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