Question

A hydrogen atom ( mass $=1.66\times {{10}^{-27}}kg,$ ionisation potential $=13.6eV$ ), moving with a velocity of $6.24\times {{10}^{4}}m{{s}^{-1}}$ makes a completely inelastic head-on collision with another stationary hydrogen atom. Both atoms are in the ground state before collision. Up to what state either of one atom may be excited?

Answer 1

- Hint: To solve this question find the initial total kinetic energy of the system. Since, this is an inelastic collision it follows only the law of conservation of momentum. Using the law of conservation of momentum write the final velocity in terms of the initial velocity. Find the final kinetic energy and then the loss of energy of the system. This loss of energy will be equal to the energy used to excite the atom. From this we can find the excitation state of the atom.

Complete step-by-step solution -
we are given that the mass of hydrogen is $m=1.66\times {{10}^{-27}}kg$
The velocity of the moving hydrogen atom is $v=6.24\times {{10}^{4}}m{{s}^{-1}}$
So, we can say that the initial kinetic energy of the system of the two hydrogen atoms is,
${{E}_{i}}=\dfrac{1}{2}m{{v}^{2}}$
After the inelastic collision, the two hydrogen atoms will combine to each other and the mass will be 2m.
Let the velocity of the combined system after collision will be ${{v}_{f}}$.
Applying the law of conservation of momentum on the system,
$\begin{align}
  & mv+0=2m{{v}_{f}} \\
 & {{v}_{f}}=\dfrac{v}{2} \\
\end{align}$
The velocity of the final system will be $\dfrac{v}{2}$
The final kinetic energy of the system will be,
$\begin{align}
  & {{E}_{f}}=\dfrac{1}{2}\times 2m\times v_{f}^{2} \\
 & {{E}_{f}}=\dfrac{1}{2}\times 2m\times {{\left( \dfrac{v}{2} \right)}^{2}} \\
 & {{E}_{f}}=\dfrac{1}{4}m{{v}^{2}} \\
\end{align}$
So, the change in energy or loss of energy of the system is,
$\begin{align}
  & \Delta E={{E}_{i}}-{{E}_{f}} \\
 & \Delta E=\dfrac{1}{2}m{{v}^{2}}-\dfrac{1}{4}m{{v}^{2}} \\
 & \Delta E=\dfrac{1}{4}m{{v}^{2}} \\
\end{align}$
Putting the values of mass and velocity,
$\begin{align}
  & \Delta E=\dfrac{1}{4}\times 1.66\times {{10}^{-27}}\times {{\left( 6.24\times {{10}^{4}} \right)}^{2}} \\
 & \Delta E=16.16\times {{10}^{-19}}J \\
\end{align}$
Converting into eV,
$\Delta E=\dfrac{16.16\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}=10.1eV$
Now let this lost energy be used to excite the hydrogen atom to a higher excited state from ground state.
So, we can write that,
$\begin{align}
  & \Delta E=13.6\left( \dfrac{1}{{{1}^{2}}}-\dfrac{1}{n_{f}^{2}} \right) \\
 & 10.1=13.6\left( 1-\dfrac{1}{n_{f}^{2}} \right) \\
 & \left( 1-\dfrac{1}{n_{f}^{2}} \right)=\dfrac{10.1}{13.6} \\
 & \dfrac{1}{n_{f}^{2}}=1-\dfrac{10.1}{13.6} \\
 & \dfrac{1}{n_{f}^{2}}=\dfrac{3.5}{13.6} \\
 & n_{f}^{2}=3.89 \\
 & {{n}_{f}}=\sqrt{3.89} \\
 & {{n}_{f}}\approx 2 \\
\end{align}$
So, the hydrogen atom will excite to the second state from the ground state.
Note: In inelastic collision only momentum of the system is conserved. If we consider an elastic collision both the momentum and kinetic energy of the system will be conserved. In inelastic collision, kinetic energy of the system does not conserve.