NCERT Exemplar Q & A

Hint

(b)

\(
M=m_{\text {proton }}+m_{\text {eloctrou }}-\frac{E_K}{c^2}
\)
And \(E_k\) is just the binding energy. That’s why we have to subtract off the binding energy. Hence the effective mass of the atom can be written as
\(
M=m_{\text {proton }}+m_{\text {electron }}-\frac{B}{c^2}
\)
Or you could look at the problem the other way round. Start with a hydrogen atom of mass \(m_{\text {proton }}+m_{\text {eloctron. }}\). To split the electron and proton apart we have to add energy. In fact the amount of energy we have to add is just the binding energy, and this adds a mass of \(\frac{B}{c^2}\)
Having split the atom we now just have a separate proton and electron, with a combined mass of \(m_{\text {proton }}+m_{\text {election }}\),
so we have: \(M+\frac{B}{c^2}=m_{\text {proton }}+m_{\text {electron }}\) and a quick rearrangement once again gives us:
\(
M=m_{\text {proton }}+m_{\text {electron }}-\frac{B}{c^2}
\)
From above discussion it is clear that Effective mass of hydrogen atom
\(
M=m_{\text {proton }}+m_{\text {electron }}-\frac{B}{c^2} \text {, where } B \text { is } \mathrm{BE} \text { of hydrogen atom }=13.6 \mathrm{eV} \text {. }
\)

Hint

(b) \(\beta\)-particle carries one unit of negative charge and \(\alpha\)-particle carries 2 units of positive charge and \(\gamma\)-photon carries no charge, therefore electronic energy levels of the atom changes for \(\alpha\) and \(\beta\) decay, but not for \(\gamma\)-decay.

Hint

(a) Let the nucleus be \({ }_Z X^A\). \(\beta^{-}\)decay is represented as: \({ }_z X^A \rightarrow{ }_{z+1} A^Y+{ }_{-1} e^0+\bar{v}+Q_1\)
\(
Q_1=\left[m_n\left({ }_z X^A\right)-m_n\left({ }_{z+1} Y^A\right)-m_e\right] c^2
\)
\(
\begin{aligned}
& =\left[m_n\left({ }_z X^A\right)+Z m_e-m_n\left({ }_{z+1} Y^A\right)-(Z+1) m_e\right] c^2 \\
& =\left[m\left({ }_z X^A\right)-m\left({ }_{z-1} Y^A\right)\right] c^2
\end{aligned}
\)
\(
\Rightarrow \quad Q_1=\left(M_x-M_y\right) c^2
\)
\(
\beta^{+} \text {decay is represented as; } { }_z X^A \rightarrow_{z-1} Y^4+{ }_{+1} e^0+v+Q_2
\)
\(
Q_2=\left[m_n\left({ }_z X^A\right)-m_n\left({ }_{z-1} Y^A\right)-m_e\right] c^2
\)
\(
\begin{aligned}
& =\left[m_n\left({ }_z X^A\right)+Z m_e-m_n\left({ }_{z-1} Y^A\right)-(Z-1) m_e-2 m_e\right] c^2 \\
& =\left[m\left({ }_z X^A\right)-m\left({ }_{z-1} Y^A\right)-2 m_e\right] c^2
\end{aligned}
\)
\(
\Rightarrow \quad Q_2=\left(M_x-M_y-2 m_e\right) c^2
\)

Hint

(b) Electrostatic force between protons are repulsive.

Explanation:
In a nucleus, protons are positively charged and repel each other due to the electrostatic force. This repulsive force can destabilize the nucleus. Neutrons, on the other hand, have no charge and can act as a buffer between protons, mitigating the repulsive forces and stabilizing the nucleus. Therefore, heavier nuclei, with a larger number of protons, need more neutrons to maintain stability.

