Question

An electromagnetic wave of wavelength $\lambda$ is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de Brogue wavelength ${\lambda }^{\prime }$, then

• A. $\lambda =\frac{mc}{h}{\lambda }^{\prime 2}$
• B. $\lambda =\frac{3mc}{2h}{\lambda }^{\prime 2}$
• C. $\lambda =\frac{2mc}{h}{\lambda }^{\prime 2}$
• D. $\lambda =\frac{5mc}{h}{\lambda }^{\prime 2}$

Answer 1

Answer: (C)

$\lambda =\frac{2mc}{h}{\lambda }^{\prime 2}$

Kinetic energy of emitted electron = Energy of incident photon
i.e. $\frac{1}{2}m{v}^2=h\nu$

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

$[\because mv=p,\nu =\frac{c}{\lambda }]$
or $p=\sqrt{\frac{2mhc}{\lambda }}$

de-Broglie wavelength of emitted electrons

${\lambda }^{\prime }=\frac{h}{p}=\frac{h}{\sqrt{\frac{2mhc}{\lambda }}}$ or

${\lambda }^{\prime }=\sqrt{\frac{h\lambda }{2mc}}$ $\therefore \lambda =\frac{2mc}{h}{\lambda }^{\prime 2}$