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
A) $3 \times 10^{- 31}ms^{- 1}$
B) $2.7 \times 10^{- 21}ms^{- 1}$
C) $2.7 \times 10^{- 18}ms^{- 1}$
D) $9 \times 10^{- 2}ms^{- 1}$
Explanation
(c) De broglie wave length $\lambda = \frac{h}{mv}$
As both particle and electron having same wave length therefore their momentum will be equal
$m_{p}\ v_{p} = m_{e}v_{e} \Rightarrow v_{p} = \frac{m_{e}v_{e}}{m_{p}} = \frac{9.1 \times 10^{- 31} \times 3 \times 10^{6}}{10^{- 6}}$
$\Rightarrow v_{p} = 2.7 \times 10^{- 18}\ m/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.