Light of wavelength 400nm hits metal — find maximum KE of electrons

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

Question

Light of wavelength 400 nm is incident on a metal surface whose work function is 2.0 eV. Find the maximum kinetic energy of the emitted photoelectrons. (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 maximum kinetic energy of emitted photoelectrons is given by Einstein’s photoelectric equation:

KEmax=hν−ϕKE_{max} = h\nu - \phi

where hνh\nu is the energy of one photon and ϕ\phi is the work function (minimum energy needed to eject an electron from the surface).

In terms of wavelength: ν=c/λ\nu = c/\lambda, so hν=hc/λh\nu = hc/\lambda.

 Ephoton=hcλ=6.626×10−34×3×108400×10−9E_{photon} = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} =1.9878×10−254×10−7=4.97×10−19 J= \frac{1.9878 \times 10^{-25}}{4 \times 10^{-7}} = 4.97 \times 10^{-19} \text{ J}

Convert to eV: Ephoton=4.97×10−191.6×10−19=3.10E_{photon} = \frac{4.97 \times 10^{-19}}{1.6 \times 10^{-19}} = 3.10 eV

A useful shortcut: for any wavelength λ\lambda in nm, the photon energy in eV is E=1240λ (nm)E = \frac{1240}{\lambda \text{ (nm)}} eV. This comes from hc=1240hc = 1240 eV·nm. Here: E=1240400=3.10E = \frac{1240}{400} = 3.10 eV.

 KEmax=Ephoton−ϕ=3.10−2.0=1.10 eVKE_{max} = E_{photon} - \phi = 3.10 - 2.0 = 1.10 \text{ eV}

In Joules: KEmax=1.10×1.6×10−19=1.76×10−19KE_{max} = 1.10 \times 1.6 \times 10^{-19} = 1.76 \times 10^{-19} J.

KEmax=1.10 eV=1.76×10−19 J\boxed{KE_{max} = 1.10 \text{ eV} = 1.76 \times 10^{-19} \text{ J}}

Why This Works

Einstein’s model says light consists of discrete packets called photons, each carrying energy E=hνE = h\nu. When a photon hits a metal surface, it can eject an electron — but only if its energy exceeds the work function ϕ\phi (the binding energy of the electron to the metal).

The excess energy (hν−ϕh\nu - \phi) goes into the kinetic energy of the ejected electron. The word “maximum” is important: not all electrons are at the surface. Electrons deeper inside the metal need extra energy to reach the surface, so they emerge with less kinetic energy. Only surface electrons get the maximum kinetic energy.

This model elegantly explained the photoelectric effect, which classical wave theory couldn’t — it predicted (correctly) that KE depends on frequency, not intensity, and that emission is instantaneous. This work earned Einstein the 1921 Nobel Prize in Physics.

Alternative Method

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

E=1240400=3.10E = \frac{1240}{400} = 3.10 eV

KEmax=3.10−2.0=1.10KE_{max} = 3.10 - 2.0 = 1.10 eV.

Memorising hc=1240hc = 1240 eV·nm (or 1240012400 eV·Å) saves significant calculation time in JEE and NEET MCQs.

Common Mistake

Students sometimes write KEmax=hν−ϕKE_{max} = h\nu - \phi but then substitute ϕ\phi in eV and hνh\nu in Joules (or vice versa), leading to a wrong answer. Keep both quantities in the SAME units before subtracting. Either convert everything to Joules, or use the shortcut to work entirely in eV.

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