91Ó°ÊÓ

Electron moving at ${10}^6\text{m}/\text{s}$.

Final speed of each mass ${\mathrm{m}}_0$ is ${\mathrm{u}}_{\mathrm{f}}=0.6\text{c}$ at ${60}^{°}$ to the photons original direction of motion

Wavelength of the incoming photon$=\mathrm{λ}$.

The classical formula for the momentum and kinetic energy to find the wavelength of the photon is de Broglie relation and equate its momentum to that of the electron.

a)

By the de brogile relation find the wavelength of the photon as:

$\begin{array}{c}{\mathrm{p}}_{\mathrm{photon}}={\mathrm{p}}_{\mathrm{electron}}\\ \mathrm{λ}=\frac{\mathrm{h}}{{\mathrm{m}}_{\mathrm{e}}\mathrm{v}}\\ \mathrm{λ}=\frac{6.63×{10}^{−34}}{9.1×{10}^{−31}\text{ k²µ}×{10}^6\text{ }\frac{\text{m}}{\mathrm{s}}}\\ \mathrm{λ}=7.28×{10}^{−10}\text{″¾}\end{array}$

Therefore, the wavelength of the photon $\mathrm{λ}=\mathbf{7}.\mathbf{28}×{\mathbf{10}}^{−10}\text{″¾}$.

b)

Consider $\mathrm{hc}=1240$eVnm and solve for the ration of the energy as:

$\begin{array}{c}\frac{{\mathrm{E}}_{\text{photon}}}{{\mathrm{KE}}_{\text{electron}}}=\frac{\frac{\mathrm{hc}}{\mathrm{λ}}}{\frac{1}{2}{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^2}\\ =\frac{\frac{1240\text{ e³Õ}â‹\dots \text{nm}}{0.728\text{ }\text{nm}}×1.6×{10}^{−19}\frac{\text{J}}{\text{eV}}}{\frac{1}{2}×9.1×{10}^{−31}\text{kg}×{\left({10}^6\frac{\text{m}}{\text{s}}\right)}^2}\\ =\frac{2.72×{10}^{−16}\text{J}}{4.55×{10}^{−19}\text{J}}\\ =\mathbf{597}.\mathbf{8}\end{array}$

Therefore, the ratio of the two kinetic energy is $597.8$.