JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2016
- Hard
- MCQ
A neutron moving with a speed \('v'\) makes a head on collision with a stationary hydrogen atom in ground state . The minimum kinetic energy of the neutron for which inelastic collision will take place is....\(eV\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(20.4\)
Step-by-step Solution
Detailed explanation
For inelastic collision \(\mathrm{v}^{\prime}=\frac{\mathrm{m}_{1}}{\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)} \mathrm{v}\)
\(=\frac{1}{(1+1)} v=\frac{v}{2}\)
\(\mathrm{n} \rightarrow \mathrm{v}(\mathrm{H}) \quad\) Before
\((n)(H) \rightarrow \frac{v}{2} \quad\) After Loss in \(K.E.\)
\(=\frac{1}{2} m v^{2}-\frac{1}{2}(2 m)\left(\frac{v}{2}\right)^{2}=\frac{1}{4} m v^{2}\)
\(K.E.\) lost is used to jump from \(1\,st\) orbit to \(2\,nd\) orbit \(\Delta \mathrm{K} . \mathrm{E}=10.2 \mathrm{eV}\)
\(\Rightarrow \frac{1}{4} m v^{2}=10.2\)
Minimum \(K.E.\) of neutron for inelastic collision
\(\frac{1}{2} m v^{2}=2 \times 10.2=20.4 \mathrm{eV}\)
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