Question

A particle is moving $5$ times as fast as an electron. The ratio of the de-Broglie wavelength of the particles to that of the electron is $1.87\times {10}^{-4}.$The mass of the particle is close to:

• A. $4.8\times {10}^{-27}\mathrm{kg}$
• B. $9.1\times {10}^{-31}\mathrm{kg}$
• C. $9.7\times {10}^{-28}\mathrm{kg}$
• D. $1.2\times {10}^{-28}\mathrm{kg}$

Answer 1

The correct option is C

$9.7\times {10}^{-28}\mathrm{kg}$

Answer:

Explanation for correct option:

Given:

$$\begin{gathered} {\mathrm{V}}_{\mathrm{p}}=5{\mathrm{V}}_{\mathrm{e}} \\ \frac{{\mathrm{V}}_{\mathrm{p}}}{{\mathrm{V}}_{\mathrm{e}}}=5 \\ \frac{{\mathrm{\lambda }}_{\mathrm{p}}}{{\mathrm{\lambda }}_{\mathrm{e}}}=1.878\times {10}^{-4} \\ {\mathrm{m}}_{\mathrm{p}}=? \end{gathered}$$

From the de-Broglie wavelength,

$$\begin{gathered} \mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{mv}} \\ {\mathrm{\lambda }}_{\mathrm{p}}=\frac{\mathrm{h}}{{\mathrm{m}}_{\mathrm{p}}.{\mathrm{v}}_{\mathrm{p}}},{\mathrm{\lambda }}_{\mathrm{e}}=\frac{\mathrm{h}}{{\mathrm{m}}_{\mathrm{e}}.{\mathrm{v}}_{\mathrm{e}}} \\ \frac{{\mathrm{\lambda }}_{\mathrm{p}}}{{\mathrm{\lambda }}_{\mathrm{e}}}=\frac{{\displaystyle \frac{\mathrm{h}}{{\mathrm{m}}_{\mathrm{p}}.{\mathrm{v}}_{\mathrm{p}}}}}{{\displaystyle \frac{\mathrm{h}}{{\mathrm{m}}_{\mathrm{e}}.{\mathrm{v}}_{\mathrm{e}}}}}=\frac{{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}_{\mathrm{e}}}{{\mathrm{m}}_{\mathrm{p}}{\mathrm{v}}_{\mathrm{p}}} \\ \frac{{\mathrm{\lambda }}_{\mathrm{p}}}{{\mathrm{\lambda }}_{\mathrm{e}}}=\frac{{\mathrm{m}}_{\mathrm{e}}}{{\mathrm{m}}_{\mathrm{p}}}\left\{\frac{{\mathrm{v}}_{\mathrm{e}}}{{\mathrm{v}}_{\mathrm{p}}}\right\} \\ 1.878\times {10}^{-4}=\frac{{\mathrm{m}}_{\mathrm{e}}}{{\mathrm{m}}_{\mathrm{p}}}\left\{\frac{1}{5}\right\} \\ {\mathrm{m}}_{\mathrm{p}}=\frac{9.1\times {10}^{-31}\times 1}{1.878\times {10}^{-4}\times 5} \\ =0.97\times {10}^{-27}\mathrm{kg} \\ {\mathbf{m}}_{\mathbf{p}}\mathbf{=}\mathbf{9}\mathbf{}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{28}}\mathbf{kg} \end{gathered}$$

Hence option (C) is the correct option.