Question

Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of (a) and (b).

Given = 226.02540 u, = 222.01750 u,

= 220.01137 u, = 216.00189 u.

Answer 1

(a) Alpha particle decay of 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.

Q-value of

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

Where,

c = Speed of light

It is given that:

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 α-particle

(b) Alpha particle decay of is shown by the following nuclear reaction.

It is given that:

Mass of = 220.01137 u

Mass of = 216.00189 u

∴Q-value =

≈ 641 MeV

Kinetic energy of the α-particle

= 6.29 MeV