Question

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

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

Answer 1

Answer: (A)

$\lambda =\frac{h}{mv}$

${\lambda }_p=\frac{h}{m_p{v}_p},{\lambda }_e=\frac{h}{m_e{v}_e}$

$\Rightarrow \frac{{\lambda }_p}{{\lambda }_e}=\frac{m_e{v}_e}{m_p{v}_p}$ As ${\lambda }_p={\lambda }_e$

thus, $m_p{v}_p=m_e{v}_e$

$1840m_e{v}_p=m_e{v}_e$

$\Rightarrow v_e=1840v_p\left(\because m_p=1840m_e\right)$

$v=1840v_p\Rightarrow v_p=\frac{v}{1840}$