When the momentum of a proton is changed by an amount p0, then the corresponding change in the de-Broglie wavelength is found to be 0.25 %. Then, the original momentum of the proton was

Question

When the momentum of a proton is changed by an amount p0, then the corresponding change in the de-Broglie wavelength is found to be 0.25 %. Then, the original momentum of the proton was (A) p0 (B) 100p0 (C) 400p0 (D) 4p0

Tardigrade Admin · Accepted Answer

We have λ=(h/p) λ propto (1/p) ⇒ (Δ p/p) =(Δ λ/λ) ⇒ |(Δ p/p)| =|(Δ λ/λ)| (p0/p) =(0.25/100) (p0/p) =(1/400) p =400 p0

Q. When the momentum of a proton is changed by an amount $p_{0}$ , then the corresponding change in the de-Broglie wavelength is found to be $0.25%$ . Then, the original momentum of the proton was

$p_{0}$

$100 p_{0}$

$400 p_{0}$

$4 p_{0}$

We have $λ = \frac{h}{p}$
$λ \propto \frac{1}{p}$
$\Rightarrow \frac{Δ p}{p} = \frac{Δ λ}{λ}$
$\Rightarrow ∣ ∣ \frac{Δ p}{p} ∣ ∣ = ∣ ∣ \frac{Δ λ}{λ} ∣ ∣$
$\frac{p _{0}}{p} = \frac{0.25}{100}$
$\frac{p _{0}}{p} = \frac{1}{400}$
$p = 400 p_{0}$

Q. When the momentum of a proton is changed by an amount p0​, then the corresponding change in the de-Broglie wavelength is found to be 0.25%. Then, the original momentum of the proton was

p0​

100p0​

400p0​

4p0​