Photoelectric effect — work function, threshold frequency, and Einstein's equation

medium CBSE JEE-MAIN • NCERT Class 11 • 3 min read
Tags Atomic Structure

Question

Light of wavelength 300 nm strikes a metal surface with work function 2.5 eV. Find (a) the energy of each photon, (b) the maximum kinetic energy of the emitted photoelectrons, and (c) the threshold wavelength. Given: h=6.63×10−34h = 6.63 \times 10^{-34} J s, c=3×108c = 3 \times 10^8 m/s, 1 eV=1.6×10−191\text{ eV} = 1.6 \times 10^{-19} J.

(NCERT Class 11, Chapter 2)

Solution — Step by Step

E=hν=hcλ=6.63×10−34×3×108300×10−9E = h\nu = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9}} E=19.89×10−263×10−7=6.63×10−19 JE = \frac{19.89 \times 10^{-26}}{3 \times 10^{-7}} = 6.63 \times 10^{-19} \text{ J}

Converting to eV: E=6.63×10−191.6×10−19=4.14 eVE = \frac{6.63 \times 10^{-19}}{1.6 \times 10^{-19}} = \mathbf{4.14 \text{ eV}}

 KEmax=hν−ϕ=E−ϕKE_{max} = h\nu - \phi = E - \phi KEmax=4.14−2.5KE_{max} = 4.14 - 2.5 KEmax=1.64 eV\boxed{KE_{max} = 1.64 \text{ eV}}

Since KEmax>0KE_{max} > 0, photoelectrons are emitted.

At threshold, KEmax=0KE_{max} = 0, so all photon energy goes into overcoming the work function:

ϕ=hcλ0\phi = \frac{hc}{\lambda_0} λ0=hcϕ=6.63×10−34×3×1082.5×1.6×10−19\lambda_0 = \frac{hc}{\phi} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{2.5 \times 1.6 \times 10^{-19}} λ0=19.89×10−264.0×10−19=4.97×10−7 m\lambda_0 = \frac{19.89 \times 10^{-26}}{4.0 \times 10^{-19}} = 4.97 \times 10^{-7} \text{ m} λ0≈497 nm\boxed{\lambda_0 \approx 497 \text{ nm}}

Light with wavelength longer than 497 nm (lower energy) will not eject electrons from this metal.

Why This Works

Einstein’s equation is an energy conservation statement: the photon’s energy (hνh\nu) is used partly to free the electron from the metal (work function ϕ\phi) and the remainder appears as kinetic energy. This one-photon-one-electron model explained all the features of the photoelectric effect that classical wave theory could not:

  • Why there is a threshold frequency (photon must have at least energy ϕ\phi)
  • Why KE depends on frequency, not intensity
  • Why emission is instantaneous (energy is delivered in discrete packets)

Alternative Method

A quick shortcut: E(eV)=1240λ(nm)E(\text{eV}) = \frac{1240}{\lambda(\text{nm})}. For λ=300\lambda = 300 nm: E=1240/300=4.13E = 1240/300 = 4.13 eV. This avoids all the unit conversions and is very useful in competitive exams.

Memorise hc=1240hc = 1240 eV⋅\cdotnm. This single number lets you convert between wavelength and energy instantly: E(eV)=1240/λ(nm)E(\text{eV}) = 1240/\lambda(\text{nm}). For JEE, this saves at least 1-2 minutes per question on photoelectric effect and atomic spectra problems.

Common Mistake

Students often confuse “increasing intensity” with “increasing energy per photon.” Increasing intensity means more photons per second (more electrons emitted, higher photocurrent) but does NOT change the energy of each photon. Only increasing frequency (decreasing wavelength) increases the energy per photon and thus the maximum KE of photoelectrons. This distinction is the entire point of the photoelectric effect.

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