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: $m{v}_0=3mv$
$\Rightarrow v=\frac{v_0}{3}$
Conservation of mechanical energy: $\frac{1}{2}m{v}_0^2=\frac{1}{2}3m{\left(\frac{v_0}{3}\right)}^2+U_0$
Here $U_0$ is the minimum energy required for the reaction to happen
$\Rightarrow U_0=K_0-\frac{K_0}{3}=\frac{2K_0}{3}$
In case 2:
Conservation of linear momentum: $2m{v}_0^{{}^{\prime }}=3m{v}^{{}^{\prime }}\Rightarrow v^{\prime }=\frac{2}{3}{v}_0^{{}^{\prime }}$
Conservation of mechanical energy: $\frac{1}{2}2m{\left(v_0^{{}^{\prime }}\right)}^2=\frac{1}{2}3m{\left(\frac{2}{3}{v}_0^{{}^{\prime }}\right)}^2+\frac{2K_0}{3}$
$\Rightarrow m{v}_0^{{}^{\prime }2}-\frac{2}{3}m{v}_0^{{}^{\prime }2}=\frac{2K_0}{3}$
$\Rightarrow \frac{1}{3}m{v}_0^{{}^{\prime }2}=\frac{2K_0}{3}$
$\Rightarrow \frac{1}{2}2m{v}_0^{{}^{\prime }2}=2K_0$