Photoelectric Effect: Tricky Questions Solved (4)

Question

The work function of a metal is $\phi = 2.5\text{ eV}$. Light of wavelength $\lambda = 400\text{ nm}$ falls on the surface. (a) Find the maximum kinetic energy of the emitted photoelectrons. (b) Find the stopping potential. (c) What is the threshold wavelength for this metal? Use $hc = 1240\text{ eV·nm}$.

Solution — Step by Step

Why This Works

Einstein’s view: light energy comes in quanta of $hf$. Each photon either ejects one electron (if it has enough energy) or does nothing — intensity controls how many photons arrive per second, not their individual energies.

The work function is the minimum energy needed to liberate an electron from the metal. Anything beyond $\phi$ becomes kinetic energy of the ejected electron.

Alternative Method

Using SI: $E_{\text{photon}} = hf$ with $h = 6.63 \times 10^{-34}\text{ J·s}$ and $f = c/\lambda$. Convert eV to joules at the end. The $hc = 1240\text{ eV·nm}$ shortcut saves a unit-conversion step and is universally accepted in JEE/NEET.

Common Mistake