When a hydrogen atom going from n = 2 to n = 1 emits a photon, its recoil speed is x 5 m s - 1 . Where x = _______. (Use: mass of hydrogen atom = 1 . 6 × 10 - 27 kg , charge of electron e = 1 . 6 × 10 - 19 C )

Step-by-step solution

According to the above diagram, the difference in energy between the two states is given by ΔE = - 3 . 4 eV - - 13 . 6 eV = 10 . 2 eV The formula to calculate the recoil speed is given by v = ∆ E m c . . . 1 From equation (1), it follows that v = 10 . 2 eV × 1 . 6 × 10 - 19 J 1 eV 1 . 6 × 10 - 27 kg × 3 × 10 8 m s - 1 = 3 . 4 m s - 1 = 17 5 m s - 1 Therefore, x = 17 .