JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
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....\(eV\)
- A \(20.4\)
- B \(10.2\)
- C \(12.1\)
- D \(16.8\)
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…
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