Question

The maximum kinetic energy of photoelectrons emitted from a metal surface when light of wavelength $\lambda$ is incident on it is $1\text{eV}$. When a light of wavelength $\frac{\lambda }{3}$ is incident on the surface, Maximum kinetic energy becomes $4\text{eV}$. The work function of the metal is

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

Answer 1

The correct option is B $\frac{hc}{3\lambda }$


From Einstein's photoelectric equation,
$K.E_{max}=\frac{hc}{\lambda }-\varphi$

For case 1:
$1\text{eV}=\frac{hc}{\lambda }-\varphi ...\left(1\right)$

For case 2:
$4\text{eV}=\frac{hc}{\left(\frac{\lambda }{3}\right)}-\varphi ...\left(2\right)$

Using $\left(1\right)$ and $\left(2\right)$,

$\frac{4}{1}=\frac{(\frac{3hc}{\lambda }-\varphi )}{(\frac{hc}{\lambda }-\varphi )}$

$\Rightarrow \frac{4hc}{\lambda }-4\varphi =\frac{3hc}{\lambda }-\varphi$

$\Rightarrow 3\varphi =\frac{hc}{\lambda }$

$\Rightarrow \varphi =\frac{hc}{3\lambda }$

Hence, option $\left(B\right)$ is correct.