Question

The momentum of a particle having a de-Broglie wavelength of ${10}^{-17}m$ is:

$(Givenh=6.625\times {10}^{-34}m)$

• A. $3.3125\times {10}^{-7}kgm{S}^{-1}$
• B. $26.5\times {10}^{-7}kgm{S}^{-1}$
• C. $6.625\times {10}^{-17}kgm{S}^{-1}$
• D. $13.25\times {10}^{-17}kgm{S}^{-1}$

Answer 1

Answer: (C)

$6.625\times {10}^{-17}kgm{S}^{-1}$

$\lambda =\frac{h}{p}$

$p=\frac{h}{\lambda }$

$p=\frac{6.625\times {10}^{-34}}{{10}^{-17}}$

$p=6.625\times {10}^{-17}$ kg m $S^{-1}$