NCERT Exercise Q & A

Exercise Section

You may find the following data useful in solving the exercises:
\(
\begin{array}{rlrl}
e & =1.6 \times 10^{-19} \mathrm{C} & N & =6.023 \times 10^{23} \text { per mole } \\
1 /\left(4 \pi \varepsilon_0\right) & =9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 / \mathrm{C}^2 & & k=1.381 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}
\end{array}
\)
\(
\begin{array}{rlrl}
1 \mathrm{~MeV} & =1.6 \times 10^{-13} \mathrm{~J} & & 1 \mathrm{~u}=931.5 \mathrm{MeV} / \mathrm{c}^2 \\
1 \text { year } & =3.154 \times 10^7 \mathrm{~s} & & \\
m_{\mathrm{H}} & =1.007825 \mathrm{u} & & m_{\mathrm{n}}=1.008665 \mathrm{u} \\
m\left({ }_2^4 \mathrm{He}\right) & =4.002603 \mathrm{u} & & m_{\mathrm{e}}=0.000548 \mathrm{u}
\end{array}
\)

Q13.1: Obtain the binding energy (in MeV ) of a nitrogen nucleus \(\left({ }_7^{14} \mathrm{~N}\right)\), given \(m\left({ }_7^{14} \mathrm{~N}\right)=14.00307 \mathrm{~u}\)

Answer: Step 1: Understand the Binding Energy Formula
The binding energy (BE) of a nucleus can be calculated using the mass defect ( \(\Delta M\) ) and the formula:
\(
B E=\Delta M \cdot c^2
\)
where \(c\) is the speed of light. In nuclear physics, we often use the conversion factor:
\(
1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^2
\)
Thus, we can express the binding energy as:
\(
B E=\Delta M \cdot 931.5 \mathrm{MeV}
\)
Step 2: Calculate the Mass Defect ( \(\Delta M\) )
The mass defect is calculated using the formula:
\(
\Delta M=Z \cdot m_H+N \cdot m_n-m_{\text {nucleus }}
\)
where:
\(Z\) is the number of protons (for nitrogen, \(Z=7\) ),
\(N\) is the number of neutrons (for nitrogen, \(N=7\) ),
\(m_H\) is the mass of a proton \((1.007825 \mathrm{u})\),
\(m_n\) is the mass of a neutron ( 1.008665 u ),
\(m_{\text {nucleus }}\) is the mass of the nitrogen nucleus \((14.00307 \mathrm{u})\).
Substituting the values:
\(
\Delta M=7 \cdot 1.007825+7 \cdot 1.008665-14.00307=0.11236 \mathrm{u}
\)
Step 3: Calculate the Binding Energy
Now, substituting \(\Delta M\) into the binding energy formula:
\(
B E=0.11236 \cdot 931.5 \mathrm{MeV}
\)
Calculating this gives:
\(
B E \approx 104.66334 \mathrm{MeV}
\)
The binding energy of the nitrogen nucleus \({ }_7^{14} N\) is approximately:
\(
\text { Binding Energy } \approx 104.66 \mathrm{MeV}
\)

Q13.2: Obtain the binding energy of the nuclei \({ }_{26}^{56} \mathrm{Fe}\) and \({ }_{83}^{209} \mathrm{Bi}\) in units of MeV from the following data:
\(
m\left({ }_{26}^{56} \mathrm{Fe}\right)=55.934939 \mathrm{u} \quad m\left({ }_{83}^{209} \mathrm{Bi}\right)=208.980388 \mathrm{u}
\)

Answer: Atomic mass of \({ }_{26}^{56} \mathrm{Fe}, m_1=55.934939 u\)
\({ }_{26}^{56} \mathrm{Fe}\) nucleus has 26 protons and \((56-26)=30\) neutrons
Hence, the mass defect of the nucleus, \(\Delta m=26 \times m_H+30 \times m_n-m_1\) where,
Mass of a proton, \(m_H=1.007825 u\)
Mass of a neutron, \(m_n=1.008665 u\)
\(
\begin{aligned}
& \therefore \triangle m=26 \times 1.007825+30 \times 1.008665 \\
& -55.934939 \\
& =26.20345+30.25995-55.934939 \\
& =0.528461 u
\end{aligned}
\)
But \(\quad 1 u=931.5 \mathrm{MeV} / C^2\)
The binding energy of this nucleus is given as:
\(
E_{b 1}=\Delta m c^2
\)
Where,
\(c\) = Speed of light
\(
\begin{aligned}
& \therefore E_{b 1}=0.528461 \times 931.5\left(\frac{\mathrm{MeV}}{c^2}\right) \times c^2 \\
& =492.26 \mathrm{MeV}
\end{aligned}
\)
Average binding energy per nucleon \(=\frac{492.26}{56}=8.79 \mathrm{Mev}\)
Atomic mass of \(m\left({ }_{83}^{209} \mathrm{Bi}\right)=208.980388 \mathrm{u}\)
\(\left({ }_{83}^{209} \mathrm{Bi}\right)\) nucleus has 83 protons and \(209-83 =126\) neutrons
Hence, the mass defect of this nucleus is given as:
\(
\Delta m=83 \times m_H+126 \times m_n-m_2
\)
Where,
Mass of a proton, \(m_H=1.007825 u\)
Mass of a neutron, \(m_n=1.008665 u\)
\(
\begin{aligned}
& \triangle m=83 \times 1.007825+126 \times 1.008665 \\
& -208.980388 \\
& =83.649475+127.091790-208.980388 \\
& =1.760877 \quad u
\end{aligned}
\)
\(
\text { But } 1 u=931.5 \mathrm{MeV} / c^2
\)
\(
\therefore \Delta m=1.760877 \times 931.5 \mathrm{MeV} / c^2
\)
Hence, the binding energy of this nucleus is given as:
\(
\begin{aligned}
& E_{b 2}=\Delta m c^2 \\
& =1.760877 \times 931.5\left(\frac{\mathrm{MeV}}{c^2}\right) \times c^2 \\
& =1640.26 \mathrm{Mev}
\end{aligned}
\)
Average bindingenergy per nucleon \(=\frac{1640.26}{209}=7.848 \mathrm{Mev}\)

Q13.3: 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 \({ }_{29}^{63} \mathrm{Cu}\) atoms (of mass 62.92960 u ).