Hint

(b) Key concept: A moderator is a material used in a nuclear reactor to slow down the neutrons produced from fission. By slowing the neutrons down the probability of a neutron interacting with Uranium-235 nuclei is greatly increased thereby maintaining the chain reaction. Moderators are made from materials with light nuclei which do not absorb the neutrons but rather slow them down by a series of collisions.
The moderator only slows neutrons down in order to increase the interaction with Uranium nuclei. They do not give any protection if the reaction goes out of control. 1 fa chain reaction is heading out of control the reactors need to be able to reduce the concentration of neutrons. For this, the reactor uses control rods. Control rods are matte from material with the ability to absorb neutrons. Cadmium and Boron are examples of suitable materials. By inserting control rods between the fuel rods the chain reaction can be slowed dowp-or shut down. Withdrawing the control rods can restart or speed up the reaction. In our given question, the moderator used have light nuclei (like proton). When protons undergo perfectly elastic collision with the neutron emitted their velocities are exchanged, i.e., neutrons come to rest and protons move with the velocity of neutrons.
Heavy nuclei will not serve the purpose because elastic collisions of neutrons with heavy nuclei will not slow them down.

Hint

(a, b) Fusion processes are impossible at ordinary temperatures and pressures. The reason is that nuclei are positively charged and nuclear forces are short range strongest forces. In order to force two hydrogen nuclei together, we need to have a very high pressure, or a very high temperature, or both. A high pressure helps because it causes all the hydrogen nuclei in the sun to squeeze into a smaller space. Then there is more chance of one hydrogen bumping into another. A high temperature helps because it makes the hydrogen nuclei move faster. They need this extra speed so that they can get close together and join. It is as if the nucleus has to break through a barrier, and so the faster it is moving, the greater chance it has. So, at the “normal” temperature and pressure on earth, a hydrogen nucleus has basically no chance of ever joining with another hydrogen nucleus.
Important point: We know that in the middle of the sun, where the temperature is about 16 million degrees, and the pressure is 250 billion atmospheres, hydrogen nuclei will sometimes have enough energy to join together. (An atmosphere is the “normal”, pressure of the air here on earth. A pressure of 250 billion atmospheres is like having a large mountain piled on top of you!)

Hint

(a) We have,
\(
R=R_0 A^{1 / 3}=(1 \cdot 1 \mathrm{fm})(70)^{1 / 3}
\)
\(
=(1.1 \mathrm{fm})(4.12)=4.53 \mathrm{fm} .
\)

Hint

(c) The alpha particle contains 2 protons and 2 neutrons. The binding energy is
\(
\begin{aligned}
B & =(2 \times 1.007825 \mathrm{u}+2 \times 1.008665 \mathrm{u}-4.00260 \mathrm{u}) c^2 \\
& =(0.03038 \mathrm{u}) c^2 \\
& =0.03038 \times 931 \mathrm{MeV}=28.3 \mathrm{MeV}
\end{aligned}
\)

Note:

\(
B=\left(Z m_p+N m_n-M\right) c^2
\)

We can also use the atomic masses in place of nuclear masses. The above equation then becomes
\(
B=\left[Z m\left({ }_1^1 \mathrm{H}\right)+N m_n-m\left({ }_Z^{Z+N} \mathrm{X}\right)\right] c^2
\)
Here, \(m\left({ }_1^1 \mathrm{H}\right)\) is the mass of a hydrogen atom and \(m\left({ }_Z^{Z+N} \mathrm{X}\right)\) is the mass of an atom with \(Z\) protons and \(N\) neutrons.

Hint

(d) The mass excess of hydrogen is \(931(m-A) \mathrm{MeV}\) \(=931(1 \cdot 00783-1) \mathrm{MeV}=7 \cdot 29 \mathrm{MeV}\).

