Question

The de-Broglie wavelength associated with a proton changes by 0.25% if its momentum is changed by $P_o$. The initial momentum of proton is

• A. 100$P_o$
• B. 400$P_o$
• C. $P_o$/400
• D. $P_o$

Answer 1

The correct option is B 400$P_o$

de-Broglie wavelength is given by $\lambda =\frac{h}{P}$ ...(1)
where $P$ is the initial momentum of the particle.
We get $\Delta \lambda =\frac{h}{P^2}\Delta P$
Given : $\Delta \lambda =0.0025\lambda$ $\Delta P=P_o$
We get $0.0025\Lambda =\frac{h}{P^2}{P}_o$
Or $0.0025=\frac{P_o}{P}$ (Using 1)
$⟹P=400P_o$