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(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 .
(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%.
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.
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:
Obtain the binding energy (in MeV) of a nitrogen nucleus
, given 
=14.00307 u
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.
Obtain the binding energy of the nuclei 
and 
in units of MeV from the following data:
= 55.934939 u 
= 208.980388 u
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 = 
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).
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 × 1023atoms /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.
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 
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:
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?
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.
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.
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.
Obtain the amount of 
necessary to provide a radioactive source of 8.0 mCi strength. The half-life of is 5.3 years.
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.
The half-life of 
is 28 years. What is the disintegration rate of 15 mg of this isotope?
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.
Obtain approximately the ratio of the nuclear radii of the gold isotope 
and the silver isotope
.
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.
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.
(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
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
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 of
and 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.
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.
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.
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
.
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.
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?
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.
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.
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
How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as
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:
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.)
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.
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).
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.
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.
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.
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
(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.
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