Question

An electron is accelerated through a potential v, its de broglie wavelength is lambda. if accelerating voltage is increased to 4v, its de broglie wavelength will become
A)4 lambda
B)2 lambda
C)lambda
D)lambda/2

Answer 1

answer : option (d)

explanation : we know, from De-broglie wavelength equation ,

$\lambda=\frac{h}{p}$

where , p is linear momentum of particle and h is plank's constant.

we also know, kinetic energy = p²/2m = electrostatic energy = qV

or, p² = 2qVm

or, P = √{2qVm}

hence, $\lambda=\frac{h}{\sqrt{2qVm}}$

here it is clear that wavelength is inversely proportional to square root of accelerating voltage.

so, $\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{V_2}{V_1}}$

or, $\frac{\lambda}{\lambda_2}=\sqrt{\frac{4V}{V}}=2$

or, $\lambda_2=\frac{\lambda}{2}$

hence, De-broglie wavelength will be $\frac{\lambda}{2}$

Answer 2

Answer:

D)

Explanation: