Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of (a) ²²⁶₈₈Ra and (b) ²²⁰₈₆Rn

Given m (²²⁶₈₈Ra) = 226.02540 u, m (²²²₈₆Rn) = 222.01750 u,

m (²²²₈₆Rn) = 220.01137 u, m (²¹⁶₈₄Po) = 216.00189 u.

a) Q value of emitted α particle = (Total Initial mass-Total final mass) × c²

∴ The α particle decay of ²²⁶Ra₈₈ is given by:

Hence the Q value will be = [226.02540-(222.01750 + 4.002603)] × c²

= 0.005297 u × c²

= 0.005297 × 931.5 Mev

= 4.94MeV

Kinetic energy is given by =

× 4.94 Mev

= 4.85 MeV

b) Similarly, the decay of ²²⁰Rn₈₆ is given by :

Hence, Q value = [220.01137-(216.00189 + 4.0026)]u × c² = 0.00688u × 931.5 MeV = 6.41MeV

Kinetic energy =

= 6.41 × MeV

= 6.293 MeV