Question

The ratio of the de Broglie wavelength of a particle to that of an electron is $1.813\times 10^{- 4}$ . The mass of the particle, if it moves three times faster than the electron is $Y\times 10^{- 29} \, kg$ . Then the value of $Y$ is: $\left(m_{e}=9.1 \times 10^{-31} kg \right)$ (Approximate the answer to nearest integer)

Answer 1

de Broglie wavelength of a moving particle having mass $m$ and velocity $v \, $ is given by
$\lambda =\frac{h}{p}=\frac{h}{m v}$ or $m=\frac{h}{\lambda v}$
For and 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 $\frac{\lambda }{\lambda _{e}}=1.813\times 10^{- 4}$
Mass of the particle, $m=m_{e}\left(\frac{\lambda_{e}}{\lambda}\right)\left(\frac{v_{e}}{U}\right)$
Substituting the values, we get
$m=9.1\times 10^{- 31}\times \frac{1}{1.813 \times 10^{- 4}}\times \frac{1}{3}$
or $m=167\times 10^{- 29} \, kg$