A proton collides with a stationary Deutron and a $^{3} H e$ nucleus is formed. For this reaction to take place proton must have minimum kinetic energy of 1.4 MeV. If instead a Deutron collided with a stationary proton to make $^{3} H e$ nucleus, minimum kinetic energy that this Deutron must possess in MeV would be (write upto two digits after the decimal point).

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Solution

Here, $\frac{1}{2} m V_{p}^{2} = 1.4 M e V$

By conservation of momentum, $\Rightarrow V_{H e} = \frac{V_{p}}{3}$
and by conservation of energy, $Δ E = \frac{1}{2} m V_{p}^{2} - \frac{1}{2} 3 m {(\frac{V_{p}}{3})}^{2} \Rightarrow Δ E = \frac{2}{3} \frac{m V_{p}^{2}}{2} = \frac{2}{3} \times 1.4 M e V$

In second case by conservation of momentum, $\Rightarrow V_{H e} = \frac{2}{3} V_{D}$

$Δ E = \frac{1}{2} 2 m V_{D}^{2} - \frac{1}{2} 3 m {(\frac{2}{3} V_{D})}_{D}^{2} = \frac{2 m V_{D}^{2}}{2} \times \frac{1}{3}$

However, $Δ E = \frac{2}{3} \times 1.4 M e V$ , hence

$\frac{2}{/ 3} \times 1.4 M e V = \frac{(2 m) V_{D}^{2}}{2} \times \frac{1}{/ 3} \Rightarrow \frac{1}{2} (2 m) V_{D}^{2} = 2.8 M e V$

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