Question

A particle is moving 3 times faster than the speed of electron. If the ratio of wavelength of particle and electron is $1.8 \times 10^{-4}$, then particle is

• A. Neutron
• B. $\alpha$-particle
• C. Deuteron
• D. Tritium

Answer 1

Answer: (A)

Given : $\frac{\lambda_{\text {particle }}}{\lambda_{\text {electron }}}=1.8 \times 10^{-4}$

and $\frac{v_{\text {particle }}}{v_{\text {electron }}}=3$

According to de-Broglie equation,

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

$\frac{\lambda_{\text {particle }}}{\lambda_{\text {electron }}}=\frac{h}{m_{\text {particle }} \times v_{\text {particle }}} \times \frac{m_{\text {electron }} \times v_{\text {electron }}}{h}$

$=\frac{m_{\text {electron }}}{m_{\text {particle }}} \times \frac{v_{\text {electron. }}}{v_{\text {particte }}}$

$\Rightarrow 1.8 \times 10^{-4}=\frac{9.1 \times 10^{-31} kg }{m_{\text {particle }}} \times \frac{1}{3}$

$m_{\text {particle }} =\frac{9.1 \times 10^{-31}}{1.8 \times 10^{-4} \times 3}$

$=1.6852 \times 10^{-27}\, kg$

Actual mass of neutron is $1.67493 \times 10^{-27}\, kg.$

Hence, the particle is neutron.