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When the momentum of proton is changed by an amount $P_{0}$ , the corresponding change in the de-Broglie wavelength is found to be 0.25%. Find the original momentum of the proton.

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$\frac{P_{0}}{400}$

$P_{0}$

$4 P_{0}$

$400 P_{0}$

answer is D.

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Detailed Solution

$λ = \frac{h}{p} \Rightarrow |\frac{Δ P}{P}| = |\frac{- Δ λ}{λ}|$

$\frac{P_{0}}{P} = \frac{0.25}{100} \Rightarrow P = 400 P_{0}$

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