Question

An electromagnetic wave of wavelength $\lambda$ is incident on a photosensitive surface of negligible work function. If $m$ is mass of the photo-electron emitted from the surface and has de-Broglie wavelength ${\lambda }_d$, then -

• A. $\lambda =\left(\frac{2h}{mc}\right){\lambda }_d^2$
• B. $\lambda =\left(\frac{2m}{hc}\right){\lambda }_d^2$
• C. ${\lambda }_d=\left(\frac{2mc}{h}\right){\lambda }^2$
• D. $\lambda =\left(\frac{2mc}{h}\right){\lambda }_d^2$

Answer 1

The correct option is D $\lambda =\left(\frac{2mc}{h}\right){\lambda }_d^2$

As per Einstein's photoelectric equation,

$\frac{hc}{\lambda }={\varphi }_0+k$

Given, $\varphi \approx 0$

$\therefore \frac{hc}{\lambda }=k=\frac{p^2}{2m}$

$\Rightarrow p=\sqrt{\frac{2mhc}{\lambda }}$

Now de Broglie wavelength,

${\lambda }_d=\frac{h}{p}=\frac{h}{\sqrt{\frac{2mhc}{\lambda }}}$

$\Rightarrow \sqrt{\lambda }={\lambda }_d\sqrt{\frac{2mc}{h}}$

$\Rightarrow \lambda =\left(\frac{2mc}{h}\right){\lambda }_d^2$

Hence, option $\left(D\right)$ is the correct answer.