reverse voltage makes the anode negative relative to the emitting cathode and retards photoelectrons. As reverse potential grows, fewer electrons reach the collector. At V₀ even the fastest electron is stopped, so current becomes zero.
For electron energies in eV, V₀ in volts has the same numerical value.
Slope h/e is independent of metal; intercept depends on work function.
Any reverse potential opposing electron collection is retarding. The special value that reduces photocurrent to zero is the stopping potential.
V₀ measures the fastest photoelectron’s energy without needing to know electron speed directly.
Start with Kmax = hν − Φ.
At the stopping potential, electric work eV₀ equals Kmax.
Write eV₀ = hν − Φ.
Divide by e and compare with the straight-line form y = mx + c.
Slope = h, x-intercept = ν₀ and extrapolated y-intercept = −Φ. A larger work function shifts the threshold to the right.
Slope = h/e, x-intercept = ν₀ and y-intercept = −Φ/e. Parallel lines correspond to different metals.
The graph passes through the origin because zero frequency corresponds to zero photon energy in the relation E = hν.
The inverse curve falls steeply at short wavelength and gradually approaches zero as wavelength increases.
A photon of energy hν is completely absorbed by one electron.
The electron spends at least Φ to escape the surface.
The remaining energy appears as kinetic energy.
For the most energetic electron, the remainder is Kmax.
Begin with hν = Φ + Kmax.
Subtract Φ from both sides.
For a non-relativistic electron use Kmax = ½mvmax².
At V₀ the fastest electron loses all kinetic energy against the electric field.
The electric work is eV₀.
Set eV₀ = Kmax and substitute Einstein’s equation.
At threshold the fastest electron has Kmax = 0.
Einstein’s equation becomes hν₀ = Φ.
Divide by h.
Use c = ν₀λ₀.
Replace ν₀ by Φ/h.
Solve for λ₀.
Open each question to reveal its answer and explanation.
Radiation of frequency 5.20 × 10¹⁴ Hz falls on a metal of work function 1.90 eV. Find Kmax and V₀.
Answer: Kmax = 0.25 eV; V₀ = 0.25 V.
Complete solution: Use Kmax = hν − Φ with h = 4.136 × 10⁻¹⁵ eV s. Thus hν = 2.15 eV and Kmax = 2.15 − 1.90 = 0.25 eV. Since eV₀ = Kmax, V₀ has the same numerical value in volts.
The work function of a metal is 2.15 eV. Calculate threshold frequency and threshold wavelength.
Answer: ν₀ = 5.198e+14 Hz; λ₀ = 576.7 nm.
Complete solution: Threshold condition is Φ = hν₀, so ν₀ = Φ/h = 2.15/4.136e-15 = 5.198e+14 Hz. Also λ₀ = hc/Φ = 1240/2.15 = 576.7 nm.
Light of wavelength 270 nm illuminates a surface with Φ = 2.2 eV. Find photon energy, Kmax and stopping potential.
Answer: E = 4.59 eV, Kmax = 2.39 eV, V₀ = 2.39 V.
Complete solution: E = hc/λ = 1240/270 = 4.59 eV. Subtract the work function: Kmax = 4.59 − 2.2 = 2.39 eV. Therefore V₀ = 2.39 V.
A metal of work function 2.10 eV has stopping potential 1.25 V. Find the incident frequency.
Answer: ν = 8.100e+14 Hz.
Complete solution: From hν = Φ + eV₀, energies in eV give hν = 2.10 + 1.25 = 3.35 eV. Hence ν = 3.35/4.136e-15 = 8.100e+14 Hz.
Threshold frequency is 4.4 × 10¹⁴ Hz. Find V₀ for incident frequency 5.9 × 10¹⁴ Hz.
Answer: V₀ = 0.62 V.
Complete solution: Subtract threshold form hν₀ = Φ from hν = Φ + eV₀. Then eV₀ = h(ν − ν₀). In eV units V₀ = 4.136e-15 × 1.5 × 10¹⁴ = 0.62 V.
A Kmax–ν graph has slope 4.136 × 10⁻¹⁵ eV s and y-intercept −2.2 eV. Determine h, Φ and ν₀.
Answer: h = 4.136 × 10⁻¹⁵ eV s, Φ = 2.2 eV, ν₀ = 5.319e+14 Hz.
