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When the momentum of a photon is changed by an amount p^' then the corresponding change in the de-Broglie wavelength is found to be 0.2%. Then, the original momentum of the photon was

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$300 p^{'}$

$600 p^{'}$

$400 p^{'}$

$500 p^{'}$

answer is B.

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Detailed Solution

$∵ λ = \frac{h}{p} ∴ λ \propto \frac{1}{p}$

$\Rightarrow \frac{Δp}{p} = - \frac{Δλ}{λ}$

$∴ |\frac{Δp}{p}| = |\frac{Δλ}{λ}| \Rightarrow \frac{p^{'}}{p} = \frac{0 .20}{100} = \frac{1}{500} ∴ p = 500 p^{'}$

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