Answer: Mass of a copper coin, \(\mathrm{m}^{\prime}=3 \mathrm{~g}\)
Atomic mass of \({ }_{29}^{63} \mathrm{Cu}\) atom, \(\mathrm{m}=62.92960 \mathrm{u}\)
The total number \({ }_{29}^{63} \mathrm{Cu}\) of atoms in the coin, \(\mathrm{N}=\frac{\mathrm{N}_{\mathrm{A}} \times \mathrm{m}^{\prime}}{\text { Mass number }}\)
Where,
\(
\begin{aligned}
& \mathrm{N}_{\mathrm{A}}=\text { Avogadro’s number }=6.023 \times 10^{23} \text { atoms } / \mathrm{g} \\
& \text { Mass number }=63 \mathrm{~g} \\
& \therefore \mathrm{~N}=\frac{6.023 \times 10^{23} \times 3}{63} \\
& =2.868 \times 10^{22} \text { atoms } / \mathrm{g}
\end{aligned}
\)
\({ }_{29}^{63} \mathrm{Cu}\) nucleus has 29 protons and \((63-29) 34\) neutrons
\(\therefore\) Mass defect of this nucleus, \(\Delta \mathrm{m}^{\prime}=29 \times \mathrm{m}_{\mathrm{H}}+34 \times \mathrm{m}_{\mathrm{n}}-\mathrm{m}\)
Where,
Mass of a proton, \(\mathrm{m}_{\mathrm{H}}=1.007825 \mathrm{u}\)
Mass of a neutron, \(\mathrm{m}_{\mathrm{n}}=1.008665 \mathrm{u}\)
\(
\begin{aligned}
& \therefore \Delta \mathrm{m}^{\prime}=29 \times 1.007825+34 \times 1.008665-62.9296 \\
& =0.591935 \mathrm{u}
\end{aligned}
\)
Mass defect of all the atoms present in the coin, \(\Delta \mathrm{m}=0.591935 \times 2.868 \times 10^{22}\)
\(
=1.69766958 \times 10^{22} \mathrm{u}
\)
Hence, the binding energy of the nuclei of the coin is given as:
\(
\begin{aligned}
& \mathrm{E}_{\mathrm{b}}=\Delta \mathrm{mc}^2 \\
& =1.69766958 \times 10^{22} \times 931.5\left(\frac{\mathrm{MeV}}{\mathrm{c}^2}\right) \times \mathrm{c}^2 \\
& =1.581 \times 10^{25} \mathrm{MeV}
\end{aligned}
\)
But \(1 \mathrm{MeV}=1.6 \times 10^{-13} \mathrm{~J}\)
\(
\begin{aligned}
& E_b=1.581 \times 10^{25} \times 1.6 \times 10^{-13} \\
& =2.5296 \times 10^{12} \mathrm{~J}
\end{aligned}
\)
This much energy is required to separate all the neutrons and protons from the given coin.

Q13.4: Obtain approximately the ratio of the nuclear radii of the gold isotope \({ }_{79}^{197} \mathrm{Au}\) and the silver isotope \({ }_{47}^{107} \mathrm{Ag}\).

Answer: The nuclear radius of the gold isotope
\(
{ }_{79} \mathrm{Au}^{197}=\mathrm{R}_{\mathrm{Au}}
\)
The nuclear radius of the silver isotope
\(
{ }_{47} \mathrm{Ag}^{107}=\mathrm{R}_{\mathrm{Ag}}
\)
Mass number of gold, \(A_{A u}=197\)
Mass number of silver, \(A_{A g}=107\)
The ratio of the radii of the two nuclei is related to their mass numbers as:
\(
\begin{aligned}
\frac{R_{\text {Au }}}{R_{\text {Ag }}} & =\left(\frac{R_{\text {Au }}}{R_{\text {Ag }}}\right)^{\frac{1}{3}} \\
& =\left(\frac{197}{107}\right)^{\frac{1}{3}}=1.2256
\end{aligned}
\)
Hence, the ratio of the nuclear radii of the gold and silver isotopes is about 1.23.

