A particle is moving 5 times as fast as an electron. The ratio of the de-Broglie wavelength of the particle to that of the electron

Let mass of particle $=\mathrm{m}$

Let speed of $\mathrm{e}^{-}=\mathrm{V}$

$\Rightarrow$ speed of particle $=5 \mathrm{~V}$

Debroglie wavelength $\quad \lambda_{\mathrm{d}}=\frac{\mathrm{h}}{\mathrm{P}}=\frac{\mathrm{h}}{\mathrm{mv}}$

$\Rightarrow\left(\lambda_{\mathrm{d}}\right)_{\mathrm{P}}=\frac{\mathrm{h}}{\mathrm{m}(5 \mathrm{~V})}$ ..................(i)

$\Rightarrow\left(\lambda_{\mathrm{d}}\right)_{\mathrm{e}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{e}} \cdot \mathrm{V}}$ ..........(ii)

According to question

$\frac{(1)}{(2)}=\frac{\mathrm{m}_{e}}{5 \mathrm{~m}}=1.878 \times 10^{-4}$

$\Rightarrow \mathrm{m}=\frac{\mathrm{m}_{\mathrm{e}}}{5 \times 1.878 \times 10^{-4}}$

$\Rightarrow \mathrm{m}=\frac{9.1 \times 10^{-31}}{5 \times 1.878 \times 10^{-4}}$

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