Photoelectric Effect: A 3-Step Check for MHT-CET Numericals

The photoelectric effect is the ejection of electrons from a metal surface by light. Light arrives as photons, each carrying energy E = hf. One photon gives its energy to one electron. The electron spends φ (the work function, the minimum energy needed to leave the metal) and keeps the rest as kinetic energy: K_max = hf − φ.

The word "max" matters. Electrons that sit deeper in the metal lose some energy on the way out, so hf − φ is the most any electron can have.

Three observations broke the wave theory of light, and the OpenStax University Physics textbook lists them:

1. There is no time lag, even at very low intensity.
2. The maximum kinetic energy does not depend on intensity.
3. Below a cut-off frequency (the threshold frequency f₀, where hf₀ = φ) nothing is emitted.

Heinrich Hertz first noticed the effect in 1887. Philipp Lenard found in 1902 that electron energy did not depend on intensity. Einstein explained everything with photons in 1905 (Wikipedia: Photoelectric effect), and received the 1921 Nobel Prize in Physics for his discovery of the law of the photoelectric effect (Nobel Prize: Physics 1921).

In the Class 12 syllabus this sits in the chapter on Dual Nature of Radiation and Matter. Our MHT-CET physics syllabus checklist shows where it sits in your revision order.

Solve it by converting the light to energy in electron-volts (eV), comparing that with φ, and only then subtracting. This order stops you wasting time on questions where no electron is emitted at all. One electron-volt is the energy an electron gains across a potential difference of 1 volt.

Step 1: Convert to eV. Use E = 1240 / λ, with λ in nanometres. The shortcut comes from hc = 1240 eV·nm (OpenStax). If the question gives frequency, use E = hf with the values it gives.

Pass: you have a photon energy in eV. Fail: you are carrying joules and 10⁻¹⁹ into the next step for no reason.

Step 2: Compare with φ. If E is smaller than φ, write "no photoelectrons" and stop. Equivalently, if λ is longer than the threshold wavelength λ₀ = 1240 / φ, nothing comes out.

Pass: E ≥ φ, so you continue. Fail: E < φ and you calculated a negative kinetic energy anyway.

Step 3: Subtract. K_max = E − φ in eV. The stopping potential V₀ (the reverse voltage that just stops the fastest electrons) is the same number in volts, because eV₀ = K_max.

Pass: K_max is positive and V₀ in volts equals K_max in eV. Fail: you wrote V₀ in joules, or you subtracted φ from a K_max you had already computed.

What are the threshold wavelengths for common metals?

For sodium, aluminium, copper, silver and platinum the threshold wavelengths are 504, 304, 264, 262 and 195 nm. Light with a longer wavelength than λ₀ = 1240 / φ ejects no electrons from that metal.

Work functions are from OpenStax; λ₀ = 1240 / φ is our calculation from the OpenStax values, rounded to the nearest nanometre:

Worked example. Light of 200 nm falls on silver. E = 1240 / 200 = 6.20 eV. This is above φ = 4.73 eV, so electrons are emitted. K_max = 6.20 − 4.73 = 1.47 eV, and V₀ = 1.47 V. The same silver surface under 300 nm light gives E = 4.13 eV, which is below 4.73 eV, so there is no emission.

Brighter light gives more photons per second, not more energetic ones, so it increases the number of electrons and the current but leaves K_max unchanged. Each electron is hit by a single photon, and the energy of that photon depends only on frequency.

This also answers the other common doubt: why does very bright light below the threshold frequency eject nothing? A wave could in theory pile up energy over time. A photon cannot be pooled with others. If one photon has less energy than φ, a thousand of them still fail, one at a time. Emission above threshold is also effectively instant, in under 10⁻⁹ s (Wikipedia).

Use this table when a question changes one thing:

The saturation current is the plateau the photocurrent reaches once every emitted electron is collected. If a question holds the number of photons per second constant instead of the intensity, the saturation current stays the same when the frequency changes.

What happens to kinetic energy when the frequency doubles?

K_max does not double when the frequency doubles, because φ is subtracted from the photon energy, not from K_max. Take sodium (φ = 2.46 eV) with 400 nm light. E = 3.10 eV, so K_max = 3.10 − 2.46 = 0.64 eV. Double the frequency by halving the wavelength to 200 nm. E = 6.20 eV, so K_max = 6.20 − 2.46 = 3.74 eV. The ratio is 3.74 / 0.64 ≈ 5.8.

Only when φ is negligible would doubling the frequency double K_max. This is the trap to expect in a "frequency doubled, KE becomes…" option set. Always redo E − φ.

Read each graph by turning it back into K_max = hf − φ, then matching slope and intercepts. Because eV₀ = hf − φ, V₀ = (h/e)f − φ/e.

Two practical points:

• Two metals give parallel straight lines on a K_max–f or V₀–f graph. Parallel lines mean the same slope (h or h/e), so the same constant, and different φ.
• To find h from two readings (V₁ at f₁, V₂ at f₂), use h = e(V₂ − V₁) / (f₂ − f₁). The work function cancels out.

Practise it by repeating the three-step check on mixed problems until you stop reaching for the formula before the comparison. We recommend practising the intensity-versus-frequency trap, the doubled-frequency trap and the graph slope question first, because each one tests a different step of the check.

• Use our checked PYQ library to find previous-year questions on this chapter, and write down which of the three traps each one tests.
• The general method behind Step 1 to Step 3 is in How to Solve Physics Numericals: 4 Steps for HSC and MHT-CET.
• Keep the formulas for the whole of Class 12 in one place with our HSC physics formula sheet.

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