Question

An electron travels with a velocity of $v\, m\,s ^{-1}$. For a proton to have the same de Broglie wavelength, the velocity will be approximately

• A. $\frac{v}{1840}$
• B. $\frac{1840}{v}$
• C. $1840 v$
• D. $v$

Answer 1

Answer: (A)

$\lambda=\frac{h}{m v}$

$\lambda_{p}=\frac{h}{m_{p} v_{p}}, \lambda_{e}=\frac{h}{m_{e} v_{e}}$

$\Rightarrow \frac{\lambda_{p}}{\lambda_{e}}=\frac{m_{e} v_{e}}{m_{p} v_{p}}\,\,\,\,$ As $\lambda_{p}=\lambda_{e}$

thus, $m_{p} v_{p}=m_{e} v_{e}$

$1840 m_{e} v_{p}=m_{e} v_{e}$

$\Rightarrow v_{e}=1840 v_{p}\,\,\,\, \left(\because m_{p}=1840 m_{e}\right)$

$v=1840 v_{p} \Rightarrow v_{p}=\frac{v}{1840}$