Question

The momentum of a particle which of 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)

$6.62\times {10}^{-24}kg{ms}^{-1}$

de-Broglie wavelength $=\frac{h}{mv}=\frac{h}{p}$

$⟹0.1$ $nm=\frac{6.62\times {10}^{-34}}{P}$

$⟹P$(momentum) $=\frac{6.626\times {10}^{-34}}{{10}^{-10}m}=6.626\times {10}^{-24}$ $kgm{s}^{-1}$