Explanation:

Consider a nucleus of mass number \(A\). Let the mass of the neutral atom containing this nucleus be \(m\) atomic mass units. Also, let \(A^{\prime}\) represent the mass \(A\) atomic mass units. Thus,
mass of the atom \(=m \mathrm{u}\)
\(
A^{\prime}=A \text { u. }
\)
We define mass excess as
\(
\begin{aligned}
& \text { (mass of the atom } \left.-A^{\prime}\right) c^2 \\
& \quad=(m \mathrm{u}-A \mathrm{u}) c^2 \\
& \quad=(m-A)\left(\frac{931 \mathrm{MeV}}{\mathrm{c}^2}\right) c^2 \\
& \quad=931(m-A) \mathrm{MeV} .
\end{aligned}
\)

Hint

(c) 63.5 g of copper has \(6 \times 10^{23}\) atoms. Thus, the number of atoms in \(1 \mu \mathrm{~g}\) of Cu is
\(
N=\frac{6 \times 10^{23} \times 1 \mu \mathrm{~g}}{63.5 \mathrm{~g}}=9.45 \times 10^{15}
\)
\(
\begin{aligned}
\text { The activity } & =\lambda N \\
& =\left(1.516 \times 10^{-5} \mathrm{~s}^{-1}\right) \times\left(9.45 \times 10^{15}\right) \\
& =1.43 \times 10^{11} \text { disintegrations } \mathrm{s}^{-1} \\
& =\frac{1.43 \times 10^{11}}{3.7 \times 10^{10}} \mathrm{Ci}=3.86 \mathrm{Ci}
\end{aligned}
\)

Hint

(b) We have
\(
\frac{t}{t_{1 / 2}}=\frac{40 \text { hours }}{20 \text { hours }}=2
\)
Thus,
\(
\begin{aligned}
A & =\frac{A_0}{2^{t / t_{1 / 2}}}=\frac{A_0}{2^2}=\frac{A_0}{4} \\
\frac{A}{A_0} & =\frac{1}{4} .
\end{aligned}
\)
So one fourth of the original activity will remain after 40 hours.

Hint

(a) The radius of a \({ }^{12} \mathrm{C}\) nucleus is
\(
\begin{aligned}
R & =R_0 A^{1 / 3} \\
& =(1.1 \mathrm{fm})(12)^{1 / 3}=2.52 \mathrm{fm}
\end{aligned}
\)
The separation between the centres of the nuclei is \(2 R=5.04 \mathrm{fm}\). The potential energy of the pair is
\(
\begin{aligned}
U & =\frac{q_1 q_2}{4 \pi \varepsilon_0 r} \\
& =\left(9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2}\right) \frac{\left(6 \times 1.6 \times 10^{-19} \mathrm{C}\right)^2}{5.04 \times 10^{-15} \mathrm{~m}} \\
& =1.64 \times 10^{-12} \mathrm{~J}=10.2 \mathrm{MeV}
\end{aligned}
\)

Hint

(b) The number of protons in \({ }_{26}^{56} \mathrm{Fe}=26\) and the number of neutrons \(=56-26=30\). The binding energy of \({ }_{26}^{56} \mathrm{Fe}\) is
\(
\begin{aligned}
& =[26 \times 1.00783 \mathrm{u}+30 \times 1.00867 \mathrm{u}-55.9349 \mathrm{u}] c^2 \\
& =(0.52878 \mathrm{u}) c^2 \\
& =(0.52878 \mathrm{u})\left(931 \mathrm{MeV} u^{-1}\right)=492 \mathrm{MeV} .
\end{aligned}
\)

Hint

(d) The kinetic energy available for the beta particle and the antineutrino is
\(
\begin{aligned}
Q & =\left[m\left({ }^{176} \mathrm{Lu}\right)-m\left({ }^{176} \mathrm{Hf}\right)\right] c^2 \\
& =(175 \cdot 942694 \mathrm{u}-175 \cdot 941420 \mathrm{u})\left(931 \mathrm{MeVu}^{-1}\right) \\
& =1 \cdot 182 \mathrm{MeV}
\end{aligned}
\)
This energy is shared by the beta particle and the antineutrino. The maximum kinetic energy of a beta particle in this decay is, therefore, 1.182 MeV when the antineutrino practically does not get any share.

