Question

${\lambda }_0$ is the threshold wavelegth of a metal for photoelectron mission. The metal is exposed to the light of wavelength $\lambda$. I the minimum velocity of ejected electron is $\sqrt{\frac{2h}{m}({\lambda }_0-\lambda )p}$ then the value of $p$ is:

• A. $\lambda {\lambda }_0$
• B. $\frac{1}{\lambda {\lambda }_0}$
• C. $\frac{c}{\lambda {\lambda }_0}$
• D. none of the above

Answer 1

Answer: (C)

$\frac{c}{\lambda {\lambda }_0}$

As per Einstein's photoelectric equation,

$K.E_{max}=h(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})$

Here $\lambda$ is the wavelength of incident light and ${\lambda }_0$ is the threshold wavelength.

$\frac{1}{2}m{v}^2=hc\left(\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right)$

$v=\sqrt{\frac{2hc({\lambda }_0-\lambda )}{m\left(\lambda {\lambda }_0\right)}}$

Therefore $p=\frac{c}{\lambda {\lambda }_0}$