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and (b)
. Given 
= 226.02540 u, 
= 222.01750 u,
= 220.01137 u, 
= 216.00189 u.
Asked by Awdhesh Last Modified
Answer 
Shivam Sharma
(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
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