A moving neutron (of mass m) is about to collide with a stationary singly ionized helium atom (of mass 4m) in ground state. The speed of the incident neutron is equal to the minimum speed required such that the collision may be inelastic. If the collision is head-on perfectly elastic then the energy of neutron after collision is $\frac{n}{5} (10.2 e V)$ where n is a positive integer. Find n? (Assume Bohr’s quantization principle to be valid for energy states of singly ionized helium atom and 1 Rydberg = 13.6eV)

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answer is 9.

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Detailed Solution

Loss in KE = $\frac{1}{2} \times \frac{m \times 4 m}{5 m} {(4 - 0)}^{2} (1 - e^{2})$
(Loss in KE) =    $13.6 e V [\frac{1}{1^{2}} - \frac{1}{2^{2}}] 2^{2} = 10.2 \times 4 e V$
$\Rightarrow B u t u_{\min} \Rightarrow e = 0 \Rightarrow \frac{1}{2} m u_{\min}^{2} \times \frac{4}{5} = 10.2 \times 4 e V$
$\Rightarrow \frac{1}{2} m u_{\min}^{2} = 10.2 \times 5 e V$
Also, in HOEC:    $V_{n e u t r o n} = \frac{2 \times 4 m \times 0 + u (m - 4 m)}{5 m} = - \frac{3 u}{5}$
$K E_{a f t e r c o l l . o f n e u t r o n} = \frac{1}{2} m {(- \frac{3 u}{5})}^{2} = \frac{2}{25} \times 10.2 \times 5 = \frac{9}{5} \times 10.2 e V$
Ans. $n = 9$