If the momentum of an electron is changed by Δ p, then the de-Broglie wavelength associated with it changes by 0.50 %. The initial momentum of the electron will be

Question

If the momentum of an electron is changed by Δ p, then the de-Broglie wavelength associated with it changes by 0.50 %. The initial momentum of the electron will be (A)  (Δ p/200)  (B)  (Δ p/199)  (C) 199 Δ p  (D)  400 Δ p

Tardigrade Admin · Accepted Answer

λ=(h/p) ⇒ λ-(0.5/100) λ=(h/p+Δ p) or (199 λ/200)=(h/p+Δ p) =(199/200) (h/p) ⇒ p+Δ p=(200/199) p ⇒ p=199 Δ p

Q. If the momentum of an electron is changed by $Δ p$ , then the de-Broglie wavelength associated with it changes by $0.50%$ . The initial momentum of the electron will be

$\frac{Δ p}{200}$

$\frac{Δ p}{199}$

$199 Δ p$

$400 Δ p$

$λ = \frac{h}{p}$
$\Rightarrow λ - \frac{0.5}{100}$
$λ = \frac{h}{p + Δ p}$
or $\frac{199 λ}{200} = \frac{h}{p + Δ p}$
$= \frac{199}{200} \frac{h}{p}$
$\Rightarrow p + Δ p = \frac{200}{199} p$
$\Rightarrow p = 199Δ p$

Q. 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 the electron will be

200Δp​

199Δp​

199Δp

400Δp