First slide

Matter waves (DC Broglie waves)

Question

If the kinetic energy of the particle is increased to 16 times its previous vlaue, the percentage change in the de Broglie wavelength of the particle is

Easy

A

25
B

75
C

60
D

50

Solution

de Broglie wavelength,
$λ = \frac{h}{\sqrt{2 m K}}$
where m is the mass and K is the kinetic energy of the particle.

When kinetic energy of the particle is increased to 16 times, then its de Broglie wavelength becomes,
$λ^{'} = \frac{h}{\sqrt{2 m (16 K)}} = \frac{1}{4} \frac{λ}{\sqrt{2 m K}} = \frac{λ}{4}$ (Using (i))
% change in the de Broglie wavelength
$= \frac{λ - λ^{'}}{λ} \times 100 = (1 - \frac{λ^{'}}{λ}) \times 100 = (1 - \frac{1}{4}) \times 100 = 75 %$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App