If the momentum of an electron is changed by Δp, then the de Broglie wavelength associated with it changes by 0.50%. The initial momentum of electron will be:
Answer & explanation
Correct answer: option 3
de Broglie wavelength, $λ=\frac{h}{p}$
$∴dλ=-\frac{h}{p^2}dp;\frac{dλ}{λ}=\frac{-\frac{h}{p^2}dp}{\frac{h}{p}}=-\frac{dp}{p}$
$\frac{dλ}{λ}=\frac{dp}{p}$ (in magnitude)
Given $\frac{dλ}{λ}=\frac{0.5}{100},dp=Δp$
Let the initial momentum be $p_0$
Then $\frac{Δp}{p_0}=\frac{dλ}{λ}=\frac{0.5}{100}$
or $p_0=\frac{100}{0.5}Δp=200Δp$