Hint

(a) If the product nucleus \({ }^{198} \mathrm{Hg}\) is formed in its ground state, the kinetic energy available to the electron and the antineutrino is
\(
Q=\left[m\left({ }^{198} \mathrm{Au}\right)-m\left({ }^{198} \mathrm{Hg}\right)\right] c^2 .
\)
As \({ }^{198} \mathrm{Hg}\) * has energy 1.088 MeV more than \({ }^{198} \mathrm{Hg}\) in ground state, the kinetic energy actually available is
\(
\begin{aligned}
Q & =\left[m\left({ }^{198} \mathrm{Au}\right)-m\left({ }^{198} \mathrm{Hg}\right)\right] c^2-1.088 \mathrm{MeV} \\
& \left.=(197.968233 \mathrm{u}-197.966760 \mathrm{u})(931 \mathrm{MeV} \mathrm{u})^{-1}\right) \\
& \quad-1.088 \mathrm{MeV} \\
& =1.3686 \mathrm{MeV}-1.088 \mathrm{MeV}=0.2806 \mathrm{MeV}
\end{aligned}
\)
This is also the maximum possible kinetic energy of the electron emitted.

Hint

(a) The half-life and the decay constant are related as
\(
\begin{aligned}
t_{1 / 2} & =\frac{\ln 2}{\lambda}=\frac{0.693}{\lambda} \\
\lambda & =\frac{0.693}{t_{1 / 2}}=\frac{0.693}{2.7 \text { days }} \\
& =\frac{0.693}{2.7 \times 24 \times 3600 \mathrm{~s}}=2.9 \times 10^{-6} \mathrm{~s}^{-1}
\end{aligned}
\)
(b) The average-life is \(t_{a v}=\frac{1}{\lambda}=3.9\) days.
(c) The activity is \(A=\lambda N\). Now, 198 g of \({ }^{198} \mathrm{Au}\) has \(6 \times 10^{23}\) atoms. The number of atoms in 1.00 mg of \({ }^{198} \mathrm{Au}\) is
\(
N=6 \times 10^{23} \times \frac{1.0 \mathrm{mg}}{198 \mathrm{~g}}=3.03 \times 10^{18}
\)
Thus,
\(
\begin{aligned}
A & =\lambda N \\
& =\left(2.9 \times 10^{-6} \mathrm{~s}^{-1}\right)\left(3.03 \times 10^{18}\right) \\
& =8.8 \times 10^{12} \text { disintegrations } \mathrm{s}^{-1} \\
& =\frac{8.8 \times 10^{12}}{3.7 \times 10^{10}} \mathrm{Ci}=240 \mathrm{Ci}
\end{aligned}
\)

Hint

(d) In one half-life the number of active nuclei reduces to half the original number. Thus, in two halflives the number is reduced to \(\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\) of the original number. The number of remaining active nuclei is, therefore,
\(
\begin{aligned}
& 6.0 \times 10^{18} \times\left(\frac{1}{2}\right) \times\left(\frac{1}{2}\right) \\
& =1.5 \times 10^{18}
\end{aligned}
\)

Hint

(d) We have,
\(
A=A_0 e^{-\lambda t}
\)
\(
\begin{aligned}
500 \mathrm{~s}^{-1} & =\left(600 \mathrm{~s}^{-1}\right) e^{-\lambda t} \\
e^{-\lambda t} & =\frac{5}{6} \\
\lambda t & =\ln (6 / 5) \\
\lambda & =\frac{\ln (6 / 5)}{t}=\frac{\ln (6 / 5)}{40 \mathrm{~min}} .
\end{aligned}
\)
\(
\begin{aligned}
&\text { The half-life is }\\
&\begin{aligned}
t_{1 / 2} & =\frac{\ln 2}{\lambda} \\
& =\frac{\ln 2}{\ln (6 / 5)} \times 40 \mathrm{~min} \\
& =152 \mathrm{~min}
\end{aligned}
\end{aligned}
\)

