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

Answer: 167

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 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 $m=167\times 10^{-29}kg$