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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
,
*m*_{1}
= 6.01512 u

Mass
of lithium isotope
,
*m*_{2}
= 7.01600 u

Abundance
of
,
*η*_{1}= 7.5%

Abundance
of
,
*η*_{2}= 92.5%

The atomic mass of lithium atom is given as:

**(b)** Mass
of boron isotope
,
*m*_{1}
= 10.01294 u

Mass
of boron isotope
,
*m*_{2}
= 11.00931 u

Abundance
of
,
*η*_{1}
= *x*%

Abundance
of
,
*η*_{2}=
(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
,
*m*_{1}=
19.99 u

Abundance
of
,
*η*_{1
}= 90.51%

Atomic
mass of
,*
m*_{2 }=
20.99 u

Abundance
of
,
*η*_{2
}= 0.27%

Atomic
mass of
,
*m*_{3 }=
21.99 u

Abundance
of
,
*η*_{3}
= 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, Δ*m
*= 7*m*_{H}
+ 7*m*_{n}
− *m*

Where,

Mass
of a proton, *m*_{H}
= 1.007825 u

Mass
of a neutron, *m*_{n}= 1.008665 u

∴Δ*m*
= 7 × 1.007825 + 7 × 1.008665 − 14.00307

= 7.054775 + 7.06055 − 14.00307

= 0.11236 u

But
1 u = 931.5 MeV/c^{2}

∴Δ*m
*= 0.11236 × 931.5
MeV/*c*^{2}

Hence, the binding energy of the nucleus is given as:

*E*_{b}
= Δ*mc*^{2}

Where,

*c*
= Speed of light

∴*E*_{b
}= 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,
*m*_{1}
= 55.934939 u

nucleus has 26 protons and (56 − 26) = 30 neutrons

Hence,
the mass defect of the nucleus, Δ*m
*= 26 × *m*_{H}
+ 30 × *m*_{n}
− *m*_{1}

Where,

Mass
of a proton, *m*_{H}
= 1.007825 u

Mass
of a neutron, *m*_{n}
= 1.008665 u

∴Δ*m
*= 26 × 1.007825 + 30 ×
1.008665 − 55.934939

= 26.20345 + 30.25995 − 55.934939

= 0.528461 u

But
1 u = 931.5 MeV/c^{2}

∴Δ*m
*= 0.528461 × 931.5
MeV/*c*^{2}

The binding energy of this nucleus is given as:

*E*_{b}_{1}
= Δ*mc*^{2}

Where,

*c*
= Speed of light

∴*E*_{b}_{1}
= 0.528461 × 931.5

= 492.26 MeV

Average binding energy per nucleon

Atomic
mass of,
*m*_{2}
= 208.980388 u

nucleus has 83 protons and (209 − 83) 126 neutrons.

Hence, the mass defect of this nucleus is given as:

Δ*m*'
= 83 × *m*_{H}
+ 126 × *m*_{n}
− *m*_{2}

Where,

Mass
of a proton, *m*_{H}
= 1.007825 u

Mass
of a neutron, *m*_{n}
= 1.008665 u

∴Δ*m*'
= 83 × 1.007825 + 126 × 1.008665 − 208.980388

= 83.649475 + 127.091790 − 208.980388

= 1.760877 u

But
1 u = 931.5 MeV/*c*^{2}

∴Δ*m*'
= 1.760877 × 931.5 MeV/*c*^{2}

Hence, the binding energy of this nucleus is given as:

*E*_{b}_{2}
= Δ*m*'*c*^{2}

= 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,

N_{A}
= Avogadro’s number = 6.023 × 10^{23}_{}atoms /g

Mass number = 63 g

nucleus has 29 protons and (63 − 29) 34 neutrons

∴Mass
defect of this nucleus, Δ*m*'
= 29 × *m*_{H}
+ 34 × *m*_{n}
− *m*

Where,

Mass
of a proton, *m*_{H}
= 1.007825 u

Mass
of a neutron,* m*_{n}
= 1.008665 u

∴Δ*m*'
= 29 × 1.007825 + 34 × 1.008665 − 62.9296

= 0.591935 u

Mass
defect of all the atoms present in the
coin, Δ*m*
= 0.591935 × 2.868 × 10^{22}

=
1.69766958 × 10^{22}
u

But
1 u = 931.5 MeV/*c*^{2}

∴Δ*m
*= 1.69766958 × 10^{22}
× 931.5 MeV/*c*^{2}

Hence, the binding energy of the nuclei of the coin is given as:

*E*_{b}=
Δ*mc*^{2}

=
1.69766958 × 10^{22}
× 931.5

=
1.581 × 10^{25
}MeV

But
1 MeV = 1.6 × 10^{−13}
J

*E*_{b}*
= *1.581 × 10^{25
}× 1.6 × 10^{−13
}

=
2.5296 × 10^{12}
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 = *N*_{0}

