Problem 7: Energy Released in Dissociation
Conservation of Momentum:
Let the molecule move along the x-axis initially. Atom 1 moves perpendicular to the x-axis (y-direction). $$ \vec{P}_{H_2} = P_{2x} \hat{i} $$ $$ 0 = P_{1y} + P_{2y} \implies |P_{1y}| = |P_{2y}| $$ Since masses are equal ($m$), equal momenta implies equal kinetic energies for the y-components: $K_{1} = K_{2y}$.
From x-momentum conservation: $P_{H_2} = P_{2x}$.
Since $m_{H2} = 2m$, we have $K = P^2/2m$.
$$ K_{2x} = \frac{P_{2x}^2}{2m} = \frac{P_{H2}^2}{2m} = 2 \left( \frac{P_{H2}^2}{2(2m)} \right) = 2 K_{H_2} $$
Total Kinetic Energy Calculation:
Final Energy $K_f = K_1 + K_2 = K_1 + (K_{2x} + K_{2y}) = K_1 + 2K_{H_2} + K_1 = 2K_1 + 2K_{H_2}$.
Energy Released ($Q$):
$$ Q = K_f – K_i = (2K_1 + 2K_{H_2}) – K_{H_2} = 2K_1 + K_{H_2} $$
Using $K_1 = 0.8 \text{ eV}$ (from text) and $K_{H_2} = 1.0 \text{ eV}$: