JEE Mains Physics · Dual Nature of Radiation and Matter

Einstein's Equation and Stopping Potential

Each photon's energy pays the work function and the rest becomes the fastest electron's kinetic energy: hν = φ + K_max with K_max = eV₀; two readings on one metal let the work function be eliminated.

All Dual Nature of Radiation and Matter notes JEE Mains Physics strategy

Why this matters

Thirty-two PYQs, twenty-nine of them multiple choice, and five from 2026. Twelve shine one light on one metal: six go straight from the photon energy to the work function or the stopping potential, two give the light as an electric field, three send the fastest electron into a magnetic field, and one makes the photon from a hydrogen atom capturing an electron. Fourteen use two lights on the same metal and remove the work function by subtracting. Six compare the fastest electrons' speeds.

Concept 1 of 3: Einstein's photoelectric equation for one light

A photon gives all its energy to one electron. Part of it pays the work function to get the electron out; what is left is the kinetic energy of the fastest electron. The stopping potential is that kinetic energy measured in volts: an electron with 3 eV to spare is stopped by 3 V. Working in electron-volts keeps every step to one line.

Definition

$h\nu = \dfrac{hc}{\lambda} = \phi + K_{\max}$ , and $K_{\max} = eV_0$ . In eV, $V_0$ in volts equals $K_{\max}$ in eV.
Light given as $E = E_0[\sin\omega_1 t + \sin\omega_2 t]$ holds two frequencies. The fastest electrons come from the larger ω, with $\nu = \omega/2\pi$ .
An electron moving at right angles to a magnetic field B goes round a circle of radius $r = \dfrac{mv}{eB} = \dfrac{\sqrt{2mK}}{eB}$ , so $K = \dfrac{(eBr)^{2}}{2m}$ . A half circle brings it back to the plate a diameter, 2r, from where it left.
The photon may come from hydrogen: a jump from $n_2$ to $n_1$ gives $13.6\left(\dfrac{1}{n_1^{2}} - \dfrac{1}{n_2^{2}}\right)$ eV. An electron of kinetic energy K captured into level n gives $K + \dfrac{13.6}{n^{2}}$ eV.

Einstein's photoelectric equation

h\nu = \frac{hc}{\lambda} = \phi + K_{\max}, \qquad K_{\max} = eV_0

Worked example

Light of wavelength 248 nm falls on a metal of work function 2.3 eV. Find the stopping potential and the maximum speed of the electrons. (hc = 1240 eV nm, m = 9.1 × 10⁻³¹ kg, 1 eV = 1.6 × 10⁻¹⁹ J)

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The same idea in a real exam question:

JEE Mains · 2025 · 23 January 2025 · Q81Moderate

Example 1 · Dual Nature of Radiation and Matter · Einstein's Equation and Stopping Potential

In photoelectric effect an em-wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is 2.14 eV and stopping potential is 2 V , what is the wavelength of the em-wave? (Given hc

= 1242eVnm

where h is the Planck's constant and c is the speed of light in vaccum.)

Divide ω by 2π

A light wave written as sin(ωt) gives the angular frequency. The photon energy is hω/2π. Using hω makes it 6.28 times too large.

Two frequencies, use the higher

When the field holds two sine terms, the maximum kinetic energy comes from the higher frequency. The lower one gives slower electrons and is never the answer to a maximum.

Volts and electron-volts are the same number

A stopping potential of 2 V means the fastest electron had 2 eV. Working in eV needs no factor of e; converting to joules and back is where 1.6 × 10⁻¹⁹ errors come from.

Drill 11 more on einstein's photoelectric equation for one light

Concept 2 of 3: Two wavelengths on the same metal

When the work function is not given, the question gives two readings instead. Write Einstein's equation for each light. The work function is the same in both, so subtracting the two equations removes it. Then put the result back into either equation to find φ or the threshold.

Definition

$\dfrac{hc}{\lambda_1} = \phi + eV_1$ and $\dfrac{hc}{\lambda_2} = \phi + eV_2$ .
Subtract: $e(V_1 - V_2) = hc\left(\dfrac{1}{\lambda_1} - \dfrac{1}{\lambda_2}\right)$ . φ is gone.
When the wavelengths are λ and nλ, call $a = hc/\lambda$ . Then $a = \phi + eV_1$ and $a/n = \phi + eV_2$ .
The threshold follows from $\phi = hc/\lambda_0$ .
A longer wavelength always gives the smaller stopping potential. Check this before solving a stem.
To double the kinetic energy, the new photon must carry $\phi + 2K$ , not twice the old photon energy.

Eliminating the work function

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

Worked example

The stopping potential for a metal is 6 V with light of wavelength λ and 1 V with light of wavelength 2λ. Find the threshold wavelength in terms of λ.

