Question

The momentum of a particle which has a de-Broglie wavelength of $0.1\, nm$ is

• A. $3.2 \times 10^{-24} kg\, ms ^{-1}$
• B. $4.3 \times 10^{-22} kg\, ms ^{-1}$
• C. $5.3 \times 10^{-22} kg\, ms ^{-1}$
• D. $6.62 \times 10^{-24} kg\, ms ^{-1}$

Answer 1

Answer: (D)

$\lambda=\frac{h}{m v}$
momentum $P=m v$
$\therefore P=\frac{h}{\lambda}=\frac{6.6 \times 10^{-34} kgm ^{2} s ^{-1}}{0.1 \times 10^{-9} m }$
$P=6.6 \times 10^{-24} kg\, ms ^{-1}$