Question

A proton of mass 1.6 x 10-27 kg goes round in a circular
orbit of radius 0.10 m under a centripetal force of
4 x 10-13N then the frequency of revolution of the
proton is about​

Answer 1

Dear Student,

◆ Answer -

f = 7.958×10^6 cps

◆ Explanation -

Centripetal force applied is given by formula -

F = m.r.ω^2

4×10^-13 = 1.6×10^-27 × 0.1 × ω^2

ω^2 = 4×10^-13 / 1.6×10^-28

ω^2 = 25×10^14

ω = 5×10^7 rad/s

Frequency of revolution is calculated as -

f = ω/2π

f = 5×10^7 / 2π

f = 7.958×10^6 cps

Therefore, frequency of revolution should be 7.958×10^6 cps.

Thanks dear...