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and
have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium.
(b) Boron has two stable isotopes,
and
. Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of
=
= 6.940934 u.
(b) Mass of boron isotope
x = 19.821/0.99637 = 19.89%
And 100 − x = 80.11%
Hence, the abundance of
,
and
have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.
Solution :
Atomic mass of
= 20.1771u
Question
3. Obtain the binding energy (in MeV) of a nitrogen nucleus
, given
and
in units of MeV from the following data:
m
atoms (of mass 62.92960 u).
Solution :
Mass of a copper coin, m’ = 3 g
Atomic mass of
Where,
NA = Avogadro’s number = 6.023 × 1023 atoms /g
Mass number = 63 g
(ii) α-decay of
(iii) β−-decay of
(iv) β−-decay of
(v) β+-decay of
(vi) β+-decay of
(vii) Electron capture of
Solution :
α is a nucleus of helium
and β is an electron (e− for β− and e+ for β+). In every α-decay, there is a loss of 2 protons and 4 neutrons. In every β+-decay, there is a loss of 1 proton and a neutrino is emitted from the nucleus. In every β−-decay, there is a gain of 1 proton and an antineutrino is emitted from the nucleus.
For the given cases, the various nuclear reactions can be written as:
Question
7. A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value?
Solution :
Half-life of the radioactive isotope = T years
Original amount of the radioactive isotope = N0
(a) After decay, the amount of the radioactive isotope = N
It is given that only 3.125% of N0 remains after decay. Hence, we can write:
Where,
λ = Decay constant
t = Time
Hence, the isotope will take about 5T years to reduce to 3.125% of its original value.
(b) After decay, the amount of the radioactive isotope = N
It is given that only 1% of N0 remains after decay. Hence, we can write:
Hence, the isotope will take about 6.645T years to reduce to 1% of its original value.
present with the stable carbon isotope
. When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life (5730 years) of
= 5730 years
The decay rate of the specimen obtained from the Mohenjodaro site:
R' = 9 decays/min
Let N' be the number of radioactive atoms present in the specimen during the Mohenjodaro period.
Therefore, we can relate the decay constant, λand time, t as:
Hence, the approximate age of the Indus-Valley civilisation is 4223.5 years.
necessary to provide a radioactive source of 8.0 mCi strength. The half-life of
For
is 28 years. What is the disintegration rate of 15 mg of this isotope?
Solution :
Half life of
Rate of disintegration,
Where,
Hence, the disintegration rate of 15 mg of the given isotope is
7.878 × 1010 atoms/s.]
and the silver isotope
.
Solution :
Nuclear radius of the gold isotope
Nuclear radius of the silver isotope
Mass number of gold, AAu = 197
Mass number of silver, AAg = 107
The ratio of the radii of the two nuclei is related with their mass numbers as:
Hence, the ratio of the nuclear radii of the gold and silver isotopes is about 1.23.
Question
12. Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of (a)
and (b)
.
Given m
= 220.01137 u,
= 216.00189 u.
Solution :
(a) Alpha particle decay of
Q-value of
emitted α-particle = (Sum of initial mass − Sum of final mass) c2
Where,
c = Speed of light
It is given that:
Q-value = [226.02540 − (222.01750 + 4.002603)] u c2
= 0.005297 u c2
But 1 u = 931.5 MeV/c2
∴Q = 0.005297 × 931.5 ≈ 4.94 MeV
Kinetic energy of the α-particle =
= (222/226) x 4.94 = 4.85 MeV
(b) Alpha particle decay of
It is given that:
Mass of
The maximum energy of the emitted positron is 0.960 MeV.
Given the mass values:
calculate Q and compare it with the maximum energy of the positron emitted
Solution :
The given nuclear reaction is:
Atomic mass of m (
) = 11.011434 u
Atomic mass of
= 11.009305 u
Maximum energy possessed by the emitted positron = 0.960 MeV
The change in the Q-value (ΔQ) of the nuclear masses of the
....(i)
Where,
me = Mass of an electron or positron = 0.000548 u
c = Speed of light
m’ = Respective nuclear masses
If atomic masses are used instead of nuclear masses, then we have to add 6 me in the case of
11
Cand 5 me in the case of
11
B.
