Q. A proton of mass $1.6 \times 10^{- 27}kg$ goes round in a circular orbit of radius 0.10$\ m$ under a centripetal force of $4 \times 10^{- 13}\ N.$ then the frequency of revolution of the proton is about
A) $0.08 \times 10^{8}\ cycles\ per\ sec$
B) $4 \times 10^{8}\ cycles\ per\ sec$
C) $8 \times 10^{8}\ cycles\ per\ sec$
D) $12 \times 10^{8}\ cycles\ per\ sec$
Explanation
(a) $m4\pi^{2}n^{2}r = 4 \times 10^{- 13} \Rightarrow n = 0.08 \times 10^{8}\ cycles/sec.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.