Modern Physics 5 Question 32

35. The element curium 96248Cm has a mean life of 1013s. Its primary decay modes are spontaneous fission and α-decay, the former with a probability of 8 and the latter with a probability of 92, each fission releases 200MeV of energy. The masses involved in decay are as follows :

(1997,5 M)

96248Cm=248.072220u,

94244Pu=244.064100u and 24He=4.002603u. Calculate the power output from a sample of 1020Cm atoms.

(1u=931MeV/c2)

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Solution:

  1. The reaction involved in α-decay is

96248Cm94244Pu+24He

Mass defect

Δm= mass of 96248Cmmass of 94244Pu mass of 24He

=(248.072220244.0641004.002603)u

=0.005517u

Therefore, energy released in α-decay will be

Eα=(0.005517×931)MeV=5.136MeV

Similarly, Efission =200MeV (given)

Mean life is given as tmean =1013s=1/λ

Disintegration constant λ=1013s1

Rate of decay at the moment when number of nuclei are 1020

=λN=(1013)(1020)=107 disintegration per second 

Of these disintegrations, 8 are in fission and 92 are in α-decay.

Therefore, energy released per second

=(0.08×107×200+0.92×107×5.136)MeV=2.074×108MeV

Power output (in watt)

= energy released per second (J/s)=(2.074×108)(1.6×1013)

Power output =3.32×105W



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