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Page No 462:
Question 13.1:
(a) Two stable isotopes
of lithium
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
and
.
Answer:
(a) Mass
of lithium isotope
,
m1
= 6.01512 u
Mass
of lithium isotope
,
m2
= 7.01600 u
Abundance
of
,
η1= 7.5%
Abundance
of
,
η2= 92.5%
The atomic mass of lithium atom is given as:
(b) Mass
of boron isotope
,
m1
= 10.01294 u
Mass
of boron isotope
,
m2
= 11.00931 u
Abundance
of
,
η1
= x%
Abundance
of
,
η2=
(100 − x)%
Atomic mass of boron, m = 10.811 u
The atomic mass of boron atom is given as:
And 100 − x = 80.11%
Hence,
the abundance of
is 19.89% and that of
is
80.11%.
Page No 462:
Question 13.2:
The
three stable isotopes of neon:
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.
Answer:
Atomic
mass of
,
m1=
19.99 u
Abundance
of
,
η1
= 90.51%
Atomic
mass of
,
m2 =
20.99 u
Abundance
of
,
η2
= 0.27%
Atomic
mass of
,
m3 =
21.99 u
Abundance
of
,
η3
= 9.22%
The average atomic mass of neon is given as:
Page No 462:
Question 13.3:
Obtain
the binding energy (in MeV) of a nitrogen nucleus,
given
=14.00307
u
Answer:
Atomic
mass of nitrogen,
m =
14.00307 u
A
nucleus of nitrogen
contains 7 protons and 7 neutrons.
Hence, the mass defect of this nucleus, Δm = 7mH + 7mn − m
Where,
Mass of a proton, mH = 1.007825 u
Mass of a neutron, mn= 1.008665 u
∴Δm = 7 × 1.007825 + 7 × 1.008665 − 14.00307
= 7.054775 + 7.06055 − 14.00307
= 0.11236 u
But 1 u = 931.5 MeV/c2
∴Δm = 0.11236 × 931.5 MeV/c2
Hence, the binding energy of the nucleus is given as:
Eb = Δmc2
Where,
c = Speed of light
∴Eb
= 0.11236 ×
931.5
= 104.66334 MeV
Hence, the binding energy of a nitrogen nucleus is 104.66334 MeV.
Page No 462:
Question 13.4:
Obtain
the binding energy of the nuclei
and
in
units of MeV from the following data:
= 55.934939 u =
208.980388 u
Answer:
Atomic
mass of,
m1
= 55.934939 u
nucleus has 26 protons and (56 − 26) = 30
neutrons
Hence, the mass defect of the nucleus, Δm = 26 × mH + 30 × mn − m1
Where,
Mass of a proton, mH = 1.007825 u
Mass of a neutron, mn = 1.008665 u
∴Δm = 26 × 1.007825 + 30 × 1.008665 − 55.934939
= 26.20345 + 30.25995 − 55.934939
= 0.528461 u
But 1 u = 931.5 MeV/c2
∴Δm = 0.528461 × 931.5 MeV/c2
The binding energy of this nucleus is given as:
Eb1 = Δmc2
Where,
c = Speed of light
∴Eb1
= 0.528461 × 931.5
= 492.26 MeV
Average
binding energy per nucleon
Atomic
mass of,
m2
= 208.980388 u
nucleus has 83 protons and (209 − 83) 126
neutrons.
Hence, the mass defect of this nucleus is given as:
Δm' = 83 × mH + 126 × mn − m2
Where,
Mass of a proton, mH = 1.007825 u
Mass of a neutron, mn = 1.008665 u
∴Δm' = 83 × 1.007825 + 126 × 1.008665 − 208.980388
= 83.649475 + 127.091790 − 208.980388
= 1.760877 u
But 1 u = 931.5 MeV/c2
∴Δm' = 1.760877 × 931.5 MeV/c2
Hence, the binding energy of this nucleus is given as:
Eb2 = Δm'c2
=
1.760877 × 931.5
= 1640.26 MeV
Average
bindingenergy
per nucleon =
Page No 462:
Question 13.5:
A
given coin has a mass of 3.0 g. Calculate the nuclear energy that
would be required to separate all the neutrons and protons from each
other. For simplicity assume that the coin is entirely made of
atoms
(of mass 62.92960 u).
