Answer 1

Given

r₂ = 2.14 × 10^-10 m

n = 2

v₂ = 1.09 × 10⁶ m/s

To find: Frequency of revolution (`nu_2`)

`v = romega=r(2pinu)`

`nu=v/(2pir)`

`nu_2=v_2/(2pir_2)=(1.09xx10^6)/(2xx3.142xx2.14xx10^-10)`

`nu_2=8.11xx10^14 Hz`

The frequency of revolution of electron in 2nd Bohr orbit is 8.11 × 10¹⁴ Hz.

Answer 2

`T = (2pir)/V`

`because T=1/f`

`f = V/(2pir)`

`f=(1.09 xx 10^6)/(2 xx 3.14 xx 2.14 xx 10^-10)`

`f=8.11 xx 10^14 Hz`

The frequency of revolution of electron in 2nd Bohr orbit is 8.11 × 10¹⁴ Hz.