Complete solution: Compare Kmax = hν − Φ with y = mx + c. Thus slope = h and intercept = −Φ. Finally ν₀ = Φ/h = 5.319e+14 Hz.
The threshold wavelength is 530 nm. Find the work function in eV and joules.
Answer: Φ = 2.34 eV = 3.748e-19 J.
Complete solution: Φ = hc/λ₀ = 1240/530 = 2.34 eV. Multiplying by 1.602 × 10⁻¹⁹ J/eV gives 3.748e-19 J.
Photon energies 3.4 eV and 4.6 eV illuminate the same Φ = 2.0 eV metal. Find the ratio of maximum electron speeds.
Answer: v₂/v₁ = 1.36.
Complete solution: K₁ = 3.4 − 2.0 = 1.4 eV and K₂ = 4.6 − 2.0 = 2.6 eV. Since K = ½mv², v₂/v₁ = √(K₂/K₁) = 1.36.
A V₀–ν line has slope 4.14 × 10⁻¹⁵ V s. At 7.0 × 10¹⁴ Hz, V₀ = 1.20 V. Find Φ in eV.
Answer: Φ = 1.70 eV.
Complete solution: Use V₀ = (h/e)ν − Φ/e. When h/e is written in V s and Φ/e in volts, Φ in eV equals slope × ν − V₀ = 2.90 − 1.20 = 1.70 eV.
Compare photons of wavelengths 270 nm and 540 nm. Find E₁/E₂ and the energy difference.
Answer: E₁/E₂ = 2; ΔE = 2.30 eV.
Complete solution: Because E = hc/λ, E₁/E₂ = λ₂/λ₁ = 540/270 = 2. Numerically E₁ = 4.59 eV and E₂ = 2.30 eV, so ΔE = 2.30 eV.
Radiation of frequency 6.40 × 10¹⁴ Hz falls on a metal of work function 2.10 eV. Find Kmax and V₀.
Answer: Kmax = 0.55 eV; V₀ = 0.55 V.
Complete solution: Use Kmax = hν − Φ with h = 4.136 × 10⁻¹⁵ eV s. Thus hν = 2.65 eV and Kmax = 2.65 − 2.10 = 0.55 eV. Since eV₀ = Kmax, V₀ has the same numerical value in volts.
The work function of a metal is 2.15 eV. Calculate threshold frequency and threshold wavelength.
Answer: ν₀ = 5.198e+14 Hz; λ₀ = 576.7 nm.
Complete solution: Threshold condition is Φ = hν₀, so ν₀ = Φ/h = 2.15/4.136e-15 = 5.198e+14 Hz. Also λ₀ = hc/Φ = 1240/2.15 = 576.7 nm.
Light of wavelength 290 nm illuminates a surface with Φ = 1.8 eV. Find photon energy, Kmax and stopping potential.
Answer: E = 4.28 eV, Kmax = 2.48 eV, V₀ = 2.48 V.
Complete solution: E = hc/λ = 1240/290 = 4.28 eV. Subtract the work function: Kmax = 4.28 − 1.8 = 2.48 eV. Therefore V₀ = 2.48 V.
A metal of work function 2.10 eV has stopping potential 1.70 V. Find the incident frequency.
Answer: ν = 9.188e+14 Hz.
Complete solution: From hν = Φ + eV₀, energies in eV give hν = 2.10 + 1.70 = 3.80 eV. Hence ν = 3.80/4.136e-15 = 9.188e+14 Hz.
Threshold frequency is 4.4 × 10¹⁴ Hz. Find V₀ for incident frequency 5.9 × 10¹⁴ Hz.
Answer: V₀ = 0.62 V.
Complete solution: Subtract threshold form hν₀ = Φ from hν = Φ + eV₀. Then eV₀ = h(ν − ν₀). In eV units V₀ = 4.136e-15 × 1.5 × 10¹⁴ = 0.62 V.
A Kmax–ν graph has slope 4.136 × 10⁻¹⁵ eV s and y-intercept −2.2 eV. Determine h, Φ and ν₀.
Answer: h = 4.136 × 10⁻¹⁵ eV s, Φ = 2.2 eV, ν₀ = 5.319e+14 Hz.
Complete solution: Compare Kmax = hν − Φ with y = mx + c. Thus slope = h and intercept = −Φ. Finally ν₀ = Φ/h = 5.319e+14 Hz.
