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 is \(1.878 \times 10^{-4}\). The mass of the particle is close to

Let speed of \(e= V\)

\(\Rightarrow\) speed of particle \(=5 V\)

Debroglie wavelength \(\lambda_{ d }=\frac{ h }{ P }=\frac{ h }{ mv }\)

\(\Rightarrow\left(\lambda_{d}\right)_{P}=\frac{h}{m(5 V)}\) \(...(1)\)

\(\Rightarrow\left(\lambda_{d}\right)_{e}=\frac{ h }{ m _{ e } \cdot V }\)\(...(2)\)

According to question

\(\frac{(1)}{(2)}=\frac{m_{e}}{5 m}=1.878 \times 10^{-4}\)

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

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

\(\Rightarrow m =9.7 \times 10^{-28} kg\)