Question:
It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is \(p_d\); while for its similar collision with carbon nucleus at rest, fractional loss of energy is \(p_c\). The values of \(p_d\) and \(p_c\) are respectively :
It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is \(p_d\); while for its similar collision with carbon nucleus at rest, fractional loss of energy is \(p_c\). The values of \(p_d\) and \(p_c\) are respectively :
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For a head-on elastic collision with a target at rest, the fractional energy loss is \(\frac{4m_1m_2}{(m_1+m_2)^2}\); use masses 1, 2 and 12 u.
Updated On: Nov 11, 2026
- (0, 1)
- (0.89, 0.28)
- (0.28, 0.89)
- (0, 0)
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The Correct Option is B
Solution and Explanation
Elastic head-on collision with the target at rest: fractional energy loss \(p=\frac{4m_1m_2}{(m_1+m_2)^2}\).
Neutron mass \(\approx1\) u, deuterium \(\approx2\) u: \(p_d=\frac{4\times1\times2}{(1+2)^2}=\frac89\approx0.89\).
Carbon \(\approx12\) u: \(p_c=\frac{4\times1\times12}{(1+12)^2}=\frac{48}{169}\approx0.28\).
Answer: (2) (0.89, 0.28)
Neutron mass \(\approx1\) u, deuterium \(\approx2\) u: \(p_d=\frac{4\times1\times2}{(1+2)^2}=\frac89\approx0.89\).
Carbon \(\approx12\) u: \(p_c=\frac{4\times1\times12}{(1+12)^2}=\frac{48}{169}\approx0.28\).
Answer: (2) (0.89, 0.28)
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