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,

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

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

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

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

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

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

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

Hence, the particle is neutron.