Question

In a photo electric effect experiment, the maximum kinetic energy of the emitted electrons is $ 1\,eV $ for incoming radiation of frequency $ {{v}_{0}} $ and $ \text{3 eV} $ for incoming radiation of frequency $ \text{3}{{\text{V}}_{0}}/2. $ What is the maximum kinetic energy of the electrons emitted for incoming radiations of frequency $ \text{9}{{\text{V}}_{0}}/4? $

• A. 3 eV
• B. 4.5 eV
• C. 6 eV
• D. 9 eV

Answer 1

Answer: (C)

$ {{(KE)}_{\max }}=hv-\text{o}{{|}_{0}} $
So, $ 1\,eV=h{{v}_{0}}-\text{o}{{|}_{0}} $ ..(i)
and $ 3\,eV=\frac{h{{v}_{0}}}{2}-\text{o}{{|}_{0}} $ ..(ii)
$ \Rightarrow $ $ 3eV-1eV=\frac{h{{v}_{0}}}{2} $ or
$ h{{v}_{0}}=4\,eV $
From Eq.(i), $ \text{o}{{|}_{0}}=h{{v}_{0}}-1\,eV $
$ =4eV-1\,eV $ $ =3\,eV $
$ \therefore $ $ {{(KE)}_{\max }}=h\times \frac{9{{v}_{0}}}{4}-3\,eV $
$ =\frac{9}{4}(4\,eV)-3\,eV $
$ =6\,eV $