Hence, equation (1) reduces to:
∴ΔQ = [11.011434 − 11.009305 − 2 × 0.000548] c2
= (0.001033 c2) u
But 1 u = 931.5 Mev/c2
∴ΔQ = 0.001033 × 931.5 ≈ 0.962 MeV
The value of Q is almost comparable to the maximum energy of the emitted positron.
Question
14. The nucleus
decays by
emission. Write down the
decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
= 22.994466 u
= 22.989770 u.
Solution :
(ii)
Atomic masses are given to be
Solution :
(i) The given nuclear reaction is:
It is given that:
According to the question, the Q-value of the reaction can be written as:
The negativeQ-value of the reaction shows that the reaction is endothermic.
(ii) The given nuclear reaction is:
The positive Q-value of the reaction shows that the reaction is exothermic.
Question 16:
Solution :
The fission of
can be given as:
It is given that:
The Q-value of this nuclear reaction is given as:
The Q-value of the fission is negative. Therefore, the fission is not possible energetically. For an energetically-possible fission reaction, the Q-value must be positive.
Solution :
Average energy released per fission of
,
Amount of pure
Solution :
Question
19. How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as
Solution :
The given fusion reaction is:
It can be inferred from the given reaction that when two atoms of deuterium fuse, 3.27 MeV energy is released.
∴Total energy per nucleus released in the fusion reaction:
Question
20. Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)
Solution :
When two deuterons collide head-on, the distance between their centres, d is given as:
Radius of 1st deuteron + Radius of 2nd deuteron
Radius of a deuteron nucleus = 2 fm = 2 × 10−15 m
∴d = 2 × 10−15 + 2 × 10−15 = 4 × 10−15 m
Charge on a deuteron nucleus = Charge on an electron = e = 1.6 × 10−19 C
Potential energy of the two-deuteron system:
Question
21. From the relation R = R0A1/3, where R0 is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of A).
Solution :
We have the expression for nuclear radius as:
R = R0A1/3
Where,
R0 = Constant.
A = Mass number of the nucleus
Solution :
Question
23
Solution :
Average atomic mass of magnesium, m = 24.312 u
Solution :
Question
25. A source contains two phosphorous radio nuclides
(T1/2 = 14.3d) and
(T1/2 = 25.3d). Initially, 10% of the decays come from
Question
26. Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle. Consider the following decay processes:
Calculate the Q-values for these decays and determine that both are energetically allowed.
Solution :
Take a
emission nuclear reaction:
m1 = 223.01850 u
Mass of
m2 = 208.98107 u
Mass of
emission nuclear reaction:
.png)
We know that:
Mass of
m2 = 219.00948
Mass of
by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are
and
. Calculate Q for this fission process. The relevant atomic and particle masses are
m
=238.05079 u
m
=139.90543 u
m
= 98.90594 u
Solution :
In the fission of
It is given that:
Mass of a nucleus
m4 = 1.008665 u
Q-value of the above equation,
Where,
m’ = Represents the corresponding atomic masses of the nuclei
= m1 − 92me
= m2 − 58me
= m3 − 44me
= m4
Hence, the Q-value of the fission process is 231.007 MeV.
Question
28. Consider the D−T reaction (deuterium − tritium fusion)
(a) Calculate the energy released in MeV in this reaction from the data:
= 2.014102 u
= 3.016049 u
(b)Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction? (Hint: Kinetic energy required for one fusion event =average thermal kinetic energy available with the interacting particles = 2(3kT/2); k = Boltzman’s constant, T = absolute temperature.)
Solution :
(a) Take the D-T nuclear reaction:
It is given that:
Mass of
, m1= 2.014102 u
Mass of
, m2 = 3.016049 u
Mass of
m3 = 4.002603 u
Mass of
, m4 = 1.008665 u
Q-value of the given D-T reaction is:
Q = [m1 + m2− m3 − m4] c2
= [2.014102 + 3.016049 − 4.002603 − 1.008665] c2
= [0.018883 c2] u
But 1 u = 931.5 MeV/c2
∴Q = 0.018883 × 931.5 = 17.59 MeV
(b) Radius of deuterium and tritium, r ≈ 2.0 fm = 2 × 10−15 m
Distance between the two nuclei at the moment when they touch each other, d = r + r = 4 × 10−15 m
Charge on the deuterium nucleus = e
Charge on the tritium nucleus = e
Hence, the repulsive potential energy between the two nuclei is given as:
Where,
∈0 = Permittivity of free space
Hence, 5.76 × 10−14 J or
of kinetic energy (KE) is needed to overcome the Coulomb repulsion between the two nuclei.
