Question

The de Broglie wavelength of an electron moving with a velocity of $1.5\times {10}^{8}m{s}^{-1}$ is equal to that of a photon. Find the ratio of the kinetic energy of the photon to that of the electron.

Answer 1

The de Broglie wavelength for electron = $\lambda =h/p=c/\nu$

Given ${\lambda }_e={\lambda }_p$

KE of electron = $1/2\times m\times v^2=1/2\times p\times v_e$

Substituting for p in the above equation, we get

K.E of electron : $\frac{h{v}_e}{2{\lambda }_e}$

$\therefore {\lambda }_e=\frac{h\times v_e}{2\times K{E}_e}$

The K.E of photon is given by : $\frac{hc}{{\lambda }_p}$

$\therefore {\lambda }_p=\frac{h\times c}{K{E}_p}$

Given that the wavelengths are equal .

Taking ratio we get : $\frac{v_e}{2\times K{E}_e}$ = $\frac{c}{K{E}_p}$

Then,

$\frac{K{E}_p}{K{E}_e}=\frac{2\times c}{v_e}=$ $\frac{2\times 3\times {10}^8}{1.5\times {10}^8}=4$