Structure of Atom - Result Question 29
####31. The momentum of a particle having a de Broglie wavelength of $10^{-17}$ metres is
[1996]
(Given $h=6.625 \times 10^{-34} Js$ )
(a) $3.3125 \times 10^{-7} kg ms^{-1}$
(b) $26.5 \times 10^{-7} kg ms^{-1}$
(c) $6.625 \times 10^{-17} kg ms^{-1}$
(d) $13.25 \times 10^{-17} kg ms^{-1}$
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Solution:
- (c) Acc. to de-Broglie,
$ \begin{aligned} & \lambda=\frac{h}{m v} \Rightarrow m v=\frac{h}{\lambda}=\frac{6.625 \times 10^{-34}}{10^{-17}} \\ & \Rightarrow p=6.625 \times 10^{-17} kg m / s \end{aligned} $
