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,
${\lambda }_e=\frac{h}{m_e{v}_e}=\frac{h}{m_e\left(c100\right)}=\frac{100h}{m_{e}c}\dots \left(i\right)$
$E_e=\frac{1}{2}{m}_e{v}_e^{2}or{m}_e{v}_e=\sqrt{2E_e{m}_e}$
${\lambda }_e=\frac{h}{m_e{v}_e}=\frac{h}{\sqrt{2m_e{E}_e}}$ or $E_e=\frac{h^2}{2{\lambda }_e^2{m}_e}\dots \left(ii\right)$
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