Answer:
Mass of a copper coin, m’ = 3 g
Atomic
mass of
atom, m =
62.92960 u
The
total number of
atoms
in the coin
Where,
NA = Avogadro’s number = 6.023 × 1023 atoms /g
Mass number = 63 g
nucleus
has 29 protons and (63 − 29) 34 neutrons
∴Mass defect of this nucleus, Δm' = 29 × mH + 34 × mn − m
Where,
Mass of a proton, mH = 1.007825 u
Mass of a neutron, mn = 1.008665 u
∴Δm' = 29 × 1.007825 + 34 × 1.008665 − 62.9296
= 0.591935 u
Mass defect of all the atoms present in the coin, Δm = 0.591935 × 2.868 × 1022
= 1.69766958 × 1022 u
But 1 u = 931.5 MeV/c2
∴Δm = 1.69766958 × 1022 × 931.5 MeV/c2
Hence, the binding energy of the nuclei of the coin is given as:
Eb= Δmc2
=
1.69766958 × 1022
× 931.5
= 1.581 × 1025 MeV
But 1 MeV = 1.6 × 10−13 J
Eb = 1.581 × 1025 × 1.6 × 10−13
= 2.5296 × 1012 J
This much energy is required to separate all the neutrons and protons from the given coin.
Page No 462:
Question 13.6:
Write nuclear reaction equations for
(i)
α-decay
of
(ii)
α-decay
of
(iii)
β−-decay
of
(iv)
β−-decay
of
(v)
β+-decay
of
(vi)
β+-decay
of
(vii)
Electron capture of
Answer:
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:
Page No 462:
Question 13.7:
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?
Answer:
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:
Since, λ = 0.693/T
Hence, the isotope will take about 6.645T years to reduce to 1% of its original value.
Page No 462:
Question 13.8:
The
normal activity of living carbon-containing matter is found to be
about 15 decays per minute for every gram of carbon. This activity
arises from the small proportion of radioactive
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,
and the measured activity, the age of the specimen can be
approximately estimated. This is the principle of
dating
used in archaeology. Suppose a specimen from Mohenjodaro gives an
activity of 9 decays per minute per gram of carbon. Estimate the
approximate age of the Indus-Valley civilisation.
Answer:
Decay rate of living carbon-containing matter, R = 15 decay/min
Let N be the number of radioactive atoms present in a normal carbon- containing matter.
Half
life 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.
Page No 463:
Question 13.9:
Obtain
the amount of
necessary
to provide a radioactive source of 8.0 mCi strength. The half-life of
is
5.3 years.
Answer:
The strength of the radioactive source is given as:
Where,
N = Required number of atoms
Half-life
of,
= 5.3 years
= 5.3 × 365 × 24 × 60 × 60
= 1.67 × 108 s
For decay constant λ, we have the rate of decay as:
Where,
λ
For:
Mass of 6.023 × 1023 (Avogadro’s number) atoms = 60 g
∴Mass
of
atoms
Hence,
the amount of
necessary for the purpose is 7.106 × 10−6
g.
Page No 463:
Question 13.10:
The
half-life of
is
28 years. What is the disintegration rate of 15 mg of this isotope?
Answer:
Half
life of
,
=
28 years
= 28 × 365 × 24 × 60 × 60
= 8.83 × 108 s
Mass of the isotope, m = 15 mg
90 g
of
atom
contains 6.023 × 1023
(Avogadro’s number) atoms.
Therefore,
15 mg of
contains:
Rate
of disintegration,
Where,
λ =
Decay constant
Hence,
the disintegration rate of 15 mg of the given isotope is
7.878 ×
1010
atoms/s.
