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Q. A particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is $1.813\times 10^{-4}$ The mass of the particle is $(m_{e}=9.1\times10^{-31}\,kg)$

Dual Nature of Radiation and Matter

Solution:

de Broglie wavelength of a moving particle having mass $m $ and velocity $v$ is given by
$\lambda=\frac{h}{p}=\frac{h}{mv}$ or $m=\frac{h}{\lambda v}$
For an electron $ \lambda_{e}=\frac{h}{m_{e} v_{e}}$ or $m_{e} =\frac{h}{\lambda_{e} v_{e}}$
Given: $\frac{v}{v_{e}}=3$ and $\frac{\lambda}{\lambda_{e}}=1.813\times10^{-4}$
Mass of the particle, $m=m_{e} \left(\frac{\lambda_{e}}{\lambda}\right)\left(\frac{v_{0}}{v}\right)$
Substituting the values, we get
$m=9.1\times10^{-31}\times\frac{1}{1.813\times10^{-4}}\times\frac{1}{3}$
or $m=1.67\times10^{-27}kg$