The velocity of electron in a certain Bohr's orbit of H-atom bears the ratio `1: 275` to the velocity of light. What is the number of orbit ?

Explanation

Velocity of electron `= (1)/(275) xx 3 xx 10^(10) cm s^(-1) = 1.90 xx 10^(8) cm s^(-1)`
Velocity of electron in nth orbit of H-like particles in C.G.S. units is given by `u_(n) = (2pi Ze^(2))/(nh)`
Putting Z = 1 for H
and `e = 4.80 xx 10^(-10)" esu"`
and `h = 6.626 xx 10^(-27)` erg sec,
we get
`u_(n) = (2 xx 3.14 xx (4.80 xx 10^(-10))^(2))/(n xx (6.626 xx 10^(-27))) = (2.18 xx 10^(8))/(n)`
`:. (2.18 xx 10^(8))/(n) = 1.09 xx 10^(8)` (calculated above)
`:. n = 2`