Work function of metal is 2.5 eV — find threshold wavelength

easy CBSE JEE-MAIN NEET • 3 min read
Tags Photoelectric Effect

Question

The work function of a metal is 2.5 eV. Find the threshold wavelength of light needed to eject electrons from this metal. (Given: h=6.626×10−34h = 6.626 \times 10^{-34} J·s, c=3×108c = 3 \times 10^8 m/s, 1 eV = 1.6×10−191.6 \times 10^{-19} J)

Solution — Step by Step

The threshold wavelength (λ0\lambda_0) is the maximum wavelength of light that can eject an electron from the metal surface. At exactly this wavelength, the photon’s energy equals the work function — the electron is ejected but with zero kinetic energy.

If \lambda > \lambda_0: photon energy is too low → no photoelectric effect, regardless of intensity. If \lambda < \lambda_0 (higher frequency): electrons are ejected with positive kinetic energy.

At threshold:

hν0=ϕh\nu_0 = \phi hcλ0=ϕ\frac{hc}{\lambda_0} = \phi λ0=hcϕ\lambda_0 = \frac{hc}{\phi} ϕ=2.5 eV=2.5×1.6×10−19=4.0×10−19 J\phi = 2.5 \text{ eV} = 2.5 \times 1.6 \times 10^{-19} = 4.0 \times 10^{-19} \text{ J} λ0=hcϕ=6.626×10−34×3×1084.0×10−19\lambda_0 = \frac{hc}{\phi} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4.0 \times 10^{-19}} =1.988×10−254.0×10−19=4.97×10−7 m= \frac{1.988 \times 10^{-25}}{4.0 \times 10^{-19}} = 4.97 \times 10^{-7} \text{ m} λ0=497 nm≈500 nm\boxed{\lambda_0 = 497 \text{ nm} \approx 500 \text{ nm}}

This wavelength is in the green-blue visible range.

Why This Works

The threshold wavelength is the “minimum energy photon” that can still eject an electron. A photon’s energy is E=hc/λE = hc/\lambda — shorter wavelength means higher energy. So the threshold wavelength corresponds to the minimum photon energy (= work function), which means the longest wavelength that works.

For ϕ=2.5\phi = 2.5 eV, the threshold wavelength (~497 nm) falls in the visible range. Light with wavelengths shorter than 497 nm (blue, violet, UV) can eject electrons; light longer than 497 nm (yellow, orange, red) cannot, regardless of how intense it is. This is one of the key features of the photoelectric effect that classical wave theory failed to explain — intensity doesn’t matter, only frequency does.

Alternative Method

Using the shortcut E(eV)=1240/λ(nm)E \text{(eV)} = 1240/\lambda \text{(nm)}:

λ0=1240ϕ(eV)=12402.5=496 nm\lambda_0 = \frac{1240}{\phi \text{(eV)}} = \frac{1240}{2.5} = 496 \text{ nm}

This gives the same answer instantly. The shortcut hc=1240hc = 1240 eV·nm is worth memorising for all photoelectric effect calculations.

Common Mistake

The threshold condition is hν0=ϕh\nu_0 = \phi (kinetic energy = 0 at threshold). Students sometimes write hν0=0h\nu_0 = 0 (completely wrong) or use hν0=ϕ+KEmaxh\nu_0 = \phi + KE_{max} with KEmax≠0KE_{max} \neq 0 (that’s Einstein’s equation for general cases, not threshold specifically). At threshold, KEmax=0KE_{max} = 0 by definition.

Four useful numbers to memorise for photoelectric effect problems: (1) hc=1240hc = 1240 eV·nm; (2) h=4.14×10−15h = 4.14 \times 10^{-15} eV·s; (3) 1 eV = 1.6×10−191.6 \times 10^{-19} J; (4) threshold condition: λ0=1240/ϕ(eV)\lambda_0 = 1240/\phi(\text{eV}) nm. With these, you can solve any standard photoelectric effect problem in 30 seconds.

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