Question

When the momentum of a proton is changed by an amount P₀, 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. 100 p₀
• C. 400 p₀
• D. 4 p₀

Answer 1

Answer: (C)

400 p₀

Answer 2

$\lambda \propto \frac{1}{p}$ ⇒ $\frac{\Delta p}{p} = - \frac{\Delta\lambda}{\lambda}$ ⇒ $\left| \frac{\Delta p}{p} \right| = \left| \frac{\Delta\lambda}{\lambda} \right|$

⇒ $\frac{p_{0}}{p} = \frac{0.25}{100} = \frac{1}{400}$ ⇒ p = 400 p₀ .