Q 12-11-149JEE MainJEE Main 2020 (2 Sep, Shift 2)Medium
A particle is moving $5$ times as fast as an electron. The ratio of the de-Broglie wavelength of the particle to that of the electron is $1.878 \times 10^{-4}$. The mass of the particle is close to:
Answer: (D) $9.7\times10^{-28}\ \text{kg}$
$\lambda = \dfrac{h}{mv}$, so $\dfrac{\lambda_p}{\lambda_e} = \dfrac{m_e v_e}{m_p(5v_e)} = 1.878\times10^{-4}$.
$$m_p = \frac{9.1\times10^{-31}}{5\times1.878\times10^{-4}} \approx 9.7\times10^{-28}\ \text{kg}$$
Solution by Sreeraj P, M.Sc Physics