NCERT Solutions
Class 12 Physics
Nuclei

Q.12
Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of (a) 22688 Ra and (b) 22086 Rn.
Given m (22688 Ra) = 226.02540 u, m (22286 Rn) = 222.01750 u,
m (22286 Rn) = 220.01137 u, m (21684 Po) = 216.00189 u.
(a) Alpha particle decay of (22688Ra) 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.
(22688Ra) à (22288Ra) + (42He)
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 (22688Ra) =226.0250 u, m (22286Rn) =222.01750u, m (42He) =4.002603u
Q-value = [226.02540 − (222.01750 + 4.002603)] u c2
= 0.005297 u c2
But 1 u = 931.5
Therefore Q = (0.005297 × 931.5) ≈ 4.94 MeV
Kinetic energy of the α-particle =
x Q
= ( x 4.94) = 4.85MeV.
(b) Alpha particle decay of (22086Rn)
(22086Rn) à (21684Po) + (42He)
It is given that:
Mass of (22086Rn) = 220.01137 u
Mass of (21684Po) = 216.00189 u
Therefore, Q-value = [220.01137 – (216.00189+4.002603)] x931.5
≈ 641 MeV
Kinetic energy of the α-particle = ( x 6.41)
= 6.29 MeV