**(a)** After
decay, the amount of the radioactive isotope = *N*

It
is given that only 3.125% of *N*_{0
}remains after decay. Hence, we can
write:

Where,

λ = Decay constant

*t*
= Time

Hence,
the isotope will take about 5*T*
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 *N*_{0
}remains after decay. Hence, we can
write:

Since,
*λ*
= 0.693/*T*

Hence,
the isotope will take about 6.645*T*
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, *λ*and 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 × 10^{8}
s

For
decay constant *λ*,
we have the rate of decay as:

Where,
*λ*

For:

Mass
of 6.023 × 10^{23 }(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 × 10^{8}
s

Mass
of the isotope, *m*
= 15 mg

90 g
of
atom
contains 6.023 × 10^{23
}(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 ×
10^{10}
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
= *R*_{Au}

Nuclear
radius of the silver isotope
= *R*_{Ag}

Mass
number of gold, *A*_{Au}
= 197

Mass
number of silver, *A*_{Ag}
= 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) *c*^{2}

Where,

*c*
= Speed of light

It is given that:

*Q*-value
= [226.02540 − (222.01750 + 4.002603)] u *c*^{2}

= 0.005297 u *c*^{2}

But
1 u = 931.5 MeV/*c*^{2}

∴*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 ^{11}C
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
(Δ*Q*)
of the nuclear masses of the
nucleus is given as:

Where,

*m*_{e}
= 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 *m*_{e}
in the case ofand
5 *m*_{e
}in the case of.

Hence, equation (1) reduces to:

∴Δ*Q*
= [11.011434 − 11.009305 − 2 ×
0.000548] *c*^{2}

=
(0.001033 *c*^{2})
u

But
1 u = 931.5 Mev/*c*^{2}

∴Δ*Q*
= 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 thedecay 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, *m*_{e
}= 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
*=
[ *m*_{A}+
*m*_{b}−
*m*_{C}−
*m*_{d}]*c*^{2
}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

N_{A}_{}= Avogadro number = 6.023 ×
10^{23}

Mass number of= 239 g

1
mole of
contains
N_{A}
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 ×
10^{23}
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 ×
10^{23}
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 1^{st}
deuteron + Radius of 2^{nd}
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
*=
*R*_{0}*A*^{1}/^{3},
where *R*_{0}
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
*= *R*_{0}*A*^{1}/^{3}

Where,

*R*_{0}
= 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
*Q*_{1}.
The nuclear reaction can be written as:

Let
the amount of energy released during the positron capture process be
*Q*_{2}.
The nuclear reaction can be written as:

= Nuclear mass of

= Nuclear mass of

= Atomic mass of

= Atomic mass of

*m*_{e}
= 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 *Q*_{2}
> 0, then *Q*_{1
}> 0; Also, if *Q*_{1}>
0, it does not necessarily mean that *Q*_{2}
> 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,
*m*_{1}
= 23.98504 u

Mass
of magnesium isotope,*
m*_{2
}=
24.98584 u

Mass
of magnesium isotope,*
m*_{3
}=
25.98259 u

Abundance
of,
*η*_{1}=
78.99%

Abundance
of,
*η*_{2
}= *x*%

Hence,
abundance of,
*η*_{3
}= 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:

Δ*m*
=

∴Δ*m*
= 0.008978 × 931.5 MeV/*c*^{2}

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
(*T*_{1/2
=}
14.3d) and
(*T*_{1/2}
= 25.3d). Initially, 10% of the decays come from.
How long one must wait until 90% do so?

#### Answer:

Half
life of,
*T*_{1/2
=} 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
*m*_{1}
= 223.01850 u

Mass
of
*m*_{2}
= 208.98107 u

Mass
of,
*m*_{3}
= 14.00324 u

Hence,
the *Q*-value
of the reaction is given as:

*Q*
= (*m*_{1}
− *m*_{2}
− *m*_{3})
*c*^{2}

=
(223.01850 − 208.98107 − 14.00324) *c*^{2}

=
(0.03419 *c*^{2})
u

But
1 u = 931.5 MeV/*c*^{2}

∴*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
*m*_{1}
= 223.01850

Mass
of
*m*_{2}
= 219.00948

Mass
of,
*m*_{3}
= 4.00260

*Q*-value
of this nuclear reaction is given as:

*Q*
= (*m*_{1}
− *m*_{2}
− *m*_{3})
*c*^{2}

=
(223.01850 − 219.00948 − 4.00260) C^{2}

=
(0.00642 *c*^{2})
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 nucleus*m*_{1}
= 238.05079 u

Mass of a nucleus
*m*_{2}
= 139.90543 u

Mass of a nucleus,
*m*_{3} = 98.90594 u

Mass of a neutron*m*_{4}
= 1.008665 u

*Q*-value of the above equation,

Where,

*m*’ = Represents the corresponding atomic masses of
the nuclei

=
*m*_{1} − 92*m*_{e}

=
*m*_{2} − 58*m*_{e}

=
*m*_{3} − 44*m*_{e}

_{}=
*m*_{4}

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(3*kT*/2);
*k *=
Boltzman’s constant, *T *=
absolute temperature.)