Q13.5: The \(Q\) value of a nuclear reaction \(A+b \rightarrow C+d\) is defined by \(Q=\left[m_A+m_b-m_C-m_d\right] c^2\)
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) \({ }_1^1 \mathrm{H}+{ }_1^3 \mathrm{H} \rightarrow{ }_1^2 \mathrm{H}+{ }_1^2 \mathrm{H}\)
(ii) \({ }_6^{12} \mathrm{C}+{ }_6^{12} \mathrm{C} \rightarrow{ }_{10}^{20} \mathrm{Ne}+{ }_2^4 \mathrm{He}\)
\(
\begin{aligned}
&\text { Atomic masses are given to be }\\
&\begin{aligned}
& m\left({ }_1^2 \mathrm{H}\right)=2.014102 \mathrm{u} \\
& m\left({ }_1^3 \mathrm{H}\right)=3.016049 \mathrm{u} \\
& m\left({ }_6^{12} \mathrm{C}\right)=12.000000 \mathrm{u} \\
& m\left({ }_{10}^{20} \mathrm{Ne}\right)=19.992439 \mathrm{u}
\end{aligned}
\end{aligned}
\)

Answer: (i) The given nuclear reaction is:
\(
{ }_1^1 \mathrm{H}+{ }_1^3 \mathrm{H} \rightarrow{ }_1^2 \mathrm{H}+{ }_1^2 \mathrm{H}
\)
It is given that:
Atomic mass \(\quad m\left({ }_1^1 \mathrm{H}\right)=1.007825 \mathrm{u}\)
Atomic mass \(\quad m\left({ }_1^3 \mathrm{H}\right)=3.016049 \mathrm{u}\)
Atomic mass \(\quad m\left({ }_1^2 \mathrm{H}\right)=2.014102 \mathrm{u}\)
According to the question, the Q -value of the reaction can be written as:
\(
\begin{aligned}
Q & =\left[m\left({ }_1^1 \mathrm{H}\right)+m\left({ }_1^3 \mathrm{H}\right)-2 m\left({ }_1^2 \mathrm{H}\right)\right] c^2 \\
& =[1.007825+3.016049-2 \times 2.014102] c^2 \\
Q & =\left(-0.00433 c^2\right) \mathrm{u}
\end{aligned}
\)
But \(1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^2\)
\(
\therefore Q=-0.00433 \times 931.5=-4.0334 \mathrm{MeV}
\)
The negative Q -value of the reaction shows that the reaction is endothermic.
(ii) The given nuclear reaction is:
\(
{ }_6^{12} \mathrm{C}+{ }_6^{12} \mathrm{C} \rightarrow{ }_{10}^{20} \mathrm{Ne}+{ }_2^4 \mathrm{He}
\)
It is given that:
Atomic mass of \(\quad m\left({ }_6^{12} \mathrm{C}\right)=12.0 \mathrm{u}\)
Atomic mass of \(\quad m\left({ }_{10}^{20} \mathrm{Ne}\right)=19.992439 \mathrm{u}\)
Atomic mass of \(\quad m\left({ }_2^4 \mathrm{He}\right)=4.002603 \mathrm{u}\)
The Q-value of this reaction is given as:
\(
\begin{aligned}
&\begin{aligned}
Q & =\left[2 m\left({ }_6^{12} \mathrm{C}\right)-m\left({ }_{10}^{20} \mathrm{Ne}\right)-m\left({ }_2^4 \mathrm{He}\right)\right] c^2 \\
& =[2 \times 12.0-19.992439-4.002603] c^2 \\
& =\left(0.004958 c^2\right) \mathrm{u} \\
& =0.004958 \times 931.5=4.618377 \mathrm{MeV}
\end{aligned}\\
&\text { The positive Q-value of the reaction shows that the reaction is exothermic. }
\end{aligned}
\)

Q13.6: Suppose, we think of fission of a \({ }_{26}^{56} \mathrm{Fe}\) nucleus into two equal fragments, \({ }_{13}^{28} \mathrm{Al}\). Is the fission energetically possible? Argue by working out \(Q\) of the process. Given \(m\left({ }_{26}^{56} \mathrm{Fe}\right)=55.93494 \mathrm{u}\) and \(m\left({ }_{13}^{28} \mathrm{Al}\right)=27.98191 \mathrm{u}\).

Answer: The fission reaction can be represented as:
\(
{ }_{26}^{56} \mathrm{Fe} \rightarrow 2 \times{ }_{13}^{28} \mathrm{Al}
\)
It is given that:
Atomic mass of \(\quad m\left({ }_{26}^{56} \mathrm{Fe}\right)=55.93494 \mathrm{u}\)
Atomic mass of \(\quad m\left({ }_{13}^{28} \mathrm{Al}\right)=27.98191 \mathrm{u}\)
The Q-value of this nuclear reaction is given as:
\(
\begin{aligned}
Q & =\left[m\left({ }_{26}^{56} \mathrm{Fe}\right)-2 m\left({ }_{13}^{28} \mathrm{Al}\right)\right] c^2 \\
& =[55.93494-2 \times 27.98191] c^2 \\
& =\left(-0.02888 c^2\right) \mathrm{u}
\end{aligned}
\)
But \(1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^2\)
\(
\therefore Q=-0.02888 \times 931.5=-26.902 \mathrm{MeV}
\)
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.

Q13.7: The fission properties of \({ }_{94}^{239} \mathrm{Pu}\) are very similar to those of \({ }_{92}^{235} \mathrm{U}\). 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 \({ }_{94}^{239} \mathrm{Pu}\) undergo fission?

