Question

Find the momentum of a particle with de Brogile wavelength $2\stackrel{\circ }{\text{A}},$ is:
$\text{(Planck's constant,}h=6.62\times {10}^{-34}\text{J s})$

• A. $3.31\times {10}^{-24}{\text{kg m s}}^{-1}$
• B. $3.31\times {10}^{24}{\text{kg m s}}^{-1}$
• C. $9.62\times {10}^{24}{\text{kg m s}}^{-1}$
• D. $6.62\times {10}^{22}{\text{kg m s}}^{-1}$

Answer 1

The correct option is A $3.31\times {10}^{-24}{\text{kg m s}}^{-1}$

Given:
$\text{de Brogile wavelength}\left(\lambda \right)=2\stackrel{\circ }{\text{A}}=2\times {10}^{-10}m$
$\text{Planck's constant}\left(h\right)=6.62\times {10}^{-34}\text{J s}$
Let ${}^{\prime }{p}^{\prime }$ be the momentum of the particle.
We know that de Broglie wavelength is given by,
$\Rightarrow \lambda =\frac{h}{p}$
$\therefore \text{Momentum}\left(p\right)=\frac{h}{\lambda }=\frac{6.62\times {10}^{-34}}{2\times {10}^{-10}}=3.31\times {10}^{-24}{\text{kg m s}}^{-1}$
Hence, option (b) is correct.