Given:
Parent nucleus: $_{88}^{226}$ Ra that undergoes alpha decay. When a nucleus undergoes alpha decay, it gives off an $alpha$ particle or $_{2}^{4} He$, it also loses 2 protons and 2 neutrons in the process. Due to this, its atomic number (Z) decreases by 2 and mass number (A) by 4.

Since our parent nucleus has atomic number (Z$_{parent}$) = 88, and mass number (A$_{parent}$) = 226, due to alpha decay, its daughter nucleus will have an atomic number (Z$_{daughter}$) = 86, and mass number (A$_{daughter}$) = 222.

Thus the daughter nucleus is: $boxed{_{86}^{222} Rn}$

$_{86}^{222}$ Rn

Given:

Alpha decay of $_{86}^{226}$ Ra that releases 4.871 MeV

a) Since the nucleus of the element undergoes alpha decay, then it loses 2 protons and 2 neutrons, thus $boxed{text{decreasing its mass}}$. The energy released from this decay came from the change in mass of the nucleus.

b) The change in mass can be computed using the formula:

$$
begin{align*}
E &= (text{change in mass})E_u \
text{change in mass} &= dfrac{E}{E_u} \
&= dfrac{4.871 text{MeV}}{931.5 text{MeV}} \
&= boxed{0.00523 u}
end{align*}
$$

a) mass of the system decrease

b) $0.00523$ u

In alpha decay, two protons and two neutrons are emitted. So the atomic number ($Z$) and mass number ($A$) of the parent nucleus undergoes the following changes.

$Z rightarrow Z-2$

$A rightarrow A-4$

For $^{242}_{96}$Cm, $Z=96$ and $A = 242$.

When $^{242}_{96}$Cm undergoes alpha decay, the atomic number of the daughter nucleus is $Z = 96 – 2 = 94$. It is plutonium (Pu) element. The mass number of Pu is $A = 242 – 4 = 238$.

So the daughter nucleus is $^{238}_{96}$Pu.

$^{242}_{96}$Cm $rightarrow$ $^{238}_{94}$Pu $+$ $^4_2$He $+$ energy

$^{238}_{94}$Pu