Question

A particle of mass $1\text{mg}$ has the same de-Broglie wavelength as that of an electron moving with a velocity of $3\times {10}^6{\text{ms}}^{-1}$. The velocity of the particle is,
$[$ Take mass of electron, $m_e=9.1\times {10}^{-31}\text{kg}]$

• A. $3\times {10}^{-21}{\text{ms}}^{-1}$
• B. $2.7\times {10}^{-21}{\text{ms}}^{-1}$
• C. $2.7\times {10}^{-18}{\text{ms}}^{-1}$
• D. $3\times {10}^{-18}{\text{ms}}^{-1}$

Answer 1

The correct option is C $2.7\times {10}^{-18}{\text{ms}}^{-1}$

de-Broglie wavelength is given by,
$\lambda =\frac{h}{mv}$

As both, (the particle and electron) have same wavelength, their momenta must be equal.
i.e. $m_p{v}_p=m_e{v}_e$

$\Rightarrow v_p=\frac{m_e{v}_e}{m_p}$
$=\frac{9.1\times {10}^{-31}\times 3\times {10}^6}{{10}^{-6}}$

$\Rightarrow v_p=2.7\times {10}^{-18}{\text{ms}}^{-1}$

Hence, $\left(C\right)$ is the correct answer.