CBSE Class 12 Physics 2015 Delhi Set 1 Paper

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Question : 14
Total: 26
In the study of Geiger-Marsdon experiment on scattering of α particles by a thin foil of gold, draw the trajectory of α-particles in the coulomb field of target nucleus. Explain briefly how one gets the information on the size of the nucleus from this study.
From the relation R=R0A13, where R0 is constant and A is the mass number of the nucleus, show that nuclear matter density is independent of A.
OR
Distinguish between nuclear fission and fusion. Show how in both these processes energy is released.
Calculate the energy release in MeV in the deuterium-tritium fusion reaction:
12H+13H24He+n
Using the data:
m(12H)=2.014102u
m(13H)=3.016049u
m(24He)=4.002603u
mn=1.008665u
1u=931.5MeVc2
Solution:  
(i) Drawing of trajectory
(ii) Explanation of information on the size of nucleus
(iii) Proving that nuclear density is independent of A
Only a small fraction of the incident α-particles rebound. This shows that the mass of the atom is concentrated in a small volume in the form of nucleus and gives an idea of the size of nucleus. Radius of nucleus
R=R0A13
Density =
mass
volume

=
mA
4
3
π
R3

where, m : mass of one nucleon
A: Mass number
=
mA
4
3
π
(R0A13)3

=
3m
4πR03

Nuclear matter density is independent of A
OR
Distinction between nuclear fission and nuclear fusion
Showing release of energy in both processes Calculation of release of energy
The breaking of heavy nucleus into smaller fragments is called nuclear fission; the joining of lighter nuclei to form a heavy nucleus is called nuclear fusion.
Binding energy per nucleon, of the daughter nuclei, in both processes, is more than that of the parent nuclei. The difference in binding energy is released in the form of energy. In both processes some mass gets converted into energy.
Alternatively:
In both processes, some mass gets converted into energy.
Energy Released
Q=[m(12H)+m(13H)m(24He)m(n)] ××931.5MeV
=[2.014102+3.0160494.002603
=1.008665]×931.5MeV =0.018883×931.5MeV
=17.59MeV
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