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), (C)

$\frac{E_e}{E_p}={10}^{-2}$
$\frac{P_e}{m_e^c}={10}^{-1}$

For electrons, ${\lambda }_e=\frac{h}{m_e{v}_e}=\frac{h}{m_e\left(c/100\right)}=\frac{100h}{m_{e}c}$ ...(i)

Kinetic energy, $E_e=\frac{1}{2}{m}_e{v}_e^2$ 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 $E_e=\frac{h^2}{2{\lambda }_e^2{m}_e}$ ...(ii)

For photon of wavelength ${\lambda }_p$, energy $E_p=\frac{hc}{{\lambda }_p}=\frac{hc}{2{\lambda }_e}$ $(\because {\lambda }_p=2{\lambda }_e)$

$\frac{E_p}{E_e}=\frac{hc}{2{\lambda }_e}\times \frac{2{\lambda }_e^2{m}_e}{h^2}=\frac{{\lambda }_e{m}_{e}c}{h}=\frac{100h}{m_{e}c}\times \frac{m_{e}c}{h}=100$

$\therefore \frac{E_e}{E_p}=\frac{1}{100}={10}^{-2}$

For electron, $p_e=m_e{v}_e=m_e\times c/100$

$\therefore \frac{p_e}{m_{e}c}=\frac{1}{100}={10}^{-2}$

Thus options (b) and (c) are correct.