Question

A particle of mass 1mg has the same wave length as an electron moving with velocity of $3\times {10}^{6}m{s}^{-1}$ The velocity of the particle is

• A. $2.7\times {10}^{-18}m/s$
• B. $9\times {10}^{-2}m/s$
• C. $3\times {10}^{31}m/s$
• D. $2.7\times {10}^{-21}m/s$

Answer 1

The correct option is A $2.7\times {10}^{-18}m/s$

de-Broglie wavelength is given by $\lambda =\frac{h}{mv}$
Since, the particle and electron has same wavelength i.e. ${\lambda }_e={\lambda }_p$
$\frac{h}{m_e{v}_e}=\frac{h}{m_p{v}_p}$
We get $m_e{v}_e=m_p{v}_p$
Given : $m_p=1mg={10}^{-6}kg$ $v_e=$ $3\times {10}^{6}m/s$ $m_e=9.1\times {10}^{-31}kg$
$\therefore$ $9.1\times {10}^{-31}\times$ $3\times {10}^6=$ ${10}^{-6}\times v_p$
$⟹v_p=2.7\times {10}^{-18}m/s$
Correct answer is option A.