The threshold wavelength is 530 nm. Find the work function in eV and joules.
Answer: Φ = 2.34 eV = 3.748e-19 J.
Complete solution: Φ = hc/λ₀ = 1240/530 = 2.34 eV. Multiplying by 1.602 × 10⁻¹⁹ J/eV gives 3.748e-19 J.
Photon energies 3.2 eV and 4.4 eV illuminate the same Φ = 2.0 eV metal. Find the ratio of maximum electron speeds.
Answer: v₂/v₁ = 1.41.
Complete solution: K₁ = 3.2 − 2.0 = 1.2 eV and K₂ = 4.4 − 2.0 = 2.4 eV. Since K = ½mv², v₂/v₁ = √(K₂/K₁) = 1.41.
A V₀–ν line has slope 4.14 × 10⁻¹⁵ V s. At 7.0 × 10¹⁴ Hz, V₀ = 1.20 V. Find Φ in eV.
Answer: Φ = 1.70 eV.
Complete solution: Use V₀ = (h/e)ν − Φ/e. When h/e is written in V s and Φ/e in volts, Φ in eV equals slope × ν − V₀ = 2.90 − 1.20 = 1.70 eV.
Compare photons of wavelengths 270 nm and 540 nm. Find E₁/E₂ and the energy difference.
Answer: E₁/E₂ = 2; ΔE = 2.30 eV.
Complete solution: Because E = hc/λ, E₁/E₂ = λ₂/λ₁ = 540/270 = 2. Numerically E₁ = 4.59 eV and E₂ = 2.30 eV, so ΔE = 2.30 eV.
Radiation of frequency 7.60 × 10¹⁴ Hz falls on a metal of work function 1.90 eV. Find Kmax and V₀.
Answer: Kmax = 1.24 eV; V₀ = 1.24 V.
Complete solution: Use Kmax = hν − Φ with h = 4.136 × 10⁻¹⁵ eV s. Thus hν = 3.14 eV and Kmax = 3.14 − 1.90 = 1.24 eV. Since eV₀ = Kmax, V₀ has the same numerical value in volts.
The work function of a metal is 2.15 eV. Calculate threshold frequency and threshold wavelength.
Answer: ν₀ = 5.198e+14 Hz; λ₀ = 576.7 nm.
Complete solution: Threshold condition is Φ = hν₀, so ν₀ = Φ/h = 2.15/4.136e-15 = 5.198e+14 Hz. Also λ₀ = hc/Φ = 1240/2.15 = 576.7 nm.
Light of wavelength 310 nm illuminates a surface with Φ = 2.2 eV. Find photon energy, Kmax and stopping potential.
Answer: E = 4.00 eV, Kmax = 1.80 eV, V₀ = 1.80 V.
Complete solution: E = hc/λ = 1240/310 = 4.00 eV. Subtract the work function: Kmax = 4.00 − 2.2 = 1.80 eV. Therefore V₀ = 1.80 V.
A metal of work function 2.10 eV has stopping potential 1.10 V. Find the incident frequency.
Answer: ν = 7.737e+14 Hz.
Complete solution: From hν = Φ + eV₀, energies in eV give hν = 2.10 + 1.10 = 3.20 eV. Hence ν = 3.20/4.136e-15 = 7.737e+14 Hz.
Threshold frequency is 4.4 × 10¹⁴ Hz. Find V₀ for incident frequency 5.9 × 10¹⁴ Hz.
Answer: V₀ = 0.62 V.
Complete solution: Subtract threshold form hν₀ = Φ from hν = Φ + eV₀. Then eV₀ = h(ν − ν₀). In eV units V₀ = 4.136e-15 × 1.5 × 10¹⁴ = 0.62 V.
A Kmax–ν graph has slope 4.136 × 10⁻¹⁵ eV s and y-intercept −2.2 eV. Determine h, Φ and ν₀.
Answer: h = 4.136 × 10⁻¹⁵ eV s, Φ = 2.2 eV, ν₀ = 5.319e+14 Hz.
Complete solution: Compare Kmax = hν − Φ with y = mx + c. Thus slope = h and intercept = −Φ. Finally ν₀ = Φ/h = 5.319e+14 Hz.
The threshold wavelength is 530 nm. Find the work function in eV and joules.
Answer: Φ = 2.34 eV = 3.748e-19 J.
