Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of

Solution:

(a) Alpha particle decay of ${ }_{88}^{226} \mathrm{Ra}$ emits a helium nucleus. As a result, its mass number reduces to $(226-4) 222$ and its atomic number reduces to $(88-2) 86$. This is shown in the following nuclear reaction.

${ }_{88}^{226} \mathrm{Ra} \longrightarrow{ }_{86}^{222} \mathrm{Ra}+{ }_{2}^{4} \mathrm{He}$

Q-value of

emitted α-particle = (Sum of initial mass − Sum of final mass) c2

Where,

c = Speed of light

It is given that:

$m\left({ }_{88}^{226} \mathrm{Ra}\right)=226.02540 \mathrm{u}$

$m\left({ }_{86}^{222} \mathrm{Rn}\right)=222.01750 \mathrm{u}$

$m\left({ }_{2}^{4} \mathrm{He}\right)=4.002603 \mathrm{u}$

Q-value = [226.02540 − (222.01750 + 4.002603)] u c2
= 0.005297 u c2

But 1 u = 931.5 MeV/c2

∴Q = 0.005297 × 931.5 ≈ 4.94 MeV

Kinetic energy of the $\alpha$-particle $=\left(\frac{\text { Mass number after decay }}{\text { Mass number before decay }}\right) \times Q$

$=\frac{222}{226} \times 4.94=4.85 \mathrm{MeV}$

(b) Alpha particle decay of $\left({ }_{86}^{220} \mathrm{Rn}\right)$ is shown by the following nuclear reaction.

${ }_{86}^{220} \mathrm{Rn} \longrightarrow{ }_{84}^{216} \mathrm{Po}+{ }_{2}^{4} \mathrm{He}$

It is given that:

Mass of $\left({ }_{86}^{220} \mathrm{Rn}\right)=220.01137 \mathrm{u}$

Mass of $\left({ }_{84}^{216} \mathrm{Po}\right)=216.00189 \mathrm{u}$

$\therefore Q$-value $=[220.01137-(216.00189+4.00260)] \times 931.5$

≈ 641 MeV

Kinetic energy of the $\alpha$-particle $=\left(\frac{220-4}{220}\right) \times 6.41$

$=6.29 \mathrm{MeV}$