A stationary nucleus of mass number A undergoes alpha decay by...

Let the momentum of the emitted alpha particle be $p$. By the law of conservation of linear momentum, the recoiling daughter nucleus must have an equal and opposite momentum of magnitude $p$.

The mass of the alpha particle is $4$ and the mass of the daughter nucleus is $(A-4)$.

The kinetic energy of the alpha particle is $K_{\alpha} = \dfrac{p^2}{2(4)}$ and the kinetic energy of the daughter nucleus is $K_D = \dfrac{p^2}{2(A-4)}$.

The total kinetic energy released is $Q = K_{\alpha} + K_D = \dfrac{p^2}{2(4)} + \dfrac{p^2}{2(A-4)}$.

Simplifying the expression for $Q$:
$Q = \dfrac{p^2}{2} \left( \dfrac{1}{4} + \dfrac{1}{A-4} \right) = \dfrac{p^2}{2} \left( \dfrac{A-4+4}{4(A-4)} \right) = \dfrac{p^2}{2} \left( \dfrac{A}{4(A-4)} \right)$.

We need to find $K_D$ in terms of $Q$. Taking the ratio:
$\dfrac{K_D}{Q} = \dfrac{\dfrac{p^2}{2(A-4)}}{\dfrac{p^2}{2} \left( \dfrac{A}{4(A-4)} \right)} = \dfrac{4}{A}$.

Thus, the kinetic energy of the recoiling daughter nucleus is $K_D = \dfrac{4Q}{A}$.

Answer: $\dfrac{4Q}{A}$