Page No 463:
Question 13.11:
Obtain
approximately the ratio of the nuclear radii of the gold isotope
and the silver isotope.
Answer:
Nuclear
radius of the gold isotope
= RAu
Nuclear
radius of the silver isotope
= RAg
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.
Page No 463:
Question 13.12:
Find
the Q-value and the kinetic energy of the emitted α-particle
in the α-decay of (a)
and
(b).
Given
= 226.02540 u,
= 222.01750 u,
=
220.01137 u, =
216.00189 u.
Answer:
(a) Alpha
particle decay of
emits
a helium nucleus. As a result, its mass number reduces to (226 −
4) 222 and its atomic number reduces to (88 − 2) 86. This is
shown in the following nuclear reaction.
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
(b) Alpha
particle decay of
is shown by the following nuclear reaction.
It is given that:
Mass
of
=
220.01137 u
Mass
of
=
216.00189 u
∴Q-value
=
≈ 641 MeV
Kinetic
energy of the α-particle
= 6.29 MeV
Page No 463:
Question 13.13:
The radionuclide 11C decays according to
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
Answer:
The given nuclear reaction is:
Atomic
mass of
=
11.011434 u
Atomic
mass of
Maximum energy possessed by the emitted positron = 0.960 MeV
The
change in the Q-value
(ΔQ)
of the nuclear masses of the
nucleus is given as:
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 ofand
5 me
in the case of.
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.
Page No 463:
Question 13.14:
The
nucleus
decays
byemission.
Write down the
decay
equation and determine the maximum kinetic energy of the electrons
emitted. Given that:
=
22.994466 u
=
22.989770 u.
Answer:
In
emission, the number of protons increases by 1, and one electron and
an antineutrino are emitted from the parent nucleus.
emission of the nucleus
is given as:
It is given that:
Atomic
mass of
=
22.994466 u
Atomic
mass of
=
22.989770 u
Mass of an electron, me = 0.000548 u
Q-value of the given reaction is given as:
There
are 10 electrons in
and 11 electrons in.
Hence, the mass of the electron is cancelled in the Q-value
equation.
The
daughter nucleus is too heavy as compared to
and
.
Hence, it carries negligible energy. The kinetic energy of the
antineutrino is nearly zero. Hence, the maximum kinetic energy of the
emitted electrons is almost equal to the Q-value,
i.e., 4.374 MeV.
Page No 463:
Question 13.15:
The Q value of a nuclear reaction A + b → C + d is defined by
Q = [ mA+ mb− mC− md]c2 where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic.
(i)
(ii)
Atomic masses are given to be
Answer:
(i) The given nuclear reaction is:
It is given that:
Atomic
mass
Atomic
mass
Atomic
mass
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:
It is given that:
Atomic
mass of
Atomic
mass of
Atomic
mass of
The Q-value of this reaction is given as:
The positive Q-value of the reaction shows that the reaction is exothermic.
Page No 463:
Question 13.16:
Suppose,
we think of fission of a
nucleus
into two equal fragments,.
Is the fission energetically possible? Argue by working out Q
of
the process. Given
and.
Answer:
The
fission of
can
be given as:
It is given that:
Atomic
mass of
= 55.93494 u
Atomic
mass of
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.
Page No 463:
Question 13.17:
The
fission properties of
are
very similar to those of.
The
average energy released per fission is 180 MeV. How much energy, in
MeV, is released if all the atoms in 1 kg of pure
undergo
fission?
Answer:
Average
energy released per fission of,
Amount
of pure,
m = 1 kg = 1000 g
NA= Avogadro number = 6.023 × 1023
Mass
number of=
239 g
1
mole of
contains
NA
atoms.
∴m
g of
contains
∴Total
energy released during the fission of 1 kg ofis
calculated as:
Hence,
is released if all the atoms in 1 kg of pure
undergo
fission.