Hint

(b) The activity of sample \(A\) is 2100 counts per week. After a certain time \(t\), its activity will be reduced to 1400 counts per week. This is because a fraction of the active \({ }^{14} \mathrm{C}\) nuclei will decay in time \(t\). The sample \(B\) must be a time \(t\) older than the sample \(A\).
We have,
\(
\begin{aligned}
A & =A_0 e^{-\lambda t} \\
1400 \mathrm{~s}^{-1} & =2100 \mathrm{~s}^{-1} e^{-\lambda t} \\
e^{-\lambda t} & =\frac{2}{3} \\
t & =\frac{\ln (3 / 2)}{\lambda} \\
& =\frac{\ln (3 / 2)}{0.693} t_{1 / 2} \\
& =\frac{0.4055}{0.693} \times 5730 \mathrm{y}=3352 \mathrm{y} .
\end{aligned}
\)

Hint

(a) The number of parent nuclei decaying in a short time interval \(t\) to \(t+d t\) is \(\lambda_p N_p d t\). This is also the number of daughter nuclei produced in this interval. The number of daughter nuclei decaying during the same time interval is \(\lambda_d N_d d t\). The number of the daughter nuclei will be constant if
\(
\begin{array}{ll}
& \lambda_p N_p d t=\lambda_d N_d d t \\
\text { or, } & \lambda_p N_p=\lambda_d N_d
\end{array}
\)

Hint

(d) The activity of the sample at time \(t\) is given by
\(
A=A_0 e^{-\lambda t}
\)
where \(\lambda\) is the decay constant and \(A_0\) is the activity at time \(t=0\) when the capacitor plates are connected. The charge on the capacitor at time \(t\) is given by
\(
Q=Q_0 e^{-t / C R}
\)
where \(Q_0\) is the charge at \(t=0\) and \(C=100 \mu \mathrm{~F}\) is the capacitance. Thus,
\(
\frac{Q}{A}=\frac{Q_0}{A_0} \frac{e^{-t / C R}}{e^{-\lambda t}}
\)
It is independent of \(t\) if \(\lambda=\frac{1}{C R}\)
\(
R=\frac{1}{\lambda C}=\frac{t_{a v}}{C}=\frac{20 \times 10^{-3} \mathrm{~s}}{100 \times 10^{-6} \mathrm{~F}}=200 \Omega
\)

Hint

(a) The decay constant for the first process is \(\lambda_1=\frac{\ln 2}{t_1}\) and for the second process it is \(\lambda_2=\frac{\ln 2}{t_1}\). The probability that an active nucleus decays by the first process in a time interval \(d t\) is \(\lambda_1 d t\). Similarly, the probability that it decays by the second process is \(\lambda_2 d t\). The probability that it either decays by the first process or by the second process is \(\lambda_1 d t+\lambda_2 d t\). If the effective decay constant is \(\lambda\), this probability is also equal to \(\lambda d t\). Thus,
\(
\begin{aligned}
\lambda d t & =\lambda_1 d t+\lambda_2 d t \\
\lambda & =\lambda_1+\lambda_2
\end{aligned}
\)
\(
\frac{1}{t}=\frac{1}{t_1}+\frac{1}{t_2}
\)

Hint

(c) The mass of a \({ }^{12} \mathrm{C}\) atom is exactly 12 u. The energy released in the reaction \(3\left({ }_2^4 \mathrm{He}\right) \rightarrow{ }_6^{12} \mathrm{C}\) is
\(
\begin{aligned}
& {\left[3 m\left({ }_2^4 \mathrm{He}\right)-m\left({ }_6^{12} \mathrm{C}\right)\right] c^2 } \\
= & {[3 \times 4 \cdot 002603 \mathrm{u}-12 \mathrm{u}]\left(931 \mathrm{MeV} \mathrm{u}{ }^{-1}\right)=7 \cdot 27 \mathrm{MeV} }
\end{aligned}
\)

Note: \(E=M c^2\)

Hint

(a) In nuclear physics, a unit used for measurement of mass is unified atomic mass unit, which is denoted by \(u\).
It is defined such that
\(1 \mathrm{u}=\frac{1}{12} \times(\) Mass of neutral carbon atom in its ground state \()\)
Mass of neutral carbon atom in its ground state \(=12 \times 1 \mathrm{u}=12 \mathrm{u}\)
Thus, the mass of neutral carbon atom in its ground state is exactly 12 u.

