Question

Photons of energy 2.0 eV fall on a metal plate and release photo electrons with a maximum velocity V. By decreasing $\lambda$ by 25% the maximum velocity of photo electrons is doubled. The work function of the metal of the material plate in eV is nearly :

• A. 2.22
• B. 1.985
• C. 2.35
• D. 1.8

Answer 1

The correct option is D 1.8

Let work function be W,
In 1st case,
$\to \frac{2hc}{\lambda }-2W=m{v}^2------1$
In and case new $\lambda \left({\lambda }_1\right)=\frac{3\lambda }{4}$
$\to \frac{4hc}{3\lambda }-W=\frac{1}{2}m{\left(2v\right)}^2----$
$\to \frac{4hc}{3\lambda }=W+2m{v}^2-------2$
Also given, $\frac{hc}{\lambda }=2ev.------3$
from 1 and 2,
$\frac{4hc}{3\lambda }=W+\frac{4hc}{3\lambda }-4W$
$\to 3W=\frac{8hc}{3\lambda }$
$\to W=\frac{8}{9}\frac{hc}{\lambda }\to from3,$
$W=\frac{16}{9}eV\approx 1.8eV$

So, the answer is option (D).