First slide

Wave nature of matter

Question

Find the ratio of de Broglie wavelength of a proton and an alpha-particle which have been accelerated through same potential difference.

Easy

A

$2 \sqrt{2} : 1$
B

3:2
C

$3 \sqrt{2} : 1$
D

2:1

Solution

de Broglie wavelength $= λ = \frac{h}{m v} = \frac{h}{\sqrt{2 m q V}}$
$\begin{array}{l} ∴ \frac{λ_{p}}{λ_{α}} = \sqrt{\frac{m_{α} q_{α} V_{α}}{m_{p} q_{p} V_{p}}} \\ Putting V_{α} = V_{p}, \\ \Rightarrow \frac{λ_{p}}{λ_{α}} = \sqrt{\frac{4 \times 2}{1 \times 1}} = 2 \sqrt{2} \end{array}$

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