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A proton collides with a stationary deuteron and a $^{3} H e$ nucleus is formed. For this reaction to take place, the proton must have minimum kinetic energy 1.4 MeV. If instead, a deuteron collides with a stationary proton to make a $^{3} H e$ nucleus. Then

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Minimum kinetic energy deuteron must possess = 0.7 MeV

Energy needed for the reaction is approx 0.93 MeV

Minimum kinetic energy deuteron must possess = 2.8 MeV

Energy needed for the reaction is approx 2.8 MeV

answer is A, C.

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Detailed Solution

$\frac{1}{2} \frac{M m}{M + m} V^{2} = Q \frac{1}{2} m V^{2} = K K = (1 + \frac{m}{M}) Q 1.4 = (1 + \frac{1}{2}) Q Q = \frac{2}{3} \times 1.4$

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