Question

The element curium ₉₆Cm²⁴⁸ has a mean life of 10¹³ seconds. Its primary decay modes are spontaneous fission and α-decay, the former with a probability of 8% and the latter with probability of 92%. Each fission releases 200 MeV of energy. The masses involved in α-decay are as follows: 

₉₆Cm²⁴⁸ = 248.07220 u

₉₄Pu²⁴⁴ = 244.064100 u

and ₂He⁴ = 4.002603 u

Calculate the power output from a sample of 10²0 Cm atoms. 

(1 AMU = 931 mEv/C²)

Answer 1

α-decay of Cm takes place as follows:

₉₆Cm²⁴⁸ → ₉₄Pu²⁴⁴ + ₂He⁴

Mass defect = Δm

Δm = (M)_cm –[(M)_Pu + M_α]

Δm = (248.07220) – [244.064100 + 4.002603] 

Δm = 0.005517 u

Energy released per α-decay 

= (0.005517) (931) MeV 

= 5.136 MeV 

Probability of spontaneous fission = 8%

Probability of α-decay = 92% 

Energy released in each ₉₆Cm²⁴⁸ transformation

= (0.08 × 200 + 0.92 × 5.136) MeV 

= 20.725 MeV 

Energy released by 10²⁰ atoms = 20.752 X 10²⁰ MeV 

Mean life time = 10¹³ sec