Question

Two hydrogen atoms collide in a head on collision and end up with zero kinetic energy. Each then emits a photon of 121.6 nm (n=2 to n=1 transition). At what speed were the atoms moving before the collision?

Answer 1

Explanation:

Expression for the kinetic energy is as follows.

         K.E = $\frac{1}{2}mv^{2}$

Now, total kinetic energy will be as follows.

    K.E = $2 \times \frac{1}{2}mv^{2}$ = $m \times v^{2}$

Since, this energy converts into electromagnetic radiation of wavelength 121.6 nm.

Relation between energy and photon is as follows.

   Energy of photon = $\frac{hc}{\lambda}$

                                = $\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{121.6 \times 10^{-9}}$

                                 = $1.63 \times 10^{-18} J$

    $m \times v^{2} = 1.63 \times 10^{-18}$

          v = $\sqrt{\frac{1.63 \times 10^{-18}}{1.67 \times 10^{-27}}$

             = $3.12 \times 10^{4}$ m/s

Thus, we can conclude that atoms were moving at a speed of $3.12 \times 10^{4}$ m/s before the collision.