Question

How many revolutions does an electron perform before making transition from n = 2 to n = 1 state in hydrogen atom? The average life time of first excited state in H-atoms is 10⁻⁸ s, R = 1.097*10⁷ m⁻¹.[Ans: 8.23*10⁶]

Answer 1

Speed of electron in any orbit = vn = vo/n, (vo = 2.18*106m/s),

so , v2 = vo/2 = 1.09*106m/s,

radius of orbit = rn = aon2 ,

(ao = 52.9 * 10-12m),

so ,

r2 = 2.12*10-10m,

v = X/T, x = @r so ( r is radius of second orbit)v = @r/T,

( @ is angular displacement) @ = vT/r,

@ = 2pi(n), (here n is number of revolutions),

2pi(n) = vT/r,

substituting values of v,r,T we get n = 8.18*106 revolution and solution in a fine manner.