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If the momentum of an electron is changed by $Δ p,$ then the de-Broglie wavelength associated with it changes by 0. 50%. The initial momentum of the electron will be :

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$\frac{Δ P}{199}$

$200 Δ p$

$400 Δ p$

$\frac{Δ P}{200}$

answer is C.

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Detailed Solution

$P = \frac{h}{λ} \Rightarrow dp = - \frac{hdλ}{λ^{2}} \Rightarrow dp = - \frac{pdλ}{λ}$

$P = \frac{ΔP}{|\frac{dλ}{λ}|} = ΔP (200)$

$\frac{d λ}{λ} \times 100 = \frac{1}{2}$

$\frac{d λ}{λ} = \frac{1}{200}$

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