Question

An electron moving with velocity 'v' is found to have a certain value of de-Broglie wavelength. The velocity to be possessed by the neutron to have the same de-Broglie wavelength is

• A. $\frac{1840}{v}$
• B. $1840 \,v$
• C. $\frac{v} {1840}$
• D. $v$

Answer 1

Answer: (C)

$\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{e}} \mathrm{V}} \lambda_{\mathrm{m}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{n}} \mathrm{v}_{\mathrm{n}}}$

$\therefore \lambda_{\mathrm{e}}=\lambda_{\mathrm{n}}$

$\therefore \frac{1}{\mathrm{~m}_{\mathrm{e}} \mathrm{v}}=\frac{1}{\mathrm{~m}_{\mathrm{n}} \mathrm{v}_{\mathrm{n}}}$

or $\frac{\mathrm{V}_{\mathrm{n}}}{\mathrm{v}}=\frac{\mathrm{m}_{\mathrm{c}}}{\mathrm{m}_{\mathrm{n}}}=\frac{9.1 \times 10^{-28}}{1.67 \times 10^{-24}}=\frac{1}{1840}$

or $\mathrm{v}_{\mathrm{n}}=\frac{1}{1840} \times \mathrm{v}=\frac{\mathrm{v}}{1840}$