Question

The wave length of a photon is twice the de Broglie wave length of the electron. If the speed of the electron is $v_e=\frac{c}{100}$. then the ratio of the energy of electron to that of photon $\frac{E_e}{E_p}$ is

• A. ${10}^{-1}$
• B. ${10}^{-2}$
• C. ${10}^{-3}$
• D. ${10}^{-5}$

Answer 1

The correct option is B ${10}^{-2}$

${\lambda }_e=\frac{h}{m_e{v}_e}$

$=\frac{h}{m_e\left(c/100\right)}$

$=\frac{100}{m_{e}c}$

$K.E,$ $E_e=\frac{1}{2}{m}_e{v}_e^2$

$m_e{v}_e=\sqrt{2E_e{m}_e}$

$\lambda =\frac{h}{m_e{v}^2}$

$=\frac{h}{2m_e{E}_e}$

$E_e=\frac{h^2}{2{\lambda }^2{m}_e}$

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

$=\frac{100h}{m_{e}c}\times \frac{m_{e}c}{h}=100$

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

For electron $P_e=m_e{v}^2=m_e\times \frac{c}{100}={10}^{-2}$

Hence,

option $\left(B\right)$ is correct answer.