Question:
A particle is travelling 4 times as fast as an electron. Assuming the ratio of
de-Broglie wavelength of a particle to that of electron is $2: 1$, the mass of
the particle is :-
Correct Option: , 4
Solution:
(4)
$\lambda=\frac{\mathrm{h}}{\mathrm{p}}$
$\frac{\lambda_{\mathrm{p}}}{\lambda_{\mathrm{e}}}=\frac{\mathrm{P}_{\mathrm{e}}}{\mathrm{P}_{\mathrm{p}}}=\frac{\mathrm{m}_{\mathrm{e}} \mathrm{v}_{\mathrm{e}}}{\mathrm{m}_{\mathrm{p}} \mathrm{v}_{\mathrm{p}}}$
$2=\frac{\mathrm{m}_{\mathrm{e}}}{\mathrm{m}_{\mathrm{p}}}\left(\frac{\mathrm{v}_{\mathrm{e}}}{4 \mathrm{v}_{\mathrm{e}}}\right)$
$\therefore \mathrm{m}_{\mathrm{p}}=\frac{\mathrm{m}_{\mathrm{e}}}{8}$