Hint

(c) the number of nucleons in the nucleus.
Explanation:
Nucleon: A term that refers to both protons and neutrons within an atom’s nucleus.
Mass number: Represents the total number of protons and neutrons in a nucleus, meaning it is equal to the number of nucleons present.

Why other options are incorrect:
(a) the number of neutrons in the nucleus: While neutrons are part of the mass number, they only represent one component of it.
(b) the number of protons in the nucleus: Similar to option (a), this only represents a part of the nucleus’s composition.
(d) none of them: This is incorrect as the mass number is defined as the total number of nucleons.

Hint

(c) Carbon’s atomic number is 6 , so both \({ }^{12} \mathrm{C}\) and \({ }^{14} \mathrm{C}\) have 6 protons.
In a neutral atom, the number of electrons equals the number of protons, so both have 6 electrons.
Determine the number of neutrons for each isotope.
For \({ }^{12} \mathrm{C}\) : neutrons \(=\) mass number – protons \(=12-6=6\).
For \({ }^{14} \mathrm{C}\) : neutrons \(=\) mass number – protons \(=14-6=8\).
Compare the number of neutrons.
\({ }^{14} \mathrm{C}\) has 8 neutrons, while \({ }^{12} \mathrm{C}\) has 6 neutrons.
The difference is \(8-6=2\) extra neutrons in \({ }^{14} \mathrm{C}\).
\({ }^{14} \mathrm{C}\) atom has two extra neutrons and no extra electrons compared to \({ }^{12} \mathrm{C}\) atom.

Hint

(d) Sometimes more than and sometimes equal to its atomic number.

Explanation:
Atomic number: This represents the number of protons in a nucleus.
Mass number: This represents the total number of protons and neutrons in a nucleus.

Therefore, for an element like hydrogen, which has one proton and no neutrons, the atomic number and mass number are equal. However, for all other elements, the mass number will always be greater than the atomic number because there will always be at least one neutron present in the nucleus.

Hint

(a) The average nuclear radius \((R)\) and the mass number of the element \((A)\) has the following relation :
\(
\begin{aligned}
& \mathrm{R}=\mathrm{R}_0 \mathrm{~A}^{\frac{1}{3}} \\
& \frac{\mathrm{R}}{\mathrm{R}_0}=\mathrm{A}^{\frac{1}{3}} \\
& \operatorname{In}\left(\frac{\mathrm{R}}{\mathrm{R}_0}\right)=\frac{1}{3} \operatorname{In} \mathrm{~A}
\end{aligned}
\)
Therefore, the graph of \(\ln \left(\frac{\mathrm{R}}{\mathrm{R}_0}\right)\) versus \(\ln \mathrm{A}\) is a straight line passing through the origin with slope \(\frac{1}{3}\).

Hint

(b)

\(
F_{\mathrm{pp}}=F_{\mathrm{pn}}=F_{\mathrm{nn}}
\)
Protons and neutrons are present inside the nucleus and they exert strong attractive nuclear force on each other. These forces are equal in magnitude, irrespective of the charge present on the nucleons.
\(
\therefore F_{\mathrm{pp}}=F_{\mathrm{pn}}=F_{\mathrm{nn}}
\)
Here, \(F_{\mathrm{pp}}, F_{\mathrm{pn}} a n d F_{\mathrm{nn}}\) denote the magnitudes of the nuclear force by a proton on a proton, by a proton on a neutron and by a neutron on a neutron, respectively.

