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If the momentum of an electron is changed by P, then the de Broglie wavelength associated with it changes by 0.5%. The initial momentum of electron will be
A
100 P
B
200 P
C
400 P
D
$\frac{P}{200}$
Detailed Solution
$\lambda = \frac{h}{p}$, so for small changes $\frac{\Delta p}{p} = \frac{\Delta\lambda}{\lambda}$ (in magnitude)
Here $\Delta p = P$ and $\frac{\Delta\lambda}{\lambda} = \frac{0.5}{100}$
$\frac{P}{p} = \frac{0.5}{100} \Rightarrow p = 200P$
Here $\Delta p = P$ and $\frac{\Delta\lambda}{\lambda} = \frac{0.5}{100}$
$\frac{P}{p} = \frac{0.5}{100} \Rightarrow p = 200P$