Answer: Average energy released per fission of \({ }_{94}^{239} \mathrm{Pu}, \mathrm{E}_{\mathrm{av}}=180 \mathrm{MeV}\)
Amount of pure \({ }_{94} \mathrm{Pu}^{239} \mathrm{~m}=1 \mathrm{~kg}=1000 \mathrm{~g}\)
\(\mathrm{N}_{\mathrm{A}}=\) Avogadro number \(=6.023 \times 10^{23}\)
Mass number of \({ }_{94}^{239} \mathrm{Pu}=239 \mathrm{~g}\)
1 mole of \({ }_{94} \mathrm{Pu}^{239}\) contains \(\mathrm{N}_{\mathrm{A}}\) atoms.
\(\therefore\) mg of \({ }_{94} \mathrm{Pu}^{239}\) contain \(\left(\frac{\mathrm{N}_{\mathrm{A}}}{\text { Mass number }} \times m\right)\) atom
\(
=\frac{6.023 \times 10^{23}}{239} \times 1000=2.52 \times 10^{24} \text { atoms }
\)
\(\therefore\) Total energy released during the fission of 1 kg of \({ }_{94}^{239} \mathrm{Pu}\) is calculated as:
\(
\begin{aligned}
& \mathrm{E}=\mathrm{E}_{\mathrm{av}} \times 2.52 \times 10^{24} \\
& =180 \times 2.52 \times 10^{24}=4.536 \times 10^{26} \mathrm{MeV}
\end{aligned}
\)
Hence, \(4.536 \times 10^{26}\) is released if all the atoms in 1 kg of pure \({ }_{94} \mathrm{Pu}^{239}\) undergo fission.

Q13.8: How long can an electric lamp of 100 W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as
\(
{ }_1^2 \mathrm{H}+{ }_1^2 \mathrm{H} \rightarrow{ }_2^3 \mathrm{He}+\mathrm{n}+3.27 \mathrm{MeV}
\)

Answer: The fusion reaction given is:
\(
{ }_1 H^2+{ }_1 H^2 \rightarrow_1 H^3+n+3.17 \mathrm{MeV}
\)
Amount of deuterium, \(m=2 \mathrm{~kg}\)
1 mole, i.e., 2 g of deuterium contains \(6.023 \times 10^{23}\) atoms.
\(\therefore 2.0 \mathrm{~kg}\) of deuterium contains
\(
=\frac{6.023 \times 10^{23}}{2} \times 2000=6.023 \times 10^{26} \text { atoms }
\)
It can be inferred from the given reaction that when two atoms of deuterium fuse, 3.27 MeV energy is released.
\(\therefore\) Total energy per nucleus released in the fusion reaction:
\(
\begin{aligned}
E & =\frac{3.27}{2} \times 6.023 \times 10^{26} \mathrm{MeV} \\
& =\frac{3.27}{2} \times 6.023 \times 10^{26} \times 1.6 \times 10^{-19} \times 10^6 \\
& =1.576 \times 10^{14} \mathrm{~J}
\end{aligned}
\)
Power of the electric lamp, \(\mathrm{P}=100 \mathrm{~W}=100 \mathrm{~J} / \mathrm{s}\)
Hence, the energy consumed by the lamp per second \(=100 \mathrm{~J}\)
The total time for which the electric lamp will glow is calculated as:
\(
\begin{aligned}
& \frac{1.576 \times 10^{14}}{100} \mathrm{~s} \\
& \frac{1.576 \times 10^{14}}{100 \times 60 \times 60 \times 24 \times 365} \approx 4.9 \times 10^4 \text { years }
\end{aligned}
\)

Q13.9: 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 \(1^{\text {st }}\) deuteron + Radius of \(2^{\text {nd }}\) deuteron
Radius of a deuteron nucleus \(=2 \mathrm{fm}=2 \times 10^{-15} \mathrm{~m}\)
\(
\therefore \mathrm{d}=2 \times 10^{-15}+2 \times 10^{-15}=4 \times 10^{-15} \mathrm{~m}
\)
Charge on a deuteron nucleus \(=\) Charge on an electron \(=e=1.6 \times 10^{-19} \mathrm{C}\)
The potential energy of the two-deuteron system:
\(
\mathrm{V}=\frac{\mathrm{e}^2}{4 \pi \epsilon_0 \mathrm{~d}}
\)
Where,
\(\epsilon_0=\) Permittivity of free space
\(
\begin{aligned}
& \frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2} \\
& \therefore \mathrm{~V}=\frac{9 \times 10^9 \times\left(1.6 \times 10^{-19}\right)^2}{4 \times 10^{-15}} \mathrm{~J} \\
& =\frac{9 \times 10^9 \times\left(1.6 \times 10^{-19}\right)^2}{4 \times 10^{-15} \times\left(1.6 \times 10^{-19}\right) \mathrm{eV}} \\
& =360 \mathrm{keV}
\end{aligned}
\)
Hence, the height of the potential barrier of the two-deuteron system is 360 keV.

Q13.10: 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: To show that the nuclear matter density is nearly constant and independent of the mass number \(A\), we start with the given relation:
\(
R=R_0 A^{1 / 3}
\)
where \(R_0\) is a constant and \(A\) is the mass number of the nucleus.
The density \(\rho\) of a nucleus can be defined as the mass per unit volume. The mass of the nucleus is given by its mass number \(A\), and the volume \(V\) of the nucleus can be expressed in terms of its radius \(R\).
The volume \(V\) of a nucleus (assuming it is spherical) is given by the formula:
\(
V=\frac{4}{3} \pi R^3
\)
Substituting the expression for \(R\) into the volume formula, we have:
\(
V=\frac{4}{3} \pi\left(R_0 A^{1 / 3}\right)^3
\)
Now, simplify the expression for the volume:
\(
V=\frac{4}{3} \pi R_0^3 A
\)
Now we can express the density \(\rho\) as:
\(
\rho=\frac{\text { mass }}{\text { volume }}=\frac{A}{V}
\)
Substituting the volume we found:
\(
\rho=\frac{A}{\frac{4}{3} \pi R_0^3 A}
\)
\(
\rho=\frac{3}{4 \pi R_0^3}
\)
The expression \(\rho=\frac{3}{4 \pi R_0^3}\) shows that the density \(\rho\) is a constant value, as it does not depend on \(A\). Therefore, we conclude that the nuclear matter density is nearly constant and independent of the mass number \({A}\).

