A stationary hydrogen atom in the first excited state emits a photon. If the mass of the hydrogen atom is m and its ionization energy is E , then the recoil velocity acquired by the atom is [speed of light = c ]

Step-by-step solution

m v = h λ , here v is recoil speed of H -atom and λ is the wavelength of the photon. When the electron jumps down from the first excited state to the ground state the energy released is ∆ E 2 → 1 = E 1 - 1 2 2 = 3 E 4 3 E 4 = 1 2 m v 2 + h c λ 3 E 4 = 1 2 m v 2 + m v c m 2 v 2 + m c v - 3 E 4 = 0 ⇒ v = 3 E 2 m + c 2 - c