Note:
The nuclear force, or strong nuclear force, acts between nucleons (protons and neutrons).
It is responsible for holding the nucleus together.
A key property of the strong nuclear force is its charge independence.
This means the force’s magnitude does not depend on the electric charge of the interacting nucleons.
Since the nuclear force is charge-independent, the force between two protons \(\left(F_{p p}\right)\), a proton and a neutron \(\left(F_{p n}\right)\), and two neutrons \(\left(F_{n n}\right)\) will be equal in magnitude at the same separation distance.
This holds true at 1 fm , which is within the range where the strong nuclear force is dominant.
The magnitudes of the nuclear forces are equal due to the charge independence of the strong nuclear force.

Hint

(d) Protons and neutrons are present inside the nucleus and they exert strong attractive nuclear forces on each other, which are equal in magnitude. Due to their positive charge, protons repel each other. Hence the net attractive force between two protons gets reduced, but the nuclear force is stronger than the electrostatic force at a separation of 1 fm.
\(
\therefore {Fpp}<{Fpn}={Fnn}
\)
Here, \(F_{p p}, F_{p n}\) and \(F_{n n}\) denote the magnitudes of the net force by a proton on a proton, by a proton on a neutron and by a neutron on a neutron, respectively.

Hint

(b) Two protons exert strong attractive nuclear force and repulsive electrostatic force on each other. Nuclear forces are short range forces existing in the range of a few fms. Therefore, at a separation of 10 nm, the electromagnetic force is greater than the nuclear force, i.e. \(F_e \gg F_n\).

Hint

(d) As the mass number \(A\) increases, the binding energy per nucleon in a nucleus varies in a way that depends on the actual value of \(A\).

Explanation: The binding energy per nucleon initially increases with increasing mass number for lighter nuclei, but then it reaches a peak around iron ( \(\mathrm{A}=56\) ) and subsequently decreases for heavier nuclei. This means the relationship between binding energy per nucleon and mass number is not a simple linear increase or decrease, but rather a complex curve with a peak value.

Note: Binding energy per nucleon in a nucleus first increases with increasing mass number (A) and reaches a maximum of 8.7 MeV for \(A(50-80)\). Then, again it slowly starts decreasing with the increase in A and drops to the value of 7.5 MeV.

Hint

(d) Binding energy of a nucleus is the energy required to separate a nucleus completely into its constituent nucleons (protons and neutrons). This is also the energy released when the individual nucleons combine to form a stable nucleus. According to Einstein’s mass-energy equivalence principle ( \(E=m c^2\) ), this energy is equivalent to the mass defect, which is the difference between the sum of the rest masses of the individual nucleons and the actual rest mass of the nucleus. The sum of the rest masses of the nucleons is always greater than the rest mass of the nucleus. Therefore, options a, b, and c are correct descriptions of nuclear binding energy. Option d states that it is the sum of the kinetic energy of all the nucleons in the nucleus. While nucleons within the nucleus do possess kinetic energy due to their confinement, the binding energy is fundamentally related to the potential energy associated with the strong nuclear force holding the nucleons together and the mass defect, not the sum of their kinetic energies.

Hint

(c) The average life is the mean life time for a nuclei to decay.
It is given as \(\tau=\frac{1}{\lambda}=\frac{T_{1 / 2}}{0.693}\)
Here, \(\tau, \lambda\) and \(T_{1 / 2}\)
are the average life, decay constant and half life-time of the active nuclei, respectively. The value of the average lifetime comes to be more than the average lifetime. Therefore, in one average life, more than half the active nuclei decay.

Hint

(d) Photon The atomic number and mass number of a nucleus is defined as the number of protons and the sum of the number of protons and neutrons present in the nucleus, respectively. Since in the decay, neither the atomic number nor the mass number change, it cannot be a beta-decay (release of electron, proton or neutron). Hence, the particle emitted can only be a photon.

Hint

(c) During negative beta decay, a neutron in the nucleus decays emitting an electron.

