If the momentum of electron is changed by P then the de Broglie wavelength associated with it changes by 1%. The initial momentum of electron will be:

100 P

The correct answer is Option (2) → 100 P

The De-Broglie wavelength (λ) is,

$λ=\frac{h}{P}$

∴ Change in wavelength after momentum change ΔP.

$λ'=\frac{h}{P+ΔP}$

∴ Change in wavelength,

$Δλ=λ-λ'=\frac{h}{P}-\frac{h}{P+ΔP}$

$=\frac{h}{P}\left(1-\frac{1}{1+\frac{ΔP}{P}}\right)$

$Δλ≃\frac{h}{P}×\frac{ΔP}{P}$

$⇒\frac{Δλ}{λ}=\frac{ΔP}{P}=0.01$

$⇒P=\frac{ΔP}{0.01}=100P$  $[ΔP=P]$

∴ The initial momentum is 100 P.