Exemplar Section

Q13.11: \(\mathrm{H} e_2^3\) and \(\mathrm{H} e_1^3\) nuclei have the same mass number. Do they have the same binding energy?

Answer: \(\mathrm{He}_2^3\) (Helium-3) consists of 2 protons and 1 neutron.
\(\mathrm{He}_1^3\) (Tritium) consists of 1 proton and 2 neutrons.
Binding energy is the energy required to hold the nucleons (protons and neutrons) together in the nucleus. It is influenced by the number of protons and neutrons, as well as the forces acting between them.
Evaluate the Forces:
In \(\mathrm{He}_2^3\), there are 2 protons and 1 neutron. The two protons will experience a repulsive electrostatic force due to their positive charge.
In \(\mathrm{He}_1^3\), there is only 1 proton and 2 neutrons. The absence of a second proton means there is no repulsive force acting on the single proton.
Compare Binding Energies:
The presence of the repulsive force in \(\mathrm{He}_2^3\) means that more energy is required to bind the nucleons together compared to \(\mathrm{He}_1^3\), where the lack of repulsion allows for a more stable configuration.
Therefore, \(\mathrm{He}_2^3\) will have a higher binding energy than \(\mathrm{He}_1^3\).
Since the binding energy depends on the number of protons and the interactions between them, and given that \(\mathrm{He}_2^3\) has a repulsive force due to two protons while \(\mathrm{He}_1^3\) does not, we conclude that \(\mathrm{He}_2^3\) and \(\mathrm{He}_1^3\) do not have the same binding energy.
No, \(\mathrm{He}_2^3\) and \(\mathrm{He}_1^3\) do not have the same binding energy.

Q13.12: Draw a graph showing the variation of decay rate with number of active nuclei.

Answer: According to Rutherford and Soddy’s law for radioactive decay \(=\frac{-d N}{d t}=\lambda N\)
where decay constant ( \(\lambda\) ) is constant for a given radioactive material.
Therefore, the graph between \(N\) and \(\frac{d N}{d t}\) is a straight line as shown in the diagram.

Q13.13: Which sample, A or B shown in Fig. 13.2 has a shorter mean-life?

Answer: Key Concept: Mean (or average) life ( \(\mathbf{r}\) ): The time for which a radioactive material remains active is defined as the mean (average) life of that material.
It is defined as the sum of lives of all atoms divided by the total number of atoms.
i.e., \(r=\frac{\text { Sum of the lives of all the atoms }}{\text { Total number of atoms }}=\frac{1}{\lambda}\)
From \(N=N_0 e^{-\lambda t} \Rightarrow \operatorname{In} \frac{\frac{N}{N_0}}{t}=-\lambda\), the slope of the line shown in the graph, i.e., the magnitude of the inverse of slope of \(\frac{N}{N_0}\) vs \(t\) curve is known as mean life ( \(\tau\) ).
From \(N=N_0 e^{-\lambda t}\), if \(t=\frac{1}{\lambda}=\tau\)
\(
\begin{aligned}
& \Rightarrow N=N_0 e^{-t}=N_0\left(\frac{1}{e}\right)=0.37 \mathrm{~N}_0 \\
& =37 \% \text { of } \mathrm{N}_0
\end{aligned}
\)
i.e., mean life is the time interval in which the number of undecayed atoms (N) becomes \(\frac{1}{e}\) times or 0.37 times or \(37 \%\) of the original number of atoms
It is the time in which the number of decayed atoms \(\left(\mathrm{N}_0-\mathrm{N}\right)\) becomes \(\left(1-\frac{1}{e}\right)\) times or 0.63 times or \(63 \%\) of the original number of atoms.

Solution: From the given figure, we can say that
At \(\quad t=0,\left(\frac{d N}{d t}\right)_A=\left(\frac{d N}{d t}\right)_B\)
As \(\frac{d N}{d t}=-\lambda N\)
\(
\Rightarrow \quad\left(N_0\right)_A=\left(N_0\right)_B
\)
Now considering any instant \(t\) by drawing a line perpendicular to the time axis, we find that \(\left(\frac{d N}{d t}\right)_A>\left(\frac{d N}{d t}\right)_B\)
Hence, \(\lambda_A N_A=\lambda_B N_B\)
\(
\therefore \quad N_A>N_B \Rightarrow \lambda_B>\lambda_A
\)
[Rate of decay of \(B\) is slower]
But average life \(\tau=\frac{1}{\lambda} \Rightarrow \tau_A>\tau_B\)

Q13.14: Which one of the following cannot emit radiation and why? Excited nucleus, excited electron.

Answer: Key Concept: The energy of the internal motion of a nucleus is quantized. A typical nucleus has a set of allowed energy levels, including a ground state (state of lowest energy) and several excited states. Because of the great strength of nuclear interactions, excitation energies of nuclei are typical of the order of 1 MeV, compared with a few eV for atomic energy levels. In ordinary physical and chemical transformations, the nucleus always remains in its ground state. When a nucleus is placed in an excited state, either by bombardment with high-energy particles or by a radioactive transformation, it can decay to the ground state by emission of one or more photons called gamma rays or gamma-ray photons, with typical energies of 10 keV to 5 MeV. This process is called gamma ( \(\gamma\) ) decay.

