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)

see earlier answer

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