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If the momentum of an electron is changed by $∆$ 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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$\frac{∆ p}{199}$

199 $∆ p$

400 $∆$ p

$\frac{∆ p}{200}$

answer is C.

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Detailed Solution

$λ = \frac{h}{p}$

⇒ $λ - \frac{0.5 λ}{100} = \frac{h}{p + ∆ p}$

⇒ $\frac{199 λ}{200} = \frac{h}{p + ∆ p} = \frac{199 h}{200 p}$

⇒ $p + ∆ p = \frac{200}{199} p$

⇒ p = 199 $∆$ p

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