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

Question

When the momentum of a photon is changed by an amount p′ , the corresponding change in the de-Broglie wavelength is found to be 0.2 % . Then, the original momentum of the photon was (A) 300 p' (B) 500 p' (C) 400 p' (D) 100 p'

Tardigrade Admin · Accepted Answer

∵ λ =(h/p) ∴ λ propto (1/p) ⇒ (Δ p/p)= -(Δ λ /λ ) ∴ |(Δ p/p)|=|(Δ λ /λ )| ⇒ (p'/p)=(0.20/100)=(1/500) ⇒ p=500 p'

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

$300 p^{'}$

$500 p^{'}$

$400 p^{'}$

$100 p^{'}$

$∵ λ = \frac{h}{p}$
$∴ λ \propto \frac{1}{p}$
$\Rightarrow \frac{Δ p}{p} = - \frac{Δ λ}{λ}$
$∴ ∣ ∣ \frac{Δ p}{p} ∣ ∣ = ∣ ∣ \frac{Δ λ}{λ} ∣ ∣$
$\Rightarrow \frac{p ^{'}}{p} = \frac{0.20}{100} = \frac{1}{500}$
$\Rightarrow p = 500 p^{'}$

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

300p′

500p′

400p′

100p′