A Bohr hydrogen atom undergoes a transition n = 5 → n = 4 and emits a photon of frequency ν . Frequency of circular motion of an electron in n = 4 orbit is ν 4 . The ratio ν ν 4 is

Step-by-step solution

E n = - m z 2 e 4 8 ε 0 2 n 2 h 2 So, h ν = + m z 2 e 4 8 ε 0 2 h 2 1 16 - 1 25 ∴ ν = m z 2 e 4 8 ε 0 2 h 3 9 16 × 25 . . . ( 1 ) And frequency ν 4 = 1 T = v 2 π r = Z e 2 2 ε 0 n h 1 2 π   π m z e 2 ε 0 h 2 n 2 = z 2 e 4 m 4 ε 0 2 n 3 h 3 . . . ( 2 ) ∴   ν ν 4 = 18 25 = 0.72