KE_max = hν - φ. Calculate stopping potential for given wavelength.

Photoelectric Effect — Einstein's Equation

Question

A metal surface has a work function of 2.0 eV. Light of wavelength 400 nm falls on it. Find:

The maximum kinetic energy of emitted photoelectrons
The stopping potential

Given: $h = 6.626 \times 10^{-34}$ J·s, $c = 3 \times 10^8$ m/s, $1$ eV $= 1.6 \times 10^{-19}$ J

Solution — Step by Step

We need the energy of one photon hitting the surface. Use $E = hc/\lambda$ .

E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}}

E = \frac{1.988 \times 10^{-25}}{4 \times 10^{-7}} = 4.97 \times 10^{-19} \text{ J}

Converting to eV: $E = \dfrac{4.97 \times 10^{-19}}{1.6 \times 10^{-19}} \approx \mathbf{3.1 \text{ eV}}$

Einstein’s equation tells us that the photon’s energy goes into two parts: knocking the electron free (work function $\phi$ ) and giving it kinetic energy.

KE_{max} = h\nu - \phi = E_{photon} - \phi

KE_{max} = 3.1 \text{ eV} - 2.0 \text{ eV} = \mathbf{1.1 \text{ eV}}

The electron uses 2.0 eV just to escape the metal — the leftover 1.1 eV becomes kinetic energy.

Stopping potential $V_0$ is the voltage needed to bring the fastest electron to rest. We equate electrical work done against the electron to its kinetic energy.

eV_0 = KE_{max}

V_0 = \frac{KE_{max}}{e} = \frac{1.1 \text{ eV}}{e} = \mathbf{1.1 \text{ V}}

This step is almost trivial once you’re in eV — numerically, stopping potential in volts equals $KE_{max}$ in eV. Remember this shortcut for MCQs.

Why This Works

Einstein proposed that light comes in discrete packets called photons, each carrying energy $E = h\nu$ . When a photon hits the metal, it interacts with one electron — all or nothing. The electron needs a minimum energy $\phi$ (the work function) just to escape the surface. Whatever’s left over becomes kinetic energy.

This is why intensity doesn’t matter for the threshold — shining more light just sends more photons, but if each photon is below threshold energy, no electrons come out regardless of brightness. Frequency is what counts. This was the key experimental observation that classical wave theory could never explain, and it won Einstein the 1921 Nobel Prize (not relativity — that’s a common misconception worth knowing for general awareness).

The stopping potential concept comes from the reverse setup: apply a retarding voltage until even the fastest electrons can’t reach the collector. At exactly $V_0$ , current drops to zero.

Alternative Method

For quick MCQ solving, use the energy directly in eV without converting to joules.

For any visible/UV light, use the shortcut formula:

E_{photon} \text{ (in eV)} = \frac{1240}{\lambda \text{ (in nm)}}

For $\lambda = 400$ nm: $E = 1240/400 = 3.1$ eV — same answer, calculated in 5 seconds.

This $hc = 1240$ eV·nm is a number worth memorising. It appears in JEE Main almost every year, including the 2023 Shift 2 paper.

Then directly: $KE_{max} = 3.1 - 2.0 = 1.1$ eV and $V_0 = 1.1$ V. No unit conversions needed.

Common Mistake

Confusing stopping potential units with energy units.

Students write $V_0 = 1.1 \times 1.6 \times 10^{-19}$ V — multiplying by the charge of an electron unnecessarily. The stopping potential is 1.1 V, not $1.76 \times 10^{-19}$ V.

The relation $eV_0 = KE_{max}$ means: (charge in coulombs) × (voltage in volts) = energy in joules. If you already have $KE_{max}$ in eV, the numerical value is directly the stopping potential in volts. Don’t double-convert.

KE_{max} = h\nu - \phi = \frac{hc}{\lambda} - \phi

eV_0 = KE_{max}

\frac{hc}{\lambda_{threshold}} = \phi \quad \text{(at threshold, } KE_{max} = 0\text{)}

Shortcut: $hc = 1240$ eV·nm

This is a high-weightage topic for both CBSE Class 12 boards and JEE Main. The boards often ask for stopping potential directly; JEE likes to combine this with de Broglie wavelength of the emitted electron — so once you’re comfortable here, the next step is finding $\lambda_{dB} = h/\sqrt{2mKE_{max}}$ for the emitted electron.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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