Question

The K.E of the photoelectons is E when the incident wave length is $\frac{\lambda }{2}$. The K.E becomes 2E when the incident wave length is $\frac{\lambda }{3}$. The work function of the metal is

• A. $\frac{hc}{\lambda }$
• B. $\frac{2hc}{\lambda }$
• C. $\frac{3hc}{\lambda }$
• D. $\frac{hc}{3\lambda }$

Answer 1

The correct option is A $\frac{hc}{\lambda }$

Let work function be $W$
For first case,
$\frac{2hC}{\lambda }-W=E$ ---------------------------(I)
For 2nd case,
$\frac{3hC}{\lambda }-\frac{W}{2}=E$ --------------------(II)
From (I) and (II),
$\Rightarrow \frac{2hC}{\lambda }-W=\frac{3hC}{\lambda }-\frac{W}{2}$
$\Rightarrow \frac{4hC}{2\lambda }-\frac{3hC}{2\lambda }=\frac{W}{2}$
$\Rightarrow \frac{hC}{\lambda }=W$