Question

Light quanta with energy $4.9eV$ eject photoelectrons from metal with work function $4.5eV$. Find the maximum impulse transmitted to the metal surface when electron are ejected.

Answer 1

Einstein Photoelectric equation

$\frac{1}{2}m{v}^2=hf-W$

$\frac{1}{2}m{v}^2=4.9-4.5=0.4eV=0.4\times 16\times {10}^{-19}J$

$V=\sqrt{\frac{2\epsilon }{m}}$

The momentum charge that is equal to impulse

$I=mv=m\sqrt{\frac{2\epsilon }{m}}=\sqrt{2m\epsilon }$

$I=\sqrt{2\times 9.1\times {10}^{-31}\times 0.4\times 1.6\times {10}^{-19}}$

$I=mv=3.45\times {10}^{-25}kg-m/sec$

The de-Broglie wavelength,

$\lambda =\frac{h}{mv}$

$=\frac{6.63\times {10}^{-33}}{3.45\times {10}^{-25}}$

$=2\times {10}^{-8}m$