Page No 464:
Question 13.18:
A
1000 MW fission reactor consumes half of its fuel in 5.00 y. How much
did
it contain initially? Assume that the reactor operates 80% of the
time, that all the energy generated arises from the fission of
and
that this nuclide is consumed only by the fission process.
Answer:
Half
life of the fuel of the fission reactor,
years
= 5 × 365 × 24 × 60 × 60 s
We
know that in the fission of 1 g of
nucleus,
the energy released is equal to 200 MeV.
1
mole, i.e., 235 g of
contains
6.023 ×
1023
atoms.
∴1
g
contains
The
total energy generated per gram ofis
calculated as:
The reactor operates only 80% of the time.
Hence,
the amount of
consumed
in 5 years by the 1000 MW fission reactor is calculated as:
∴Initial
amount of
=
2 ×
1538 = 3076 kg
Page No 464:
Question 13.19:
How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as
Answer:
The given fusion reaction is:
Amount of deuterium, m = 2 kg
1 mole, i.e., 2 g of deuterium contains 6.023 × 1023 atoms.
∴2.0
kg of deuterium contains
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:
Power of the electric lamp, P = 100 W = 100 J/s
Hence, the energy consumed by the lamp per second = 100 J
The total time for which the electric lamp will glow is calculated as:
Page No 464:
Question 13.20:
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.)
Answer:
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:
Where,
= Permittivity of free space
Hence, the height of the potential barrier of the two-deuteron system is
360 keV.
Page No 464:
Question 13.21:
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).
Answer:
We have the expression for nuclear radius as:
R = R0A1/3
Where,
R0 = Constant.
A = Mass number of the nucleus
Nuclear
matter density,
Let m be the average mass of the nucleus.
Hence, mass of the nucleus = mA
Hence, the nuclear matter density is independent of A. It is nearly constant.
Page No 464:
Question 13.22:
For
the
(positron) emission from a nucleus, there is another competing
process known as electron capture (electron from an inner orbit, say,
the K−shell, is captured by the nucleus and a neutrino is
emitted).
Show
that if
emission is energetically allowed, electron capture is necessarily
allowed but not vice−versa.
Answer:
Let the amount of energy released during the electron capture process be Q1. The nuclear reaction can be written as:
Let the amount of energy released during the positron capture process be Q2. The nuclear reaction can be written as:
=
Nuclear mass of
=
Nuclear mass of
=
Atomic mass of
=
Atomic mass of
me = Mass of an electron
c = Speed of light
Q-value of the electron capture reaction is given as:
Q-value of the positron capture reaction is given as:
It can be inferred that if Q2 > 0, then Q1 > 0; Also, if Q1> 0, it does not necessarily mean that Q2 > 0.
In
other words, this means that ifemission
is energetically allowed, then the electron capture process is
necessarily allowed, but not vice-versa. This is because the Q-value
must be positive for an energetically-allowed nuclear reaction.
Page No 464:
Question 13.23:
In
a periodic table the average atomic mass of magnesium is given as
24.312 u. The average value is based on their relative natural
abundance on earth. The three isotopes and their masses are
(23.98504u),
(24.98584u)
and
(25.98259u).
The natural abundance of
is
78.99% by mass. Calculate the abundances of other two isotopes.
Answer:
Average atomic mass of magnesium, m = 24.312 u
Mass
of magnesium isotope,
m1
= 23.98504 u
Mass
of magnesium isotope,
m2
=
24.98584 u
Mass
of magnesium isotope,
m3
=
25.98259 u
Abundance
of,
η1=
78.99%
Abundance
of,
η2
= x%
Hence,
abundance of,
η3
= 100 − x
− 78.99% = (21.01 − x)%
We have the relation for the average atomic mass as:
Hence,
the abundance of
is
9.3% and that of
is
11.71%.
