MHT-CET Physics · Dual Nature of Radiation and Matter

The Photoelectric Effect

Light ejects electrons from a metal one photon at a time: each photon of energy hν gives one electron at most hν − φ of kinetic energy, so the frequency decides whether emission happens and how fast the electrons leave, while the intensity only decides how many leave.

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Why this matters

56 PYQs, 20 of them HARD. Fifteen test what intensity and frequency each control. Eighteen apply Einstein's equation to one situation — a speed, a stopping potential, a threshold wavelength. Fifteen, most of the HARD ones, give two readings at two wavelengths or frequencies and ask for the work function or the threshold; eight read a photocurrent or stopping-potential graph. Four cards.

Concept 1 of 4: What Intensity and Frequency Each Control

Each photon carries energy hν, and one photon frees at most one electron. So FREQUENCY sets the energy per electron: below the threshold ν₀ = φ/h nothing is emitted however bright the light, and above it the maximum kinetic energy rises linearly with ν. INTENSITY sets the number of photons: it changes the photocurrent and the saturation current, never the stopping potential. Moving a source farther away dims it but leaves the stopping potential alone. Doubling the frequency more than doubles the kinetic energy, because KE = hν − φ has the work function subtracted. Photons also carry momentum h/λ, so light absorbed on a surface pushes with force P/c, and light reflected with 2P/c. Diffraction shows light's wave nature and the photoelectric effect its particle nature.

Definition

Frequency decides emission ( $\nu \ge \nu_0$ , or $\lambda \le \lambda_0$ ) and $KE_{\max}$ , which is linear in ν.
Intensity decides the number of electrons: photocurrent and saturation current ∝ intensity.
Stopping potential is independent of intensity; frequency below threshold gives zero current.
Doubling ν: $KE' = 2KE + \phi > 2KE$ ; $KE' = KE + h\nu$ .
Radiation force: absorbed $\dfrac{P}{c}$ , reflected $\dfrac{2P}{c}$ (90 W, half each ⇒ $4.5\times10^{-7}$ N).

Threshold

h\nu_0 = \phi = \frac{hc}{\lambda_0}

Worked example

Light of frequency 3ν₀ falls on a metal. Its intensity is doubled and its frequency halved. What happens to the current and the stopping potential?

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The same idea in a real exam question:

MHT-CET · 2025 · 25 April Shift II · Q14Moderate

Example 1 · Dual Nature of Radiation and Matter · Photoelectric Effect — Stopping Potential, Threshold, and Work Function

If the frequency of incident light in a photoelectric experiment is doubled, then stopping potential will

Letting intensity change the stopping potential

Brighter light means more photons of the SAME energy. The number of electrons goes up; their maximum energy, and so the stopping potential, does not.

Saying doubled frequency doubles the kinetic energy

KE = hν − φ. At 2ν it is 2hν − φ = 2KE + φ — more than double. The stopping potential likewise becomes more than double.

Drill 14 more on what intensity and frequency each control

Concept 2 of 4: Einstein's Equation in One Situation

Einstein's equation balances the photon's energy against the work function and the electron's kinetic energy: hν = φ + ½mv²ₘₐₓ, and the stopping potential is that kinetic energy in electron-volts, eV₀ = ½mv²ₘₐₓ. When the photon energy is given as a multiple of φ, the kinetic energy is that multiple minus one, and speeds go as the square root: photons of 2φ and 3φ give speeds in the ratio 1 : √2. With hc = 1240 eV nm, a wavelength in nanometres converts to photon energy in eV at once.

Definition

$h\nu = \phi + \tfrac{1}{2}mv_{\max}^2$ ; $eV_0 = \tfrac{1}{2}mv_{\max}^2$ .
$E_{\text{photon}}(\text{eV}) = \dfrac{1240}{\lambda\,(\text{nm})}$ : 310 nm on φ = 1.13 eV ⇒ $V_0 = 2.87$ V.
Photons of kφ: $KE = (k - 1)\phi$ , $v \propto \sqrt{k - 1}$ (2φ and 5φ ⇒ 1 : 2).
Threshold ν, incident 4ν: $v = \sqrt{\dfrac{6h\nu}{m}}$ .
From $v$ and $e/m$ : $V_0 = \dfrac{v^2}{2(e/m)}$ .
Two frequencies on one surface: $v_1^2 - v_2^2 = \dfrac{2h}{m}(n_1 - n_2)$ .

Einstein's equation

h\nu = \phi + \tfrac{1}{2}mv_{\max}^2 = \phi + eV_0

Worked example

Light of 248 nm falls on a metal of work function 2.0 eV. Stopping potential?

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The same idea in a real exam question:

MHT-CET · 2024 · 12th May Shift 1 · Q13Moderate

Example 2 · Dual Nature of Radiation and Matter · Photoelectric Effect — Stopping Potential, Threshold, and Work Function

Radiations of two photons having energies twice and five times the work function of metal are incident successively on metal surface. The ratio of the maximum velocity of photo electrons emitted in the two cases will be

Taking the speed ratio from the photon energies

Speed goes with the kinetic energy, and KE is the photon energy MINUS φ. Photons of 2φ and 3φ give KE φ and 2φ, speeds 1 : √2 — not √2 : √3.

Scaling the work function with the frequency

φ belongs to the metal. Two metals with φ in the ratio 1 : 2 lit by f and 2f give KE hf − φ and 2hf − 2φ, a ratio of 1 : 2.

