Question

If $\lambda_{0}$ and $\lambda$ be the threshold wavelength and wavelength of incident light, the velocity of photoelectron ejected from the metal surface is :

• A. $\sqrt{\frac{2h}{m}\left(\lambda_{0}-\lambda\right)}$
• B. $\sqrt{\frac{2hc}{m}\left(\lambda_{0}-\lambda\right)}$
• C. $\sqrt{\frac{2hc}{m}\left(\frac{\lambda_{0}-\lambda}{\lambda\,\lambda_{0}}\right)}$
• D. $\sqrt{\frac{2h}{m}\left(\frac{1}{\lambda_{0}}-\frac{1}{\lambda}\right)}$

Answer 1

Answer: (C)

$E=W+\frac{1}{2} \,mv^{2}$

$\Rightarrow \frac{hc}{\lambda}=\frac{hc}{\lambda_{0}}+\frac{1}{2} mv^{2}$

$\Rightarrow v^{2}=\frac{2hc}{m} \left[\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right] $
$\Rightarrow v=\sqrt{\frac{2hc}{m}\left[\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right]}$

$\Rightarrow v=\sqrt{\frac{2hc}{m}\left[\frac{\lambda_{0}-\lambda}{\lambda \lambda_{0}}\right]}$