SATHEE: Structure of Atom - Result Question 29

Structure of Atom - Result Question 29

####31. The momentum of a particle having a de Broglie wavelength of $10^{- 17}$ metres is

[1996]

(Given $h = 6.625 \times 10^{- 34} J s$ )

(a) $3.3125 \times 10^{- 7} k g m s^{- 1}$

(b) $26.5 \times 10^{- 7} k g m s^{- 1}$

(d) $13.25 \times 10^{- 17} k g m s^{- 1}$

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Solution:

$\begin{aligned} λ = \frac{h}{m v} \Rightarrow m v = \frac{h}{λ} = \frac{6.625 \times 10^{- 34}}{10^{- 17}} \\ \Rightarrow p = 6.625 \times 10^{- 17} k g m / s \end{aligned}$

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Structure of Atom - Result Question 29

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