Drill 17 more on einstein's equation in one situation

Concept 3 of 4: Threshold From Two Readings

One reading has two unknowns, the photon energy scale and φ. Two readings at two wavelengths fix both. Write eV₁ = hc/λ₁ − φ and eV₂ = hc/λ₂ − φ, then eliminate: subtracting removes φ, and a ratio of stopping potentials lets you multiply one equation to match the other. Once φ is known as a fraction of hc/λ, the threshold wavelength is hc/φ. The same method works with frequencies or with a ratio of kinetic energies.

Definition

$eV_1 = \dfrac{hc}{\lambda_1} - \phi$ , $eV_2 = \dfrac{hc}{\lambda_2} - \phi$ ; eliminate, then $\lambda_0 = \dfrac{hc}{\phi}$ .
4.8 V at λ, 1.6 V at 2λ ⇒ $\lambda_0 = 4\lambda$ . V at λ, V/6 at 3λ ⇒ $5\lambda$ . 4V₀ at λ, V₀ at 3λ ⇒ $9\lambda$ .
Frequencies: V₀ at ν, V₀/4 at ν/2 ⇒ $\nu_0 = \dfrac{\nu}{3}$ .
KE ratio 1 : k at $\nu_1, \nu_2$ : $\nu_0 = \dfrac{k\nu_1 - \nu_2}{k - 1}$ .
Speeds V and 2V: $\phi = \dfrac{hc}{3}\left[\dfrac{4}{\lambda_1} - \dfrac{1}{\lambda_2}\right]$ .

Two readings

e(V_1 - V_2) = hc\left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right), \qquad \lambda_0 = \frac{hc}{\phi}

Worked example

Light of wavelength λ gives a stopping potential of 3 V; light of 2λ gives 1 V. Threshold wavelength?

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The same idea in a real exam question:

MHT-CET · 2024 · 13th May Shift 2 · Q8Hard

Example 3 · Dual Nature of Radiation and Matter · Photoelectric Effect — Stopping Potential, Threshold, and Work Function

When radiations of wavelength '

\lambda

' is incident on a metallic surface the stopping potential required is 4.8 volt. If same surface is illuminated with radiations of double the wavelength then required stopping potential becomes 1.6 volt. Then the value of threshold wavelength for the surface is

Dividing the two equations directly

V₁/V₂ is not (hc/λ₁)/(hc/λ₂): φ sits in both. Multiply one equation so the stopping potentials match, then subtract.

Swapping λ₁ and λ₂ in the answer

The longer wavelength gives the SLOWER electron. With speeds V at λ₁ and 2V at λ₂, the 4 goes with 1/λ₁: φ = (hc/3)(4/λ₁ − 1/λ₂).

Drill 14 more on threshold from two readings

Concept 4 of 4: Reading the Photoelectric Graphs

Photocurrent against anode potential: the current is zero at the stopping potential (on the negative side) and levels off at the saturation current (on the positive side). Where a curve meets the axis tells you the FREQUENCY — a more negative cut-off means a higher frequency — and how high it levels off tells you the INTENSITY. Stopping potential against frequency is a straight line of slope h/e for every metal; it cuts the axis at the threshold frequency, so the line further right belongs to the metal with the larger work function. Kinetic energy against frequency is the same line scaled by e, starting from the threshold.

Definition

$I$ – $V$ : cut-off ⇒ frequency (more negative = higher ν); saturation level ⇒ intensity.
$V_0$ – $\nu$ : slope $\dfrac{h}{e}$ for every metal (slope × e = h); x-intercept $\nu_0$ ; intercept further right ⇒ larger φ.
$KE$ – $\nu$ : straight line starting at $\nu_0$ , slope h.

Stopping-potential line

V_0 = \frac{h}{e}\nu - \frac{\phi}{e}

Worked example

Two curves on an I–V graph share the same saturation current, but one cuts off at −2 V and the other at −1 V. Compare them.

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The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 2 · Q45Moderate

Example 4 · Dual Nature of Radiation and Matter · Photoelectric Effect — Stopping Potential, Threshold, and Work Function

The stopping potential as a function of frequency of incident radiation is plotted for two different photoelectric surfaces A and B. The graph shows that the work function of A is

Reading intensity from the cut-off

The cut-off on the potential axis is set by frequency; the saturation height is set by intensity. Curves that meet at one stopping potential share a frequency, whatever their heights.

Drill 7 more on reading the photoelectric graphs

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

What Intensity and Frequency Each Control
Threshold
$h\nu_0 = \phi = \frac{hc}{\lambda_0}$
Einstein's Equation in One Situation
Einstein's equation
$h\nu = \phi + \tfrac{1}{2}mv_{\max}^2 = \phi + eV_0$
Threshold From Two Readings
Two readings
$e(V_1 - V_2) = hc\left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right), \qquad \lambda_0 = \frac{hc}{\phi}$
Reading the Photoelectric Graphs
Stopping-potential line
$V_0 = \frac{h}{e}\nu - \frac{\phi}{e}$

Watch out for (7)

Letting intensity change the stopping potential
→ What Intensity and Frequency Each Control
Saying doubled frequency doubles the kinetic energy
→ What Intensity and Frequency Each Control
Taking the speed ratio from the photon energies
→ Einstein's Equation in One Situation
Scaling the work function with the frequency
→ Einstein's Equation in One Situation
Dividing the two equations directly
→ Threshold From Two Readings
Swapping λ₁ and λ₂ in the answer
→ Threshold From Two Readings
Reading intensity from the cut-off
→ Reading the Photoelectric Graphs

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