Question

An electromagnetic wave of wavelength ' $\lambda$ ' is incident on a photosensitive surface of negligible work function. If ' $m$ ' mass is of photoelectron emitted from the surface has de-Broglie wavelength $\lambda_{ d }$, then :

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

Answer 1

Answer: (C)

$K _{\text {max }}=\frac{ hc }{\lambda}$
$\Rightarrow \frac{ p ^{2}}{2 m }=\frac{ hc }{\lambda}$
$ \Rightarrow p =\sqrt{\frac{2 mhc }{\lambda}}$
$\Rightarrow \lambda_{ d }=\frac{ h }{ p }= h \sqrt{\frac{\lambda}{2 mhc }}=\sqrt{\frac{\lambda h }{2 mc }} $
$\Rightarrow \lambda=\left(\frac{2 mc }{ h }\right) \lambda_{ d }^{2}$