Physics - Modern Physics Question with Solution | TestHub

de-Broglie wavelength $\lambda =\frac{h}{\sqrt{2mK}}$ The kinetic energy of the electron $$K_{\text{electron }}=\frac{1}{2m}\cdot \frac{h^2}{{\lambda }^2}$$ Where $h=$ Planck constant $\lambda =$ wavelength The photon energy $$E_{\text{phocton }}=\frac{hc}{\lambda }$$ From Eqs. (ii) and (i), we get $\frac{E_{\text{photon }}}{K_{\text{electron }}}=\frac{hc/\lambda }{h^2/2m\cdot {\lambda }^2}=\frac{hc\cdot {\lambda }^2\times 2m}{h^2\cdot \lambda }$ where $m=0.5\mathrm{MeV}=5\times 10^5\mathrm{eV}$ $\begin{array}{rl}\frac{h}{\lambda c}=50\times 10^3\mathrm{eV}=\frac{2\mathrm{m}\lambda c}{h}& \\ =\frac{2\mathrm{m}}{h/\lambda c}& =\frac{2\times 5\times 10^5}{50\times 10^3}\left(\because \mathrm{m}=\frac{h}{c\lambda }\right)\\ & =20:1\end{array}$