However, it is given that:
KE
Where,
k = Boltzmann constant = 1.38 × 10−23 m2 kg s−2 K−1
T = Temperature required for triggering the reaction
Hence, the gas must be heated to a temperature of 1.39 × 109 K to initiate the reaction.
Question
29. Obtain the maximum kinetic energy of β-particles, and the radiation frequencies of γ decays in the decay scheme shown in Fig. 13.6. You are given that
m (198Au) = 197.968233 u
m (198Hg) =197.966760 u
Solution :
It can be observed from the given γ-decay diagram that γ1 decays from the 1.088 MeV energy level to the 0 MeV energy level.
Hence, the energy corresponding to γ1-decay is given as:
E1 = 1.088 − 0 = 1.088 MeV
hν1= 1.088 × 1.6 × 10−19 × 106 J
Where,
h = Planck’s constant = 6.6 × 10−34 Js
ν1 = Frequency of radiation radiated by γ1-decay
It can be observed from the given γ-decay diagram that γ2 decays from the 0.412 MeV energy level to the 0 MeV energy level.
Hence, the energy corresponding to γ2-decay is given as:
E2 = 0.412 − 0 = 0.412 MeV
hν2= 0.412 × 1.6 × 10−19 × 106 J
Where,
ν2 = Frequency of radiation radiated by γ2-decay
It can be observed from the given γ-decay diagram that γ3 decays from the 1.088 MeV energy level to the 0.412 MeV energy level.
Hence, the energy corresponding to γ3-decay is given as:
E3 = 1.088 − 0.412 = 0.676 MeV
hν3= 0.676 × 10−19 × 106 J
Where,
ν3 = Frequency of radiation radiated by γ3-decay
.png)
Mass of
= 197.968233 u
Mass of
= 197.966760 u
1 u = 931.5 MeV/c2
Energy of the highest level is given as:
β1 decays from the 1.3720995 MeV level to the 1.088 MeV level
∴Maximum kinetic energy of the β1 particle = 1.3720995 − 1.088
= 0.2840995 MeV
β2 decays from the 1.3720995 MeV level to the 0.412 MeV level
∴Maximum kinetic energy of the β2 particle = 1.3720995 − 0.412
= 0.9600995 MeV
Question
30. Calculate and compare the energy released by a) fusion of 1.0 kg of hydrogen deep within Sun and b) the fission of 1.0 kg of 235U in a fission reactor.
Solution :
(a) Amount of hydrogen, m = 1 kg = 1000 g
1 mole, i.e., 1 g of hydrogen
contains 6.023 × 1023 atoms.
∴1000 g of
nucleus. In this process 26 MeV of energy is released.
Hence, the energy released from the fusion of 1 kg
(b) Amount of
= 1 kg = 1000 g
1 mole, i.e., 235 g of
It is known that the amount of energy released in the fission of one atom of
Therefore, the energy released in the fusion of 1 kg of hydrogen is nearly 8 times the energy released in the fission of 1 kg of uranium.
Question
31. Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of 235U to be about 200MeV.
Solution :
Amount of electric power to be generated, P = 2 × 105 MW
10% of this amount has to be obtained from nuclear power plants.
∴Amount of nuclear power,
= 2 × 104 MW
= 2 × 104 × 106 J/s
= 2 × 1010 × 60 × 60 × 24 × 365 J/y
Heat energy released per fission of a 235U nucleus, E = 200 MeV
Efficiency of a reactor = 25%
Hence, the amount of energy converted into the electrical energy per fission is calculated as:
Number of atoms required for fission per year:
1 mole, i.e., 235 g of U235 contains 6.023 × 1023 atoms.
∴Mass of 6.023 × 1023 atoms of U235 = 235 g = 235 × 10−3 kg
∴Mass of 78840 × 1024 atoms of U235
Hence, the mass of uranium needed per year is 3.076 × 104 kg.
