If the average life time of an excited state of H atom is of order 10^-8 sec, estimate how many orbits an e^- makes when it is in the state n=2 and before it suffers a transition to n=1 state

speed of electron in any orbit = v_n = v_o/n                    (v_o = 2.18*10⁶m/s)

so , v₂ = v_o/2  = 1.09*10⁶m/s

radius of  orbit = r_n = a_on²                      (a_o = 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        .......................1

substituting values of v,r,T we get

n = 8.18*10⁶ revolutions

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