Excited electrons cannot emit radiation because the energy of electronic energy levels is in the range of eV and not MeV ( mega electron volt), y-radiations have energy of the order of MeV.

SA

Q13.15: Why do stable nuclei never have more protons than neutrons?

Answer:Protons are positively charged and repel one another electrically. This repulsion becomes so great in nuclei with more than 10 protons or so, that an excess of neutrons which produce only attractive forces, is required for stability.

Important point: As you get to heavier elements, with each new proton you add, there is a larger repulsive force. The nuclear force is attractive and stronger than the electrostatic force, but it has a finite range. So you need to add extra neutrons, which do not repel each other, to add extra attractive force. You eventually reach a point where the nucleus is just too big and tends to decay via alpha decay or spontaneous fission. To view this in quantum mechanical terms, the proton potential well is not as deep as the neutron well due to the electrostatic repulsion. [Due to the Pauli exclusion principle, you only get two particles per level (spin up and spin down)]. If one well is filled higher than the other, you tend to get a beta decay to even them out. As the nuclei get larger, the neutron well gels deeper as compared to the proton well and you get more neutrons than protons.

Q13.16: Consider a radioactive nucleus A which decays to a stable nucleus \(C\) through the following sequence:
\(
\mathrm{A} \rightarrow \mathrm{~B} \rightarrow \mathrm{C}
\)
Here B is an intermediate nuclei which is also radioactive. Considering that there are \(N_o\) atoms of A initially, plot the graph showing the variation of number of atoms of \(A\) and \(B\) versus time.

Answer: Consider radioactive nucleus \(A\) have \(N_0\) atoms of \(A\) initially; or at \(t=0, N_A=N_0\) (maximum) whole \(N_B=0\). As time increases, \(N_A\) decreases exponentially, and the number of atoms of \(B\) increases. After some time \(N_B\) becomes maximum. As \(B\) is an intermediate nuclei which is also radioactive, it also start decaying and finally drop to zero exponentially by radioactive decay law. We can represent the situation as shown in the graph.

Q13.17: A piece of wood from the ruins of an ancient building was found to have a \({ }^{14} \mathrm{C}\) activity of 12 disintegrations per minute per gram of its carbon content. The \({ }^{14} \mathrm{C}\) activity of the living wood is 16 disintegrations per minute per gram. How long ago did the tree, from which the wooden sample came, die? Given half-life of \({ }^{14} \mathrm{C}\) is 5760 years.

Answer: It is given that
The \({ }^{14} \mathrm{C}\) activity of a piece of wood from the ruins of an ancient building, \(R=12 \mathrm{dis} / \mathrm{min}\) per g
The \({ }^{14} \mathrm{C}\) activity of the living wood, \(R_0=16 \mathrm{dis} / \mathrm{min}\) per g
Half-life of \({ }^{14} C T_{1 / 2}=5760 \mathrm{yr}\)
Let \(t\) be the span of the tree.
According to radioactive decay law,
\(
R=R_0 e^{-\lambda t} \text { or } \frac{R}{R_0}=e^{-\lambda t} \text { or } e^{\lambda t}=\frac{R_0}{R}
\)
Taking log on both sides,
\(
\begin{aligned}
& \lambda t \log _e e=\log _e \frac{R_0}{R} \Rightarrow \lambda t=\left(\log _{10} \frac{16}{12}\right) \times 2.303 \\
& t=\frac{2.303(\log 4-\log 3)}{\lambda}=\frac{2.303(\log 4-\log 3) T_{1 / 2}}{0.6931} \\
&= \frac{2.303(0.6020-4.771) \times 5760}{0.6931}=2391.20 \mathrm{yr}
\end{aligned}
\)

Q13.18: Are the nucleons fundamental particles, or do they consist of still smaller parts? One way to find out is to probe a nucleon just as Rutherford probed an atom. What should be the kinetic energy of an electron for it to be able to probe a nucleon? Assume the diameter of a nucleon to be approximately \(10^{-15} \mathrm{~m}\).

Answer: A nucleon is one of the particles that make up the atomic nucleus. Each atomic nucleus consists of one or more nucleons, and each atom in turn consists of a cluster of nucleons surrounded by one or more electrons. There are two known kinds of nucleons: neutron and proton. The mass number of a given atomic isotope is identical to its number of nucleons. Thus the term nucleon number may be used in place of the more common terms mass number or atomic mass number.
For resolving two objects separated by distance d, the wavelength A of the proving signal must be less than d. Therefore, to detect separate parts inside a nucleon, the electron must have a wavelength less than \(10^{-15} \mathrm{~m}\).
We know that, \(\lambda=\frac{h}{p}\) and \(\mathrm{KE}=\mathrm{PE} \dots(i)\)
Energy, \(\mathrm{E}=\frac{h c}{\lambda} \dots(ii)\)
From equation (i) and equation (ii),
The kinetic energy of an electron
\(
\begin{aligned}
& \mathrm{KE}=\mathrm{PE}=\frac{h c}{\lambda}-\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{10^{-15} \times 1.6 \times 10^{-19}} \mathrm{eV} \\
& \mathrm{KE}=10^9 \mathrm{eV}
\end{aligned}
\)

Important point: Until the 1960s, nucleons were thought to be elementary particles, each of which would not then have been made up of smaller parts. Now they are known to be composite particles, made of three quarks bound together by the so-called strong interaction. The interaction between two or more nucleons is called inter-nucleon interaction or nuclear force, which is also ultimately caused by the strong interaction. (Before the discovery of quarks, the term “strong interaction” referred to just inter-nucleon interactions.)

