A neutron with 0.6MeV kinetic energy directly collides with a stationary cartoon nucleus (in number 12). The kinetic energy of carbon nucleus after the collision is:

1.7 MeV

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0.17 MeV

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17 MeV

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Zero

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Solution

The correct option is B 0.17 MeV
Kinetic energy transferred to stationary target (carbon nucleus)

$\frac{△ k}{k} = [1 - (\frac{m_{1} - m_{2}}{m_{1} + m_{2}})^{2}]$

Where $△ K$ (delta k) is change in Kinetic energy.

$= [1 - (\frac{1 - 12}{1 + 12})^{2}] = \frac{48}{169}$

$△ K = \frac{48}{169} \times (0.6 M e V) = 0.17 M e V$

Hence (B) option is correct answer

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A neutron with 0.6MeV kinetic energy directly collides with a stationary cartoon nucleus (in number 12). The kinetic energy of carbon nucleus after the collision is:

The correct option is B 0.17 MeVKinetic energy transferred to stationary target (carbon nucleus)△kk=[1−(m1−m2m1+m2)2]Where △K (delta k) is change in Kinetic energy.=[1−(1−121+12)2]=48169△K=48169×(0.6MeV)=0.17MeVHence (B) option is correct answer