If the momentum of an electron is changed by $Δ p,$ then the de-Brogile wavelength associated with it changes by $0.05 %$ . The initial momentum of electron will be :

\frac{Δ p}{200}

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\frac{Δ p}{199}

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199 Δ p

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- 2000 Δ p

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Solution

The correct option is D $- 2000 Δ p$
The initial momentum is given as,

$\frac{d λ}{λ} = - \frac{d P}{P_{i}}$

$\frac{0.05}{100} = - \frac{Δ P}{P_{i}}$

$\frac{0.05}{100} = - \frac{Δ P}{P_{i}}$

$P_{i} = - 2000 Δ P$

The initial momentum of electron is $- 2000 Δ P$ .

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If the momentum of an electron is changed by Δp, then the de-Brogile wavelength associated with it changes by 0.05% . The initial momentum of electron will be :

The correct option is D −2000ΔpThe initial momentum is given as, dλλ=−dPPi 0.05100=−ΔPPi 0.05100=−ΔPPi Pi=−2000ΔP The initial momentum of electron is −2000ΔP.

If the momentum of an electron is changed by $Δ p,$ then the de-Brogile wavelength associated with it changes by $0.05 %$ . The initial momentum of electron will be :

The correct option is D $- 2000 Δ p$
The initial momentum is given as,

$\frac{d λ}{λ} = - \frac{d P}{P_{i}}$

$\frac{0.05}{100} = - \frac{Δ P}{P_{i}}$

$\frac{0.05}{100} = - \frac{Δ P}{P_{i}}$

$P_{i} = - 2000 Δ P$

The initial momentum of electron is $- 2000 Δ P$ .