First slide

Bohr's atom model

Question

The maximum angular speed of the electron of a hydrogen atom in a stationary orbit is

Easy

A

$6.2 \times 10^{15} {rads}^{- 1}$
B

$4.1 \times 10^{16} {rads}^{- 1}$
C

$24 \times 10^{10} {rads}^{- 1}$
D

$9.2 \times 10^{6} {rads}^{- 1}$

Solution

Maximum angular speed will be in its ground state. Hence,
$ω_{max} = \frac{v_{1}}{r_{1}} = \frac{2.2 \times 10^{6}}{0.529 \times 10^{- 10}} = 4.1 \times 10^{16} {rads}^{- 1}$

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