Question

The de Broglie wavelength of a photon is twice the de Broglie wavelength of an electron. The speed of the electron is $v_{e}=\frac{c}{100}$ then

• A. $\frac{E_{e}}{E_{p}}=10^{-4}$
• B. $\frac{E_{e}}{E_{p}}=10^{-2}$
• C. $\frac{p_{e}}{m_{e} c}=10^{-1}$
• D. $\frac{p_{e}}{m_{e} c}=10^{-4}$

Answer 1

Answer: (B)

For electron,
$\lambda_{e}=\frac{h}{m_{e}v_{e}}=\frac{h}{m_{e}\left(c 100\right)}=\frac{100 h}{m_{e}c} \ldots\left(i\right)$
$E_{e}=\frac{1}{2}m_{e}v_{e}^{2}\, or\, m_{e}v_{e}=\sqrt{2E_{e}m_{e}}$
$\lambda_{e}=\frac{h}{m_{e} v_{e}}=\frac{h}{\sqrt{2m_{e}E_{e}}}$ or $E_{e}=\frac{h^{2}}{2\lambda_{e}^{2}m_{e}} \ldots\left(ii\right)$
For photon, $E_{p}=\frac{hc}{\lambda_{p}}=\frac{hc}{2\lambda_{e}} (\because\lambda_{p}=2\lambda_{e}$ (Given)
$\therefore \frac{E_{p}}{E_{e}}=\frac{hc}{2\lambda_{e}}\times\frac{2\lambda_{e}^{2}m_{e}}{h^{2}}=\frac{\lambda_{e}m_{e}c}{h}$
$=\frac{100h}{m_{e}c}\times\frac{m_{e}c}{h}=100$
$\therefore \frac{E_{e}}{E_{p}}=\frac{1}{100}=10^{-2}$ (Using (i))
For electron, $p_{e}=m_{e}v_{e}=m_{e}\times\frac{c}{100}$
$\therefore \frac{p_{e}}{m_{e} c}=\frac{1}{100}=10^{-2}$
Thus, option $\left(b\right)$ is correct