Q13.19: A nuclide 1 is said to be the mirror isobar of nuclide 2 if \(Z_1=N_2\) and \(Z_2=N_1\). (a) What nuclide is a mirror isobar of \({ }_{11}^{23} \mathrm{Na}\)? (b) Which nuclide out of the two mirror isobars have greater binding energy and why?

Answer: Mirror nuclei are nuclei where the number of protons of element one \(\left(Z_1\right)\) equals the number of neutrons of element two \(\left(\mathrm{N}_2\right)\), the number of protons of element two \(\left(\mathrm{Z}_2\right)\) equal the number of neutrons in element one \(\left(\mathrm{N}_1\right)\) and the mass number is the same.
Pairs of mirror nuclei have the same spin and parity. If we constrain to an odd number of nucleons (A), then we find mirror nuclei that differ from one another by exchanging a proton with a neutron. Interesting to observe is their binding energy which is mainly due to the strong interaction and also due to Coulomb interaction. Since the strong interaction is invariant to protons and neutrons one can expect these mirror nuclei to have very similar binding energies.
(a) According to the question, a nuclide 1 is said to be the mirror isobar of nuclide 2, if \(Z_1=N_2\) and \(Z_2=N_1\).
Now in \({ }_{11} \mathrm{Na}^{23}, Z_1=11, N_1,=23-11=12\)
\(\therefore\) Mirror isobar of \({ }_{11} \mathrm{Na}^{23}\) is \({ }_{12} \mathrm{Mg}^{23}\), for which \(Z_2=12=N_1\) and \(N_2=23-12\) \(=11=Z_1\)
(b) We know \({ }_{12}^{23} \mathrm{Mg}\) contains even number of protons (12) against \({ }_{11}^{23} \mathrm{Na}\) which has odd number of protons (11), therefore \({ }_{11}^{23} \mathrm{Mg} \mathrm{Mg}\) has greater binding energy than \({ }_{11} \mathrm{Na}^{23}\).

LA

Q13.20: Deuteron is a bound state of a neutron and a proton with a binding energy \(\mathrm{B}=2.2 \mathrm{MeV}\). A \(\gamma\)-ray of energy \(E\) is aimed at a deuteron nucleus to try to break it into a (neutron + proton) such that the \(n\) and \(p\) move in the direction of the incident \(\gamma\)-ray. If \(E=B\), show that this cannot happen. Hence calculate how much bigger than \(B\) must \(E[latex] be for such a process to happen.

Answer: Solution: Given the binding energy of a deuteron, [latex]B=2.2 \mathrm{MeV}\) Let kinetic energy and momentum of neutron and proton be \(K_n, K_p\) and \(p_n, p_p\) respectively.
From the conservation of energy,
\(
E-B=K_n+K_p=\frac{p_n^2}{2 m}+\frac{p_p^2}{2 m} \dots(i)
\)
Now applying conservation of momentum,
\(
p_n+p_p=\frac{E}{C} \dots(ii)
\)
As \(E=B\), Eq. (i) \(p_n^2+p_p^2=0\)
It only happens if \(p_n=p_p\).
So, Eq. (ii) cannot be satisfied and the process cannot take place. Let us take \(E=B+x\), where \(x \ll B\) for the process to take place. Putting the value of \(p_n\) from Eq. (ii) in Eq. (i), we get
\(
2 p_p^2-\left(\frac{2 E}{c}\right) p_p+\left(\frac{E^2}{c^2}-2 m x\right)=0
\)
Solving quadratic equation, we get
\(
p_p=\frac{\frac{2 E}{c}+\sqrt{\frac{4 E^2}{c^2}-8\left(\frac{E^2}{c^2}-2 m x\right)}}{4}
\)
For the real value \(p_p\), the discriminant is positive
\(
\begin{aligned}
& \frac{4 E^2}{c^2}=8\left(\frac{E^2}{c^2}-2 m x\right) \\
& 16 m x=\frac{4 E^2}{c^2} \Rightarrow x=\frac{E^2}{4 m c^2}
\end{aligned}
\)
But \(x \ll B\), hence \(E \cong B\)
\(
\Rightarrow \quad x \approx \frac{B^2}{4 m c^2}
\)

Q13.21: The deuteron is bound by nuclear forces just as the H -atom is made up of \(p[latex] and [latex]e\) bound by electrostatic forces. If we consider the force between neutron and proton in deuteron as given in the form of a Coulomb potential but with an effective charge \(e^{\prime}\) :
\(
F=\frac{1}{4 \pi \varepsilon_0} \frac{e^{\prime 2}}{r}
\)
estimate the value of \(\left(e^{\prime} / e\right)\) given that the binding energy of a deuteron is 2.2 MeV.

