# Find the Q-value

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

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