Question

If photon having a wavelength of $6.2nm$ was allowed to strike a metal plate having a work function of $50eV$ then the uncertainity in wavelength of the emitted electron with $K{E}_{max}isx\times {10}^{-14}m.$ If the uncertainity in the momentum is $13.24\times {10}^{-28}kg\frac{m}{s}.$ Then value of x will be:

Answer 1

From Einstein's photoelectric effect,
$K{E}_{max}=hv-w=\frac{1240}{6.2}-50=150eV$
From de Broglie,
${\lambda }_{deBroglie}=\sqrt{\frac{150}{K{E}_{max}ineV}}\stackrel{\circ }{A}=\sqrt{\frac{150}{150}}=1\stackrel{\circ }{A}$
$p=\frac{h}{\lambda }$
then $dp=\frac{h}{{\lambda }^2}d\lambda$
$\Rightarrow \Delta \lambda =\frac{{\lambda }^2\times dp}{n}=\frac{[1\times {10}^{-10}{]}^2\times 13.24\times {10}^{-28}}{6.62\times {10}^{-34}}$
$=2\times {10}^{-14}m$
x = 2