1. (JEE Advanced – 2013) The correct statement is :-
a) The nucleus 3Li6 can emit an
alpha particle.
b) The nucleus 84PO210 can emit
a proton.
c) Deuteron and alpha particle can undergo complete
fusion.
d) The nuclei 30Zn70 and 34Se82
can undergo complete fusion.
Answer : C
1H2 + 2He4→3Li6
Sum of the masses of product is less than sum of
masses of reactant for reaction. Δm is positive so reaction is possible.
But in statement A, B and D, Δm is negative so
reaction is not possible.
2. Calculate the energy absorbed by the reaction 2He3
+ 2He4 → 4Be7. The masses of the nuclei are 2He3
= 3.016049u; 2He4 = 4.002604u; 4Be7 =7.016930u.
Solution:
Δm = {(3.016049 + 4.002604) - 7.016930} = 0.001723 u.
(1 amu = 931MeV/ c2)
E = Δm × c2 = 0.001723 u × c2 = 0.001723
(931MeV/ c2) × c2
E = 1.604 MeV
3. How long can an electric lamp of 100W be kept
glowing by fusion of 2 kg of deuterium? Take the fusion reaction as 1H2
+ 1H2 → 2He3 + n + 3.27MeV.
Solution:
2 gm of deuterium contains 6.023×1023
atoms.
2 kg of deuterium contains = (6.023×1023 × 2000)/2
= 6.023×1026 atoms
Two atoms of deuterium release 3.27 MeV.
Therefore 6.023×1026 atoms release energy
= (3.27 × 6.023×1026
×1.6×10-13)/2
= 1.576 ×1014 J
Power of lamp = 100 W = 100 J/s
In one second, the energy consumed by the lamp = 100 J
The total time for which the electric lamp will glow
is given by
Time = (1.576 ×1014 /100) second
Time = (1.576 ×1014 /100×3600×24×365) = 4.9
×104 years
4. Calculate and compare the energy released by a)
fusion of 1.0 kg of hydrogen deep with in Sun and b) the fission of 1.0 kg of U-235
in a fission reactor.
Solution:
a) 1 gm of hydrogen contains 6.023×1023
atoms.
1000 gm of hydrogen contains = (6.023×1023 ×
1000)/2 = 6.023×1026 atoms
Within the Sun, four hydrogen atoms combine and form
one 2He4. In this process 26 MeV of energy is released.
Therefore energy released from the fusion of 1.0 kg of
hydrogen is
E1 = {(6.023×1026 × 26)/4} MeV =
39.1495×1026 MeV
b) 235 gm of U235 contains 6.023×1023
atoms.
1000 gm of U235 contains = (6.023×1023
× 1000)/235
W.K.T the amount of energy released in the fission of
one atom of U235 is 200MeV.
Energy released from the fission of 1.0 kg of U235
is
E2 = {(6.023×1023 × 1000×200)/235}
MeV = 5.106×1026 MeV
(E1/ E2) = (39.1495×1026/5.106×1026)
= 7.67≈ 8
The energy released in the fusion of one kilogram of
hydrogen is nearly 8 times the energy released in the fission of one kilogram
of uranium.
5. a) Calculate the energy released in MeV in deuterium
and tritium reaction 1H2 + 1H3 → 2He4 + 0n1 from the given data:
m1(1H2)=2.014102u; m2(1H3)=3.016049u;
m3(2He4)=4.002603u;
m4(0n1)=1.008665u
b) Consider the radius of both deuterium and tritium
to be approximately 2 fm. What is the K.E needed to overcome the coulomb
repulsion between the two nuclei? To what temperature must the gas be heated to
initiate the reaction?
Solution:
a) Δm = (m1+m2)-(m3+m4)
= 0.018883 u
E = Δm × c2 = 0.018883 u × c2 =
17.59 MeV
b) Radius of D and T = 2 fm = 2 × 10-15 m
Distance between the two nuclei at the moment when
they touch each other.
Distance = r +r = 4 × 10-15 m
Charge of D and T = e
The repulsive P.E between the two nuclei (V) = Kq2/d
V={9×109×(1.6×10-19)2}/{4
× 10-15 } = 5.76×10-14 J
V=(5.76×10-14)/(1.6×10-19) = 3.6
× 105 eV = 360 MeV
Hence 360 MeV of KE is needed to overcome the coulomb
repulsion between the two nuclei. KE required for one fusion event is equal to average
thermal KE available with the interacting particles = 2(3KT/2)
Where K=Boltzmann constant = 1.38×10-23 J/K
T=Temperature required for triggering the reaction
E== 2(3KT/2)
T=(E/3K)={( 5.76×10-14)/(3×1.38×10-23)}=1.39×109
K
Hence the gas must be heated to a temperature of 1.39×109
K to initiate the reaction.
