First slide

Alpha-particle scattering experiment

Question

An alpha particle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus. The distance of the closest approach is of the order of

Moderate

A

10^-15 cm
B

10^-13 cm
C

10^-12 cm
D

10^-19 cm

Solution

$Use E = \frac{1}{4 π ε_{0}} \frac{(Z e) (e)}{r_{0}} 5 \times 1.6 \times 10^{- 13} = 9 \times 10^{9} \frac{\times 2 \times 92 \times {(1.6 \times 10^{- 19})}^{2}}{r_{0}} r_{0} = 5.3 \times 10^{- 14} m \approx 10^{- 12} c m$

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