In the bohr model of the hydrogen atom, an electron({rm mass}; m=9.1; times 10^{ - 31; }{rm kg}) orbits a proton at a distance of 5.3 times 10^{ - 11; }{rm m}. the proton pulls on the electron with an electricforce of 8.2 times 10^{ - 8}; {rm n}. how many revolutions per second does theelectron make?

n=6.56×10¹⁵Hz

Explanation:

Given Data

Mass=9.1×10⁻³¹ kg

Radius distance=5.3×10⁻¹¹m

Electric Force=8.2×10⁻⁸N

To find

Revolutions per second

Solution

Let F be the force of attraction

let n  be the number of revolutions per sec made by the electron around the nucleus then the centripetal force is given by

F=mω²r......................where ω=2π  n

F=m4π²n²r...............eq(i)

as the values given where

Mass=9.1×10⁻³¹ kg

Radius distance=5.3×10⁻¹¹m

Electric Force=8.2×10⁻⁸N

we have to find n from eq(i)

n²=F/(m4π²r)

energy can be converted from one form to another, but it cannot be created or destroyed in ordinary chemical or physical means. do not understand what you are asking but hope this