The ratio of energies of photons produced due to transition of electron of hydrogen atom from its (i) third to 2^ nd energy level and (ii) highest energy level to 3^ rd level is

Step-by-step solution

The photon energy for an electron transition from n_i to n_f in hydrogen is given by E = R_H ( 1 n_f^2 - 1 n_i^2 ) . For the n_i = 3 n_f = 2 transition: E₁ = R_H ( 1 4 - 1 9 ) = R_H ( 9 - 4 36 ) = R_H 5 36 For the n_i = n_f = 3 transition: E₂ = R_H ( 1 9 - 0 ) = R_H 1 9 The energy ratio is: E₁ E₂ = R_H 5 36 R_H 1 9 = 5 36 9 1 = 45 36 = 5 4 The ratio of photon energies is therefore 5:4 . Answer: B