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The same idea in a real exam question:

JEE Mains · 2024 · 31 January 2024 · Q20Moderate

Example 2 · Dual Nature of Radiation and Matter · Einstein's Equation and Stopping Potential

When a metal surface is illuminated by light of wavelength

\lambda

, the stopping potential is

8\text{ }V

. When the same surface is illuminated by light of wavelength

3\lambda

, stopping potential is

2\text{ }V

. The threshold wavelength for this surface is:

Subtract, do not divide

Dividing the two Einstein equations leaves the work function in both numerator and denominator. Subtracting them removes it in one step.

A longer wavelength cannot give a larger stopping potential

Longer wavelength means less energy per photon, so less left over after the work function. A stem that gives the longer wavelength the larger stopping potential describes something impossible.

Doubling K is not halving λ

To double the kinetic energy the photon must carry φ + 2K. Halving the wavelength doubles the photon energy to 2φ + 2K, which gives more than 2K.

Drill 13 more on two wavelengths on the same metal

Concept 3 of 3: Comparing the maximum speeds of photoelectrons

Speed comes from kinetic energy, and kinetic energy is what is left after the work function. So find the leftover energy in each case first, then take the square root of their ratio. Comparing the photon energies directly, or forgetting the square root, gives one of the wrong options.

Definition

$\tfrac{1}{2}mv_{\max}^{2} = h\nu - \phi = h(\nu - \nu_0)$ .
Photon energy $k\phi$ : $K = (k - 1)\phi$ , so $v \propto \sqrt{k - 1}$ .
Frequency $k\nu_0$ : $K = (k - 1)h\nu_0$ , so again $v \propto \sqrt{k - 1}$ .
Two frequencies on identical cathodes: subtract the equations, $\tfrac{1}{2}m(v_1^{2} - v_2^{2}) = h(\nu_1 - \nu_2)$ ; φ cancels.

Speed of the fastest photoelectron

\tfrac{1}{2}mv_{\max}^{2} = h(\nu - \nu_0), \qquad \frac{v_1}{v_2} = \sqrt{\frac{\nu_1 - \nu_0}{\nu_2 - \nu_0}}

Worked example

Photons of 6.0 eV and then 2.0 eV fall on a metal of work function 1.5 eV. Find the ratio of the maximum speeds of the electrons in the two cases.

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The same idea in a real exam question:

JEE Mains · 2023 · 1 February 2023 · Q104Moderate

Example 3 · Dual Nature of Radiation and Matter · Einstein's Equation and Stopping Potential

The threshold frequency of a metal is

f_{0}

. When the light of frequency

2f_{0}

is incident on the metal plate, the maximum velocity of photoelectrons is

v_{1}

. When the frequency of incident radiation is increased to

5f_{0}

, the maximum velocity of photoelectrons emitted is

v_{2}

. The ratio of

v_{1}

v_{2}

is:

Subtract the work function before the square root

Speed goes as the square root of the leftover energy, k − 1 times φ, not of the photon energy kφ. Photons of 2φ and 8φ give speeds in the ratio 1 : √7, not 1 : 2.

A speed ratio is not an energy ratio

Kinetic energy goes as v². If the energies are in the ratio 1 : 4, the speeds are 1 : 2. Stopping at the energy ratio picks a wrong option.

Drill 5 more on comparing the maximum speeds of photoelectrons

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 (3)

Einstein's photoelectric equation for one light
Einstein's photoelectric equation
$h\nu = \frac{hc}{\lambda} = \phi + K_{\max}, \qquad K_{\max} = eV_0$
Two wavelengths on the same metal
Eliminating the work function
$e(V_1 - V_2) = hc\left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right)$
Comparing the maximum speeds of photoelectrons
Speed of the fastest photoelectron
$\tfrac{1}{2}mv_{\max}^{2} = h(\nu - \nu_0), \qquad \frac{v_1}{v_2} = \sqrt{\frac{\nu_1 - \nu_0}{\nu_2 - \nu_0}}$

Watch out for (8)

Divide ω by 2π
→ Einstein's photoelectric equation for one light
Two frequencies, use the higher
→ Einstein's photoelectric equation for one light
Volts and electron-volts are the same number
→ Einstein's photoelectric equation for one light
Subtract, do not divide
→ Two wavelengths on the same metal
A longer wavelength cannot give a larger stopping potential
→ Two wavelengths on the same metal
Doubling K is not halving λ
→ Two wavelengths on the same metal
Subtract the work function before the square root
→ Comparing the maximum speeds of photoelectrons
A speed ratio is not an energy ratio
→ Comparing the maximum speeds of photoelectrons

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