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 of the material is
($h$ = Planck's constant, $c$ = speed of light)

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

Answer 1

The correct option is A $\frac{hc}{2\lambda }$

Let $\varphi$ be the work function.
We know kinetic energy for emitted electrons,
$K.E_{\lambda }=\frac{hc}{\lambda }-\varphi$
$K.E_{\lambda /2}=\frac{hc}{\lambda /2}-\varphi$
$K.E_{\lambda /2}=3(K.E_{\lambda })................\left(1\right)$Substituting values in $\left(1\right)$, we get
$\Rightarrow \varphi =\frac{hc}{2\lambda }$