A hydrogen atom de-excites from level n to level (n-1) to a lower level (n-1) The energy (E1) at level n, E1=hv1 = hme4 x 1/n2/(4π)3 x ε02 x (h/2 π)3 ......(i) The energy (E1) at level (n-1) E1=hv2 = hme4 x (1/(n-1)2/(4π)3 x ε02 x (h/2 π)3 ....(ii) Where frequency of radiation at level (n-1) Due to de-excitation energy released E=E2-E1 hAZA1/2 = E2-E1 ......(iii) Where AZA1/2 = frequency of radiation emitted V = me4 x [(1/(n-1)2 -(1/n2)]/(4π)3 x ε02 x (h/2 π)3 = me4 x (2n-1)/(4π)3 x ε02 x (h/2 π)3 x n2(n-1) If n is large the 2n-1 = 2n and (n-1)= n Putting the values of equation (i) and (ii) in equation (iii) and get Then, V = me4 /32π3 x ε02 x (h/2 π)3 x n3......(iv) The frequency of revolution of an electron Then, the velocity of an electron, Vc = v/2 πr ....(v) If the velocity of an electron in nth orbit v= e2/4π x ε0(h/2 π)n......(vi) Then the radius of the nth orbit as r =4π x ε0(h/2 π)2n2/me2....(vii) Putting the values of equation (vi) and (vii) in equation (V) and get Vc = me4/ 32 π x ε0(h/2 π)3n3.... (viii) Hence, frequency of radiation emitted by hydrogen atom is euqal to its classical orbit frequecy.