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:
$\lambda=\frac{\mathrm{h}}{\mathrm{p}}$
$\frac{\lambda_{p}}{\lambda_{e}}=\frac{p_{c}}{p_{p}}=\frac{m_{e} v_{e}}{m_{p} v_{p}}$
$2=\frac{m_{e}}{m_{p}}\left(\frac{v_{e}}{4 v_{e}}\right)$
$\therefore \mathrm{m}_{\mathrm{p}}=\frac{\mathrm{m}_{\mathrm{e}}}{8}$
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