CBSE Class 12 Physics 2018 Delhi Set 1 Paper

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Question : 19
Total: 26
(a) Explain the processes of nuclear fission and nuclear fusion by using the plot of binding energy per nucleon (BEA) versus the mass number A.
(b) A radioactive isotope has a half-life of 10 years. How long will it take for the activity to reduce to 3.125% ?
Solution:  
(a) Plot of binding energy per nucleon is shown in Figure. From BE/nucleon curve, we note that first B.E. increases rapidly and then decreases slowly and B.E is max i.e. 8.8Mev for 56Fe atom. Again by decreasing slowly B.E. become 8.5Mev for uranium atom 92238U . This shows that nucleus with mass number A<20 are less stable, but some nucleus as 4He,12C,16O (even-even) nuclei are stable. Thus the nuclei with mass number A<20 shows fusion reaction as 2H and 3H have very low BE nucleon in comparison to 4He . Thus when two very light nuclei (A10 say) fuse to form a heavy nucleus, the B.E/A of fused heavier nucleus is more than the B.E/A of lighter nuclei. This implies release of energy in nuclear fusion.
Similarly, due to fission of a very heavy nucleus, the B.E/A of the product as daughter nuclei increases which implies the release of huge amount of energy.
Thus for lighter nuclei nuclear fusion and for heavier nuclei nuclear fission takes place and huge amount of energy is released.

(b) Let the initial activity is R0 and final activity is R then we have
R
R0
=(
1
2
)
tT12

Given R=3.125% ,
R=
3.125
100
Ro
,T
1
2
=10 years.

R=0.03125R0
R
R0
=0.03125
=(
1
2
)
5

(
1
2
)
5
=(
1
2
)
t112

t
T1
=5 or t
=5T
1
2

t=5×10
t=50 years
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