A particle is moving 5 times as fast as an electron.

(4)

de Broglie wavelength

$\lambda=\frac{h}{m v} \Rightarrow m=\frac{h}{\lambda v}$

Clearly, $m \propto \frac{1}{\lambda v}$

If $\lambda$ and $v$ be the wavelength and velocity of electron and $\lambda^{\prime}$ and $v^{\prime}$ be the wavelength and velocity of the particle then

$\Rightarrow \frac{m^{\prime}}{m}=\frac{v \lambda}{v^{\prime} \lambda^{\prime}}=\frac{1}{5} \times \frac{1}{1.878} \times 10^{-4}$

$\Rightarrow m=9.7 \times 10^{-28} \mathrm{~kg}$