Q: Two hydrogen atoms collide head on and end up with zero kinetic energy. Each atom then emits a photon of wavelength 121.6 nm. Which transition leads to this wavelength? How fast were the hydrogen atoms travelling before collision ?

Solution: Wavelength is emitted in UV region and thus n_{1} = 1

For H atom , $\large \frac{1}{\lambda} = R_h (\frac{1}{1^2} – \frac{1}{n^2}) $

$\large \frac{1}{121.6 \times 10^{-9}} = 1.097 \times 10^7 (\frac{1}{1^2} – \frac{1}{n^2}) $

Hence , n = 2

Also the energy released is due to collision and all the kinetic energy is released in form of photon.

$\large \frac{1}{2} m v^2 = \frac{h c}{\lambda} $

$\large \frac{1}{2} \times 1.67 \times 10^{-27} \times v^2 = \frac{6.625 \times 10^{-34} \times 3 \times 10^8}{121.6 \times 10^{-9}} $

v = 4.43 × 10^{4} m/sec