0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

The element curium 21896Cm has a mean-life of 1013s. Its primary decay modes are spontaneous fission and αdecay, the former with a probability of 8% and the latter with a probability of 92%. Each fission releases 200 MeV energy. The masses involved in αdecay are as follows:
21898Cm=248.07220u,24494Pu=244.064100u, and 42He=4.002603u.
Calculate the power output from a sample of 1020Cm atoms. (1u=931MeVc2)

Solution
Verified by Toppr

Activity (or rate) of fission:
A=dNdt=λN=1τN=10201013=107
As probability of fission is 8% only, therefore, actual rate of fission is:
8100×107=8×105s1
Energy released per fission =200MeV
Power output of fission =8×105×200MeVs1=16×107MeVs1

Probability of αparticle decay is 92%.
Therefore, rate of decay of αparticle is: 92100×107=92×105s1
For αdecay, the equation is:
24896Cm(248.072220)24494Pu(244.064100)+42He(4.002603)
Mass of decay products is:
244.064100+4.002603=248.06673.
Mass defect, Δm=24.072220248.066703=0.005517u
Energy released per αdecay is:
92×105×5/1363MeVs1=4.725×107MeVs1
Total power output =16×107+4.725×107
=20.725×107MeVs1
=20.725×107×1.6×1013Js1
=33.16×106W=33.16μW

Was this answer helpful?
1
Similar Questions
Q1
The element curium 21896Cm has a mean-life of 1013s. Its primary decay modes are spontaneous fission and αdecay, the former with a probability of 8% and the latter with a probability of 92%. Each fission releases 200 MeV energy. The masses involved in αdecay are as follows:
21898Cm=248.07220u,24494Pu=244.064100u, and 42He=4.002603u.
Calculate the power output from a sample of 1020Cm atoms. (1u=931MeVc2)
View Solution
Q2
The element curium 96Cm248 has a mean life of 1013 s. Its primary decay modes are spontaneously fission and α decay, the former with a probability of 8% and the latter with a probability of 92%. Each fission releases 200 MeV of energy. The masses involved α decay are as follows:
96Cm248=248.072220 u, 94Pu244=244.064100 u2He4=4.002603 u.
Calculate the power output from a sample of 1020 Cm atoms (1u=931 MeVc2).
View Solution
Q3
U-235 can decay by many ways, let us here consider only two ways, A and B. In decay of U-235 by means of A, the energy released per fission is 210MeV while in B it is 186MeV. Then, the uranium-235 sample is more likely to decay by
View Solution
Q4
A sample Q has half life 20 min. It decays by emitting alpha particle and beta particle with probability of 60% and 40% respectively.
Initial sample of Q contains 1000 nuclei, then number of α particle decay after one hour will be
View Solution
Q5
Mean life of a sample is 200 years for αdecay and 300 years for β decay. If the sample decays by both α and β emission simultaneously, then time for the sample to reduce to half is (ln2=0.693)
View Solution