Question

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

Answer 1

Here $n_1=2,n_2=3$

Energy possedded by $H_a$ light

= $13.6(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

= $13.6\times (\frac{1}{4}-\frac{1}{9})$

= $\frac{13.6\times 5}{36}=1.89eV$

For $H_a$ light to be able to emit photoelec trons from a metal the work function must be greater than ot equal to 1.89 e V.