\begin{problem*}{11.41}
Two hydrogen atoms collide head-on and end up with zero kinetic
energy.  Each atom then emits light with a wavelength of $121.6\U{nm}$
($n=2$ to $n=1$ transition).  At what speed where the atoms moving
before the collision?
\end{problem*} % problem 11.41

\begin{solution}
Ending up with zero kinetic energy after the collision means the atoms
must have zero velocity and momentum after the collision as well.
Conserving momentum, we see that having zero momentum after the
collision, means their momenta before the collision must have been
equal and opposite.  Because both atoms are hydrogens, their masses
are the same, so their velocities must have also been equal and
opposite.  Thus, their initial kinetic energies must have been equal.

Conserving energy
\begin{align}
  E_i &= 2 \cdot \frac{1}{2} m v_i^2 = m v_i^2
    = E_f = 2\cdot\frac{hc}{\lambda} \\
  v_i &= \sqrt{\frac{2hc}{\lambda m}}
    = \sqrt{\frac{2\cdot6.63\E{-34}\U{J$\cdot$s}\cdot3.00\E{8}\U{m/s}}
                 {121.6\E{-9}\U{m}\cdot1.67\E{-27}\U{kg}}}
    = \ans{44.3\U{km/s}} \;. % ((2*6.63e-34*3e8)/(121.6e-9*1.67e-27))**.5
\end{align}
Note that the mass of a hydrogen atom is the sum of the masses of a
proton and electron, but $m_e\ll m_p$, which is why $m_H\approx
m_p=1.67\E{-27}\U{kg}$.
\end{solution}
