Question

X- rays of wavelength $\lambda$ falls on a photosensitive surface emitting electrons.Assuming that the work function of the surface can be neglected, prove that the de Broglie wavelength of electrons emitted will be $\sqrt{\frac{h\lambda }{2mc}}$

Answer 1

Energy of the X-rays of wavelength$=h\frac{c}{\lambda }$
As work function is neglected,

$\frac{1}{2}{m}_e{v}_e^2=h\frac{c}{\lambda }$

$v_e^2=\frac{2hc}{m_e\lambda }$

$v_e=\sqrt{\frac{2hc}{m_e\lambda }}$

De Brogile wavelength $=\frac{h}{m_e{v}_e}$

De Brogile wavelength $=\frac{h\sqrt{m\lambda }}{m_e\sqrt{2hc}}$

$=\sqrt{\frac{h\lambda }{2m_{e}c}}$