Solution: Minimum Speed for Inelastic Collision
Key Concept: For a collision to be inelastic, some kinetic energy must be converted into internal energy (excitation). The maximum energy available for excitation in a collision is the loss in kinetic energy in the Center of Mass frame, given by $\Delta K = \frac{1}{2} \mu v^2$, where $\mu$ is the reduced mass. This energy must be at least equal to the first excitation energy of the target atom ($He^+$).
Step 1: Calculate Excitation Energy of $He^+$
$He^+$ is hydrogen-like with $Z=2$. Ground state energy $E_1 = -13.6 \times \frac{Z^2}{1^2} = -13.6 \times 4 = -54.4$ eV.
First excited state ($n=2$) energy $E_2 = -13.6 \times \frac{4}{4} = -13.6$ eV.
Minimum energy required ($\Delta E$) = $E_2 – E_1 = -13.6 – (-54.4) = 40.8$ eV.
$He^+$ is hydrogen-like with $Z=2$. Ground state energy $E_1 = -13.6 \times \frac{Z^2}{1^2} = -13.6 \times 4 = -54.4$ eV.
First excited state ($n=2$) energy $E_2 = -13.6 \times \frac{4}{4} = -13.6$ eV.
Minimum energy required ($\Delta E$) = $E_2 – E_1 = -13.6 – (-54.4) = 40.8$ eV.
Step 2: Energy Loss Equation
Mass of neutron $m_n = m$. Mass of $He^+$ atom $M \approx 4m$.
Reduced mass $\mu = \frac{m \cdot 4m}{m + 4m} = \frac{4m}{5}$.
Loss in Kinetic Energy: $$ \Delta K = \frac{1}{2} \mu v^2 = \frac{1}{2} \left( \frac{4m}{5} \right) v^2 = \frac{2}{5} m v^2 $$
Mass of neutron $m_n = m$. Mass of $He^+$ atom $M \approx 4m$.
Reduced mass $\mu = \frac{m \cdot 4m}{m + 4m} = \frac{4m}{5}$.
Loss in Kinetic Energy: $$ \Delta K = \frac{1}{2} \mu v^2 = \frac{1}{2} \left( \frac{4m}{5} \right) v^2 = \frac{2}{5} m v^2 $$
Step 3: Solve for Minimum Velocity
Equate $\Delta K$ to $\Delta E$: $$ \frac{2}{5} m v^2 = 40.8 \text{ eV} $$ $$ v^2 = \frac{5 \times 40.8 \times 1.6 \times 10^{-19}}{2 \times 1.67 \times 10^{-27}} $$ $$ v^2 \approx \frac{326.4 \times 10^{-19}}{3.34 \times 10^{-27}} \approx 97.7 \times 10^8 $$ $$ v \approx \sqrt{97.7} \times 10^4 \approx 9.89 \times 10^4 \text{ m/s} $$
Equate $\Delta K$ to $\Delta E$: $$ \frac{2}{5} m v^2 = 40.8 \text{ eV} $$ $$ v^2 = \frac{5 \times 40.8 \times 1.6 \times 10^{-19}}{2 \times 1.67 \times 10^{-27}} $$ $$ v^2 \approx \frac{326.4 \times 10^{-19}}{3.34 \times 10^{-27}} \approx 97.7 \times 10^8 $$ $$ v \approx \sqrt{97.7} \times 10^4 \approx 9.89 \times 10^4 \text{ m/s} $$
Correct Option: (d) $9.89 \times 10^4$ m/s
Note: Any speed higher than this value also allows for inelastic collision, which is why multiple options might be theoretically valid in a “select all that apply” context, but this is the minimum threshold.