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Q.
A particle of mass $1 mg$ has the same wavelength as an electron moving with a velocity of $3 \times 10^{6} ms ^{-1}$. The velocity of the particle is
Dual Nature of Radiation and Matter
Solution:
de-Broglie wavelength, $\lambda=\frac{h}{m v}$
As both particle and electrons having same wavelength therefore, their momentum will be equal to
$\Rightarrow m_{p} v_{p}=m_{e} v_{e}$
$\Rightarrow v_{p}=\frac{m_{ e } v_{e}}{m_{p}}=\frac{9.1 \times 10^{-31} \times 3 \times 10^{6}}{10^{-6}}$
$\Rightarrow v_{p}=2.7 \times 10^{-18} m / s$