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If the momentum of an electron is changed by $Δp,$ then the de-Broglie wavelength associated with it changes by $0.50 %$ . The initial momentum of the electron will be.

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$\frac{Δp}{200}$

$\frac{Δp}{199}$

$199 Δp$

$400 Δp$

answer is C.

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Detailed Solution

$λ = \frac{h}{p} \Rightarrow λ - \frac{0.5}{100} λ = \frac{h}{P + ΔP}$

$\Rightarrow \frac{199 λ}{200} = \frac{h}{P + ΔP} = \frac{199}{200} \frac{h}{p}$

$\Rightarrow p + Δp = \frac{200}{199} p$

$\Rightarrow p = 199 Δp .$

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