Q: In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution

around the sun in an orbit of radius 1.5 × 10^{11} m with orbital speed 3 × 10^{4} m/s.(Mass of earth = 6.0 × 10^{24} kg.)

Sol: Given, radius of orbit r = 1.5 × 10^{11} m

Orbital speed v = 3 × 10^{4} m/s ;

Mass of earth M= 6 × 10^{24} kg

Angular momentum, mvr = nh/(2 π)

or, n = 2πvrm/h ;

[where, n is the quantum number of the orbit]

= (2 × 3.14 × 3 × 10^{4} × 1.5 × 10^{11} × 6 × 10^{24} )/(6.63 × 10^{-34} )

= 2.57 × 10^{74}

or, n = 2.6 × 10^{74}.

Thus,the quantum number is 2.6 × 10^{74}

Thus, the quantum number is 2.6 × 10^{74} which is too large.

The electron would jump from n = 1 to n = 3

E_{3} = (-13.6)/3^{2} = -1.5 eV.

So, they belong to Lyman series.