Question

If velocity of a particle is $3$ times that of electron and ratio of de-Broglie wavelength of the particle to that of electron is $1.8\times {10}^{-4}$. The particle is -

Mass of electron $=9.1\times {10}^{-31}\text{kg}$
Mass of neutron $=1.7\times {10}^{-27}\text{kg}$

• A. Neutron
• B. Deuteron
• C. Alpha
• D. Tritium

Answer 1

The correct option is A Neutron

Given:

$\frac{{\lambda }_p}{{\lambda }_e}=1.8\times {10}^{-4}$

$\frac{v_p}{v_e}=3$

We know that,

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

$\Rightarrow \lambda \propto \frac{1}{mv}$

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

$\Rightarrow m_p=\frac{{\lambda }_e}{{\lambda }_p}\times \frac{v_e}{v_p}\times m_e$

$\Rightarrow m_p=\frac{9.1\times {10}^{-31}}{1.8\times {10}^{-4}\times 3}$

$\Rightarrow m_p=1.7\times {10}^{-27}\text{kg}=\text{Mass of the neutron}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(A\right)$ is the correct answer.