Use c = 3.0x10^8 m s^-1, 1 eV=1.6x10^-19 J, E_1=-13.6 eV and m_H = 1.67x10^-27 kg. A stationary hydrogen atom emits a photon in the n=2-> n=1 transition. Neglecting the small energy given to recoil, find the recoil speed of the atom in m s^-1 (two decimals).

JEE Advanced levelNumerical answerDifficulty 3 of 3as099

Use $c = 3.0 \times 1 0^{8} m s^{- 1}$ , $1 eV = 1.6 \times 1 0^{- 19} J$ , $E_{1} = - 13.6 eV$ and $m_{H} = 1.67 \times 1 0^{- 27} kg$ . A stationary hydrogen atom emits a photon in the $n = 2 \to n = 1$ transition. Neglecting the small energy given to recoil, find the recoil speed of the atom in $m s^{- 1}$ (two decimals).

Numerical answer type, answer in m/s, tolerance 0.03.

Photon energy

E = 10.2 eV = 1.632 e - 18 J

, so momentum

p = E / c = 5.440 e - 27 kg m s^{- 1}

. By momentum conservation the atom recoils with

v = p / m_{H} = 3.26 m s^{- 1}