#### Answer:

**(a) **Take
the D-T nuclear reaction: **
**

It is given that:

Mass
of,
*m*_{1}=
2.014102 u

Mass
of,*
m*_{2 }=
3.016049 u

Mass
of*
m*_{3 }=
4.002603 u

Mass
of,
*m*_{4 }=
1.008665 u

*Q*-value
of the given D-T reaction is:

*Q*
= [*m*_{1
}+ *m*_{2}− *m*_{3}*
− m*_{4}]
*c*^{2}

= [2.014102 + 3.016049 − 4.002603 −
1.008665] *c*^{2}

= [0.018883 *c*^{2}]
u

But
1 u = 931.5 MeV/*c*^{2}

∴*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}
m^{2} kg
s^{−2}
K^{−1}

T = Temperature required for triggering the reaction

Hence,
the gas must be heated to a temperature of 1.39 × 10^{9}
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
*(^{198}Au)
= 197.968233 u

*m
*(^{198}Hg)
=197.966760 u

#### Answer:

It
can be observed from the given *γ*-decay
diagram that *γ*_{1}
decays from the 1.088 MeV energy level to the 0 MeV energy level.

Hence,
the energy corresponding to *γ*_{1}-decay
is given as:

*E*_{1}
= 1.088 − 0 = 1.088 MeV

*h**ν*_{1}=
1.088 × 1.6 × 10^{−19}
× 10^{6}
J

Where,

*h*
= Planck’s constant = 6.6 × 10^{−34}
Js

ν_{1
}= Frequency of radiation radiated by
*γ*_{1}-decay

It
can be observed from the given *γ*-decay
diagram that *γ*_{2}
decays from the 0.412 MeV energy level to the 0 MeV energy level.

Hence,
the energy corresponding to *γ*_{2}-decay
is given as:

*E*_{2}
= 0.412 − 0 = 0.412 MeV

*h**ν*_{2}=
0.412 × 1.6 × 10^{−19}
× 10^{6}
J

Where,

ν_{2
}= Frequency of radiation radiated by
*γ*_{2}-decay

It
can be observed from the given *γ*-decay
diagram that *γ*_{3}
decays from the 1.088 MeV energy level to the 0.412 MeV energy level.

Hence,
the energy corresponding to *γ*_{3}-decay
is given as:

*E*_{3}
= 1.088 − 0.412 = 0.676 MeV

*h**ν*_{3}=
0.676 × 10^{−19}
× 10^{6}
J

Where,

ν_{3
}= Frequency of radiation radiated by
*γ*_{3}-decay

Mass of = 197.968233 u

Mass of = 197.966760 u

1 u
= 931.5 MeV/*c*^{2}

Energy of the highest level is given as:

β_{1
}decays from the 1.3720995 MeV level
to the 1.088 MeV level

∴Maximum
kinetic energy of the *β*_{1
}particle = 1.3720995 − 1.088

= 0.2840995 MeV

β_{2
}decays from the 1.3720995 MeV level
to the 0.412 MeV level

∴Maximum
kinetic energy of the *β*_{2
}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 ^{235}U
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 × 10^{23}
atoms.

∴1000
g of
contains 6.023 × 10^{23}
× 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 × 10^{23}
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 ^{235}U
to be about 200MeV.

#### Answer:

Amount
of electric power to be generated, *P*
= 2 × 10^{5}
MW

10% of this amount has to be obtained from nuclear power plants.

∴Amount of nuclear power,

=
2 × 10^{4}
MW

=
2 × 10^{4}
× 10^{6}
J/s

=
2 × 10^{10}
× 60 × 60 × 24 × 365 J/y

Heat
energy released per fission of a ^{235}U
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 U^{235}
contains 6.023 × 10^{23}
atoms.

∴Mass
of 6.023 × 10^{23}
atoms of U^{235 }=
235 g = 235 × 10^{−3}
kg

∴Mass
of 78840 × 10^{24}
atoms of U^{235}

Hence,
the mass of uranium needed per year is 3.076 × 10^{4}
kg.

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