Question

A photoelectric surface is illuminated successively by monochromatic light of wavelengths $\lambda$ and $\frac{\lambda }{2}$. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface is :

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

Answer 1

Answer: (A)

$\frac{hc}{2\lambda }$

$K.E_{\lambda }=\frac{hc}{\lambda }-\varphi$
$K.E_{\lambda /2}=\frac{hc}{\lambda /2}-\varphi$
$K.E_{\lambda /2}=3(K.E_{\lambda })$
$\Rightarrow \varphi =\frac{hc}{2\lambda }$