Complete solution: Φ = hc/λ₀ = 1240/530 = 2.34 eV. Multiplying by 1.602 × 10⁻¹⁹ J/eV gives 3.748e-19 J.
Photon energies 3.4 eV and 4.6 eV illuminate the same Φ = 2.0 eV metal. Find the ratio of maximum electron speeds.
Answer: v₂/v₁ = 1.36.
Complete solution: K₁ = 3.4 − 2.0 = 1.4 eV and K₂ = 4.6 − 2.0 = 2.6 eV. Since K = ½mv², v₂/v₁ = √(K₂/K₁) = 1.36.
A V₀–ν line has slope 4.14 × 10⁻¹⁵ V s. At 7.0 × 10¹⁴ Hz, V₀ = 1.20 V. Find Φ in eV.
Answer: Φ = 1.70 eV.
Complete solution: Use V₀ = (h/e)ν − Φ/e. When h/e is written in V s and Φ/e in volts, Φ in eV equals slope × ν − V₀ = 2.90 − 1.20 = 1.70 eV.
Compare photons of wavelengths 270 nm and 540 nm. Find E₁/E₂ and the energy difference.
Answer: E₁/E₂ = 2; ΔE = 2.30 eV.
Complete solution: Because E = hc/λ, E₁/E₂ = λ₂/λ₁ = 540/270 = 2. Numerically E₁ = 4.59 eV and E₂ = 2.30 eV, so ΔE = 2.30 eV.
Radiation of frequency 8.80 × 10¹⁴ Hz falls on a metal of work function 2.10 eV. Find Kmax and V₀.
Answer: Kmax = 1.54 eV; V₀ = 1.54 V.
Complete solution: Use Kmax = hν − Φ with h = 4.136 × 10⁻¹⁵ eV s. Thus hν = 3.64 eV and Kmax = 3.64 − 2.10 = 1.54 eV. Since eV₀ = Kmax, V₀ has the same numerical value in volts.
The work function of a metal is 2.15 eV. Calculate threshold frequency and threshold wavelength.
Answer: ν₀ = 5.198e+14 Hz; λ₀ = 576.7 nm.
Complete solution: Threshold condition is Φ = hν₀, so ν₀ = Φ/h = 2.15/4.136e-15 = 5.198e+14 Hz. Also λ₀ = hc/Φ = 1240/2.15 = 576.7 nm.
Light of wavelength 250 nm illuminates a surface with Φ = 1.8 eV. Find photon energy, Kmax and stopping potential.
Answer: E = 4.96 eV, Kmax = 3.16 eV, V₀ = 3.16 V.
Complete solution: E = hc/λ = 1240/250 = 4.96 eV. Subtract the work function: Kmax = 4.96 − 1.8 = 3.16 eV. Therefore V₀ = 3.16 V.
A metal of work function 2.10 eV has stopping potential 1.55 V. Find the incident frequency.
Answer: ν = 8.825e+14 Hz.
Complete solution: From hν = Φ + eV₀, energies in eV give hν = 2.10 + 1.55 = 3.65 eV. Hence ν = 3.65/4.136e-15 = 8.825e+14 Hz.
Threshold frequency is 4.4 × 10¹⁴ Hz. Find V₀ for incident frequency 5.9 × 10¹⁴ Hz.
Answer: V₀ = 0.62 V.
Complete solution: Subtract threshold form hν₀ = Φ from hν = Φ + eV₀. Then eV₀ = h(ν − ν₀). In eV units V₀ = 4.136e-15 × 1.5 × 10¹⁴ = 0.62 V.
A Kmax–ν graph has slope 4.136 × 10⁻¹⁵ eV s and y-intercept −2.2 eV. Determine h, Φ and ν₀.
Answer: h = 4.136 × 10⁻¹⁵ eV s, Φ = 2.2 eV, ν₀ = 5.319e+14 Hz.
Complete solution: Compare Kmax = hν − Φ with y = mx + c. Thus slope = h and intercept = −Φ. Finally ν₀ = Φ/h = 5.319e+14 Hz.
The threshold wavelength is 530 nm. Find the work function in eV and joules.
Answer: Φ = 2.34 eV = 3.748e-19 J.
Complete solution: Φ = hc/λ₀ = 1240/530 = 2.34 eV. Multiplying by 1.602 × 10⁻¹⁹ J/eV gives 3.748e-19 J.
