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. $\frac{p_e}{m_{e}c}={10}^{-4}$
• D. $\frac{p_e}{m_{e}c}={10}^{-2}$
  
  

Answer 1

Answer: (B), (D)

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



Formula used: $\lambda =\frac{h}{m{v}^{{}^{\prime }}},E=\frac{hc}{\lambda }$
For electron, wave length
${\lambda }_e=\frac{h}{m_e{v}_e}=\frac{h}{m_e\left(c/100\right)}=\frac{100h}{m_{e}c}$...$\left(i\right)$

Kinetic Energy,
$E_e=\frac{1}{2}{m}_e{v}_e^2$
Or $m_e{v}_e=\sqrt{2E_e{m}_e}$
${\lambda }_e=\frac{h}{m_e{v}_e}=\frac{h}{\sqrt{2m_e{v}_e}}$

$E_e=\frac{h^2}{2{\lambda }_e^2{m}_e}$ ...$\left(ii\right)$
For photon of wavelength ${\lambda }_p$,

Energy $E_p=\frac{hc}{{\lambda }_p}=\frac{hc}{2{\lambda }_e}$ $\because {\lambda }_p=2{\lambda }_e$

$\frac{E_p}{E_e}=\frac{hc}{2{\lambda }_e}\times \frac{2{\lambda }^2{m}_e}{h^2}$
$=\frac{{\lambda }_e{m}_{e}c}{h}$
$=\frac{100h}{m_{e}c}\times \frac{m_{e}c}{h}=100$
$\frac{E_e}{E_p}=\frac{1}{100}={10}^{-2}$

For electron,
$p_e=m_e{V}_e=m_e\times \frac{c}{100}$
$\frac{p_e}{m_{e}c}=\frac{1}{100}={10}^{-2}$


Final Answer: (b),(c)