Page No 464:
Question 13.24:
The
neutron separation energy is defined as the energy required to remove
a neutron from the nucleus. Obtain the neutron separation energies of
the nuclei
and
from
the following data:
=
39.962591 u
)
= 40.962278 u
=
25.986895 u
)
= 26.981541 u
Answer:
For
For
A
neutron
is
removed from a
nucleus. The corresponding nuclear reaction can be written as:
It is given that:
Mass
=
39.962591 u
Mass)
= 40.962278 u
Mass
=
1.008665 u
The mass defect of this reaction is given as:
Δm
=
∴Δm = 0.008978 × 931.5 MeV/c2
Hence, the energy required for neutron removal is calculated as:
For,
the neutron removal reaction can be written as:
It is given that:
Mass
= 26.981541 u
Mass
=
25.986895 u
The mass defect of this reaction is given as:
Hence, the energy required for neutron removal is calculated as:
Page No 464:
Question 13.25:
A
source contains two phosphorous radio nuclides
(T1/2
=
14.3d) and
(T1/2
= 25.3d). Initially, 10% of the decays come from.
How long one must wait until 90% do so?
Answer:
Half
life of,
T1/2
= 14.3 days
Half
life of,
T’1/2
= 25.3 days
nucleus decay is 10% of the total amount of decay.
The
source has initially 10% of
nucleus
and 90% of
nucleus.
Suppose
after t
days, the source has 10% of
nucleus
and 90% of
nucleus.
Initially:
Number
of
nucleus
= N
Number
of
nucleus
= 9 N
Finally:
Number
of
Number
of
For
nucleus,
we can write the number ratio as:
For,
we can write the number ratio as:
On dividing equation (1) by equation (2), we get:
Hence,
it will take about 208.5 days for 90% decay of
.
Page No 464:
Question 13.26:
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.
Answer:
Take
a
emission nuclear reaction:
We know that:
Mass
of
m1
= 223.01850 u
Mass
of
m2
= 208.98107 u
Mass
of,
m3
= 14.00324 u
Hence, the Q-value of the reaction is given as:
Q = (m1 − m2 − m3) c2
= (223.01850 − 208.98107 − 14.00324) c2
= (0.03419 c2) u
But 1 u = 931.5 MeV/c2
∴Q = 0.03419 × 931.5
= 31.848 MeV
Hence, the Q-value of the nuclear reaction is 31.848 MeV. Since the value is positive, the reaction is energetically allowed.
Now
take a
emission
nuclear reaction:
We know that:
Mass
of
m1
= 223.01850
Mass
of
m2
= 219.00948
Mass
of,
m3
= 4.00260
Q-value of this nuclear reaction is given as:
Q = (m1 − m2 − m3) c2
= (223.01850 − 219.00948 − 4.00260) C2
= (0.00642 c2) u
= 0.00642 × 931.5 = 5.98 MeV
Hence, the Q value of the second nuclear reaction is 5.98 MeV. Since the value is positive, the reaction is energetically allowed.
Page No 465:
Question 13.27:
Consider the fission 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
Answer:
In the fission of,
10 β− particles decay from the parent
nucleus. The nuclear reaction can be written as:
It is given that:
Mass of a nucleusm1
= 238.05079 u
Mass of a nucleus
m2
= 139.90543 u
Mass of a nucleus,
m3 = 98.90594 u
Mass of a neutronm4
= 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.
Page No 465:
Question 13.28:
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.)
Answer:
(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.
Page No 465:
Question 13.29:
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
Answer:
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
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
Page No 466:
Question 13.30:
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.
Answer:
(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
contains 6.023 × 1023
× 1000 atoms.
Within
the sun, four
nuclei combine and form one
nucleus. In this process 26 MeV of energy is released.
Hence,
the energy released from the fusion of 1
kg
is:
(b) Amount
of
= 1 kg = 1000 g
1
mole, i.e., 235 g of
contains 6.023 × 1023
atoms.
∴1000
g ofcontains
It
is known that the amount of energy released in the fission of one
atom of
is
200 MeV.
Hence,
energy released from the fission of 1 kg of
is:
∴
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.
Page No 466:
Question 13.31:
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.
Answer:
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.
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