Photon energies 3.2 eV and 4.4 eV illuminate the same Φ = 2.0 eV metal. Find the ratio of maximum electron speeds.
Answer: v₂/v₁ = 1.41.
Complete solution: K₁ = 3.2 − 2.0 = 1.2 eV and K₂ = 4.4 − 2.0 = 2.4 eV. Since K = ½mv², v₂/v₁ = √(K₂/K₁) = 1.41.
A V₀–ν line has slope 4.14 × 10⁻¹⁵ V s. At 7.0 × 10¹⁴ Hz, V₀ = 1.20 V. Find Φ in eV.
Answer: Φ = 1.70 eV.
Complete solution: Use V₀ = (h/e)ν − Φ/e. When h/e is written in V s and Φ/e in volts, Φ in eV equals slope × ν − V₀ = 2.90 − 1.20 = 1.70 eV.
Compare photons of wavelengths 270 nm and 540 nm. Find E₁/E₂ and the energy difference.
Answer: E₁/E₂ = 2; ΔE = 2.30 eV.
Complete solution: Because E = hc/λ, E₁/E₂ = λ₂/λ₁ = 540/270 = 2. Numerically E₁ = 4.59 eV and E₂ = 2.30 eV, so ΔE = 2.30 eV.
Radiation of frequency 10.00 × 10¹⁴ Hz falls on a metal of work function 1.90 eV. Find Kmax and V₀.
Answer: Kmax = 2.24 eV; V₀ = 2.24 V.
Complete solution: Use Kmax = hν − Φ with h = 4.136 × 10⁻¹⁵ eV s. Thus hν = 4.14 eV and Kmax = 4.14 − 1.90 = 2.24 eV. Since eV₀ = Kmax, V₀ has the same numerical value in volts.
The work function of a metal is 2.15 eV. Calculate threshold frequency and threshold wavelength.
Answer: ν₀ = 5.198e+14 Hz; λ₀ = 576.7 nm.
Complete solution: Threshold condition is Φ = hν₀, so ν₀ = Φ/h = 2.15/4.136e-15 = 5.198e+14 Hz. Also λ₀ = hc/Φ = 1240/2.15 = 576.7 nm.
Light of wavelength 270 nm illuminates a surface with Φ = 2.2 eV. Find photon energy, Kmax and stopping potential.
Answer: E = 4.59 eV, Kmax = 2.39 eV, V₀ = 2.39 V.
Complete solution: E = hc/λ = 1240/270 = 4.59 eV. Subtract the work function: Kmax = 4.59 − 2.2 = 2.39 eV. Therefore V₀ = 2.39 V.
A metal of work function 2.10 eV has stopping potential 0.95 V. Find the incident frequency.
Answer: ν = 7.374e+14 Hz.
Complete solution: From hν = Φ + eV₀, energies in eV give hν = 2.10 + 0.95 = 3.05 eV. Hence ν = 3.05/4.136e-15 = 7.374e+14 Hz.
Threshold frequency is 4.4 × 10¹⁴ Hz. Find V₀ for incident frequency 5.9 × 10¹⁴ Hz.
Answer: V₀ = 0.62 V.
Complete solution: Subtract threshold form hν₀ = Φ from hν = Φ + eV₀. Then eV₀ = h(ν − ν₀). In eV units V₀ = 4.136e-15 × 1.5 × 10¹⁴ = 0.62 V.
A Kmax–ν graph has slope 4.136 × 10⁻¹⁵ eV s and y-intercept −2.2 eV. Determine h, Φ and ν₀.
Answer: h = 4.136 × 10⁻¹⁵ eV s, Φ = 2.2 eV, ν₀ = 5.319e+14 Hz.
Complete solution: Compare Kmax = hν − Φ with y = mx + c. Thus slope = h and intercept = −Φ. Finally ν₀ = Φ/h = 5.319e+14 Hz.
The threshold wavelength is 530 nm. Find the work function in eV and joules.
Answer: Φ = 2.34 eV = 3.748e-19 J.
Complete solution: Φ = hc/λ₀ = 1240/530 = 2.34 eV. Multiplying by 1.602 × 10⁻¹⁹ J/eV gives 3.748e-19 J.
Photon energies 3.4 eV and 4.6 eV illuminate the same Φ = 2.0 eV metal. Find the ratio of maximum electron speeds.