Answer: We know the binding energy is H -atom
\(
E=\frac{m e^4}{\pi \varepsilon_0^2 h^2}=13.6 \mathrm{eV} \dots(i)
\)
In our question, it is given that proton and neutron were governed by the same electrostatic force and had charge \(e^{\prime}\) each. Then in the above equation we would need to replace electronic mass \(m\) by the reduced mass \(m^{\prime}\) of proton-neutron and the electronic charge \(e\) by \(e^{\prime}\).
\(
m^{\prime}=\frac{m_p \times m_n}{m_p+m_n} \approx \frac{m_p}{2}=\frac{1836 \mathrm{~m}}{2}=918 \mathrm{~m}
\)
Here, \(m_p\) represents the mass of a neutron/proton. We are given the binding energy of a deuteron is 2.2 MeV.
\(
E=\frac{918 m\left(e^{\prime}\right)^4}{8 \varepsilon_0^2 h^2}=2.2 \mathrm{MeV} \dots(ii)
\)
Dividing Eqs. (ii) and (i), we get
\(
\begin{aligned}
918\left(\frac{e^{\prime}}{e}\right)^4 & =\frac{2.2 \mathrm{MeV}}{13.6 \mathrm{eV}}=\frac{2.2 \times 10^6}{13.6} \\
\left(\frac{e^{\prime}}{e}\right)^4 & =\frac{2.2 \times 10^6}{13.6 \times 918}=176.21 \\
\Rightarrow \quad \frac{e^{\prime}}{e} & =(17621)^{1 / 4}=3.64
\end{aligned}
\)

Q13.22: Before the neutrino hypothesis, the beta decay process was thought to be the transition,
\(
n \rightarrow p+\bar{e}
\)
If this were true, show that if the neutron was at rest, the proton and electron would emerge with fixed energies and calculate them. Experimentally, the electron energy was found to have a large range.

Answer: Before \(\beta\) decay, neutron is at rest. Hence \(E_n=m_n \mathrm{c}^2, p_n=0\)
After \(\beta\) decay, from conservation of momentum:
\(
\mathbf{p}_n=\mathbf{p}_p+\mathbf{p}_e
\)
Or \(\mathbf{p}_p+\mathbf{p}_e=0 \Rightarrow\left|\mathbf{p}_p\right|=\left|\mathbf{p}_e\right|=p\)
\(
\begin{aligned}
&\text { Also, } E_p=\left(m_p^2 c^4+p_p^2 c^2\right)^{\frac{1}{2}} \text {, }\\
&E_e=\left(m_e^2 c^4+p_e^2 c^2\right)^{\frac{1}{2}}=\left(m_e^2 c^4+p_p c^2\right)^{\frac{1}{2}}
\end{aligned}
\)
From the conservation of energy:
\(
\begin{aligned}
& \left(m_p^2 c^4+p^2 c^2\right)^{\frac{1}{2}}+\left(m_e^2 c^4+p^2 c^2\right)^{\frac{1}{2}}=m_n c^2 \\
& m_p c^2 \approx 936 \mathrm{MeV}, m_n c^2 \approx 938 \mathrm{MeV}, m_e c^2=0.51 \mathrm{MeV}
\end{aligned}
\)
Since the energy difference between \(n\) and \(p\) is small, \(p c\) will be small, \(p c \ll m_p c^2\), while \(p c\) may be greater than \(m_e c^2\).
\(
\Rightarrow m_p c^2+\frac{p^2 c^2}{2 m_p^2 c^4} \simeq m_n c^2-p c
\)
To first order \(p c \simeq m_n c^2-m_p c^2=938 \mathrm{MeV}-936 \mathrm{MeV}=2 \mathrm{MeV}\)
This gives the momentum.
Then,
\(
\begin{aligned}
& E_p=\left(m_p^2 c^4+p^2 c^2\right)^{\frac{1}{2}}=\sqrt{936^2+2^2} \simeq 936 \mathrm{MeV} \\
& E_e=\left(m_e^2 c^4+p^2 c^2\right)^{\frac{1}{2}}=\sqrt{(0.51)^2+2^2} \simeq 2.06 \mathrm{MeV}
\end{aligned}
\)

Q13.23: Nuclei with magic no. of proton \(Z=2,8,20,28,50,52\) and magic no. of neutrons \(N=2,8,20,28,50,82\) and 126 are found to be very stable. (i) Verify this by calculating the proton separation energy \(S_p\) for \({ }^{120} \mathrm{Sn}(Z=50)\) and \({ }^{121} \mathrm{Sb}=(Z=51)\).
The proton separation energy for a nuclide is the minimum energy required to separate the least tightly bound proton from a nucleus of that nuclide. It is given by
\(
\mathrm{S}_{\mathrm{p}}=\left(M_{Z-1},{ }_N+M_H-M_{Z, N}\right) c^2 .
\)
\(
\begin{aligned}
&\begin{aligned}
& \text { Given }{ }^{119} \mathrm{In}=118.9058 \mathrm{u},{ }^{120} \mathrm{Sn}=119.902199 \mathrm{u}, \\
& { }^{121} \mathrm{Sb}=120.903824 \mathrm{u},{ }^1 \mathrm{H}=1.0078252 \mathrm{u} .
\end{aligned}\\
&\text { (ii) What does the existence of magic number indicate? }
\end{aligned}
\)

Answer: (i) The proton separation energy is given by
\(
\begin{aligned}
S_{\rho S n} & =\left(M_{119.70}+M_H-M_{120,70}\right) c^2 \\
& =(118.9058+1.0078252-119.902199) c^2 \\
& =0.0114362 c_{\text {CBSELabs.com }}^2
\end{aligned}
\)
Similarly
\(
\begin{aligned}
S_{p S \rho} & =\left(M_{120,70}+M_H-M_{121,70}\right) c^2 \\
& =(119.902199+1.0078252-120.903822) c^2 \\
& =0.0059912 c^2
\end{aligned}
\)
Since, \(S_{p S n}>S_{p S b}, \mathrm{Sn}\) nucleus is more stable than \(S b\) nucleus.
(ii) The existence of magic numbers indicates that the shell structure of nucleus similar to the shell structure of an atom. This also explains the peaks in binding energy/nucleon curve.