A radioactive nucleus X of mass number 220 undergoes alpha decay to...

The alpha decay equation is given by:

$X^{220} \rightarrow Y^{216} + \alpha^{4}$

The energy released ($Q$-value) in the decay process is the difference between the total binding energy of the products and the total binding energy of the reactant.

Total binding energy of $X$ is:

$BE_X = 220 \times 7.4 = 1628 \text{ MeV}$

Total binding energy of $Y$ is:

$BE_Y = 216 \times 7.6 = 1641.6 \text{ MeV}$

Total binding energy of the alpha particle is:

$BE_\alpha = 4 \times 7.0 = 28.0 \text{ MeV}$

The energy released is:

$Q = BE_Y + BE_\alpha - BE_X$

$Q = 1641.6 + 28.0 - 1628$

$Q = 1669.6 - 1628 = 41.6 \text{ MeV}$

Answer: $41.6 \text{ MeV}$