Atoms And Nuclei Question 408

Question: The element curium 96Cm248 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 later with a probability of 92 . Each fission releases 200MeV of energy. The masses involved in a decay are as follows: 96Cm248=248.072220u, 94Pu244=244.064100,u and 2He4=4.002603u Calculate the power output from a sample of 1020Cm atoms. (1,u=931MeV/c2).

Options:

A) 1.6×105W

B) 2.6×103W

C) 3.3×105W

D) 5.1×103W

Show Answer

Answer:

Correct Answer: C

Solution:

  • The total energy released E= Energy released in fission process + energy released in α - decay process

=NF×200+Nα×(0.005517×931) =(8100×1020)×200+(92100×1020)

×(0.005517×931)

=20.725×1020MeV Power output P=E/t

=20.725×1020×1.6×10131013 =3.3×105W.



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