Question

The light emitted in the transition n = 3 to n = 2 in hydrogen is called H_α light. Find the maximum work function a metal can have so that H_α light can emit photoelectrons from it.

Answer 1

The H_α light can emit the photoelectrons if its energy is greater than or equal to the work function of the metal.
Energy possessed by H_α light (E) is given by
$E=13.6\left(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right)$ eV
Here, n₁ = 2, n₂ = 3​
$$\begin{gathered} \therefore E=13.6\times \left(\frac{1}{4}-\frac{1}{9}\right) \\ =\frac{13.6\times 5}{36}=1.89\mathrm{eV} \\ =1.90\mathrm{eV} \end{gathered}$$
H_α light will be able to emit electron from the metal surface for the maximum work function of metal to be 1.90 eV.