Answer: v₂/v₁ = 1.36.
Complete solution: K₁ = 3.4 − 2.0 = 1.4 eV and K₂ = 4.6 − 2.0 = 2.6 eV. Since K = ½mv², v₂/v₁ = √(K₂/K₁) = 1.36.
A V₀–ν line has slope 4.14 × 10⁻¹⁵ V s. At 7.0 × 10¹⁴ Hz, V₀ = 1.20 V. Find Φ in eV.
Answer: Φ = 1.70 eV.
Complete solution: Use V₀ = (h/e)ν − Φ/e. When h/e is written in V s and Φ/e in volts, Φ in eV equals slope × ν − V₀ = 2.90 − 1.20 = 1.70 eV.
Compare photons of wavelengths 270 nm and 540 nm. Find E₁/E₂ and the energy difference.
Answer: E₁/E₂ = 2; ΔE = 2.30 eV.
Complete solution: Because E = hc/λ, E₁/E₂ = λ₂/λ₁ = 540/270 = 2. Numerically E₁ = 4.59 eV and E₂ = 2.30 eV, so ΔE = 2.30 eV. This inverse dependence must be applied before subtracting any work function.
Radiation of frequency 11.20 × 10¹⁴ Hz falls on a metal of work function 2.10 eV. Find Kmax and V₀.
Answer: Kmax = 2.53 eV; V₀ = 2.53 V.
Complete solution: Use Kmax = hν − Φ with h = 4.136 × 10⁻¹⁵ eV s. Thus hν = 4.63 eV and Kmax = 4.63 − 2.10 = 2.53 eV. Since eV₀ = Kmax, V₀ has the same numerical value in volts.
The work function of a metal is 2.15 eV. Calculate threshold frequency and threshold wavelength.
Answer: ν₀ = 5.198e+14 Hz; λ₀ = 576.7 nm.
Complete solution: Threshold condition is Φ = hν₀, so ν₀ = Φ/h = 2.15/4.136e-15 = 5.198e+14 Hz. Also λ₀ = hc/Φ = 1240/2.15 = 576.7 nm.
Light of wavelength 290 nm illuminates a surface with Φ = 1.8 eV. Find photon energy, Kmax and stopping potential.
Answer: E = 4.28 eV, Kmax = 2.48 eV, V₀ = 2.48 V.
Complete solution: E = hc/λ = 1240/290 = 4.28 eV. Subtract the work function: Kmax = 4.28 − 1.8 = 2.48 eV. Therefore V₀ = 2.48 V.
A metal of work function 2.10 eV has stopping potential 1.40 V. Find the incident frequency.
Answer: ν = 8.462e+14 Hz.
Complete solution: From hν = Φ + eV₀, energies in eV give hν = 2.10 + 1.40 = 3.50 eV. Hence ν = 3.50/4.136e-15 = 8.462e+14 Hz.
Threshold frequency is 4.4 × 10¹⁴ Hz. Find V₀ for incident frequency 5.9 × 10¹⁴ Hz.
Answer: V₀ = 0.62 V.
Complete solution: Subtract threshold form hν₀ = Φ from hν = Φ + eV₀. Then eV₀ = h(ν − ν₀). In eV units V₀ = 4.136e-15 × 1.5 × 10¹⁴ = 0.62 V.
The following are original PYQ-style practice questions aligned with recurring exam patterns; they are not presented as verbatim copyrighted past-paper text.
E = hν = hc/λ
Φ = hν₀ = hc/λ₀
Kmax = hν − Φ
Kmax = ½mvmax² = eV₀
V₀ = (h/e)ν − Φ/e
Kmax–ν: slope h.
V₀–ν: slope h/e.
E–ν: line through origin.
E–λ: inverse curve.
Frequency controls photon energy and Kmax. Intensity controls photon number and current. Emission is impossible below ν₀ in the ordinary one-photon regime.
Find Φ from an intercept; calculate λ₀; compare two metals; determine h from graph slope; distinguish intensity and frequency effects.
Do not set intensity proportional to photon energy. Do not treat negative Kmax as physical emission. Convert nm to m or use hc = 1240 eV nm consistently.
Write the energy balance first. Use eV units to simplify stopping-potential problems. Mark slope and both intercepts clearly on graph questions.
