Solveeit Logo
Login

Question

Question: A proton of mass 1.6 × 10<sup>–27</sup>kg goes round in a circular orbit of radius 0.10 m under a ce...

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

A

0.08 × 108cyclespersec

B

4 × 108cyclesper sec

C

8 × 108cyclespersec

D

12 × 108cyclespersec

Answer

0.08 × 108cyclespersec

Explanation

Solution

F=4×10−13NF = 4 \times 10^{- 13}N;m=1.6×10−27kgm = 1.6 \times 10^{- 27}kg; r=0.1mr = 0.1m

Centripetal force F=m4π2n2r∴n=F4mπ2rF = m4\pi^{2}n^{2}r\therefore n = \sqrt{\frac{F}{4m\pi^{2}r}}

=8×106cycles/sec⁡=0.08×108cycle/sec⁡= 8 \times 10^{6}cycles/\sec = 0.08 \times 10^{8}cycle/\sec.