Question

A proton collides with a stationary deuteron to form a $\_{}^{3}He$ nucleus. For this reaction to take place, the proton must have a minimum kinetic energy $K_{0}$ . If instead, a deuteron collides with a stationary proton to make a $\_{}^{3}He$ nucleus, then it must have minimum kinetic energy equal to

• A. $2K_{0}$
• B. $1.5K_{0}$
• C. $K_{0}$
• D. $\frac{K_{0}}{2}$

Answer 1

Answer: (A)

In case 1:
Conservation of linear momentum: $mv_{0}=3mv$
$\Rightarrow \, v=\frac{v_{0}}{3}$
Conservation of mechanical energy: $\frac{1}{2}mv_{0}^{2}=\frac{1}{2}3m\left(\frac{v_{0}}{3}\right)^{2}+U_{0}$
Here $U_{0}$ is the minimum energy required for the reaction to happen
$\Rightarrow U_{0}=K_{0}-\frac{K_{0}}{3}=\frac{2 K_{0}}{3}$
In case 2:
Conservation of linear momentum: $2mv_{0}^{'}=3mv^{'}\Rightarrow v'=\frac{2}{3}v_{0}^{'}$
Conservation of mechanical energy: $\frac{1}{2}2m\left(v_{0}^{'}\right)^{2}=\frac{1}{2}3m\left(\frac{2}{3} v_{0}^{'}\right)^{2}+\frac{2 K_{0}}{3}$
$\Rightarrow \, mv_{0}^{' 2}-\frac{2}{3}mv_{0}^{' 2}=\frac{2 K_{0}}{3}$
$\Rightarrow \frac{1}{3}mv_{0}^{' 2}=\frac{2 K_{0}}{3}$
$\Rightarrow \frac{1}{2}2mv_{0}^{' 2}=2K_{0}$