Explanation:
In negative beta decay, a neutron inside the nucleus transforms into a proton, emitting an electron (also called a beta particle) and an antineutrino. This process changes the atomic number of the element by one, while the mass number remains the same.

Alternate:
Negative beta decay is given as
\(
n \rightarrow p+e^{-}+\bar{\nu}
\)
Neutron decays to produce proton, electron and anti-neutrino.

Hint

(b) From \(R=R_0\left(\frac{1}{2}\right)^n\)
we have, \(1=64\left(\frac{1}{2}\right)^n\)
\(n=6=[latex]
Number of half-lives [latex]t=n \times t_{t_{1 / 2}}=6 \times 2=12 \mathrm{~h}\)

Hint

(b)

\(
\begin{aligned}
&\text { The half-life is given by: }\\
&T_{\frac{1}{2}}=\frac{\ln 2}{\lambda}
\end{aligned}
\)
\(
\begin{aligned}
&\text { The average-life is given by: }\\
&\tau=\frac{1}{\lambda}
\end{aligned}
\)
The half-life is \(\frac{\ln 2}{\lambda}\).
The average-life is \(\frac{1}{\lambda}\).
The half-life and average-life of the sample are respectively \((\ln 2 / \lambda)\) and \(1 / \lambda\).

Hint

(b)

\(
\begin{aligned}
&\text { Write the nuclear reaction equation. }\\
&{ }_7^{14} \mathrm{~N}+{ }_2^4 \mathrm{He} \rightarrow{ }_8^{17} \mathrm{O}+{ }_Z^A \mathrm{X}
\end{aligned}
\)
Sum of mass numbers on the left: \(14+4=18\).
Sum of mass numbers on the right: \(17+A\).
Equating them: \(18=17+A\).
Solving for \(A: A=18-17=1\).
Sum of atomic numbers on the left: \(7+2=9\).
Sum of atomic numbers on the right: \(8+Z\).
Equating them: \(9=8+Z\).
Solving for \(Z: Z=9-8=1\).
\(
\text { The emitted particle is }{ }_1^1 \mathrm{x} \text {. }
\)
\(
\text { A particle with mass number } 1 \text { and atomic number } 1 \text { is a proton }\left({ }_1^1 \mathrm{P}\right) \text {. }
\)
The emitted particle is a proton.

Hint

(a) \({ }^{57} \mathrm{Co}\) is undergoing beta decay, i.e. electron is being produced. But an electron has very less mass \(\left(9.11 \times \times 10^{-31} \mathrm{~kg}\right)\) as compared to the Co atom. Therefore, after 540 days, even though the atoms undergo large beta decay, the weight of the material in the container will be nearly 10 g.

Hint

(c) When the train is stationary, the separation between the alpha particle and recoiling uranium nucleus is \(x\) in time \(t\) after the decay. Even if the decay is taking place in a moving train and the separation is measured by the passenger sitting in it, therefore the separation between the alpha particle and nucleus will be \(x\). This is because the observer is also moving with the same speed with which the alpha particle and recoiling nucleus are moving, i.e. they all are in the same frame that is moving at a uniform speed.

Hint

(c) During a nuclear fission reaction, a heavy nucleus bombarded by thermal neutrons breaks up.

Explanation: Nuclear fission is the process where a heavy nucleus, like uranium-235, splits into two lighter nuclei when bombarded by a thermal neutron, releasing a significant amount of energy.

Hint

(c) The quantity that does not change as the mass number A increases is Density.

Explanation: When a nucleus gains more nucleons (protons and neutrons), its mass and volume both increase proportionally, resulting in a constant density.

Hint

(c, d) This is because in heavy nuclei, the \(\mathrm{N} / \mathrm{Z}\) ratio becomes larger in order to maintain their stability and reduce instability caused due to the repulsion among the protons. The neutrons exert only attractive short-range nuclear forces on each other as well as on the neighbouring protons, whereas the protons exert attractive short-range nuclear forces on each other as well as the electrostatic repulsive force. Thus, the nuclei with high mass number, in order to be stable, have large neutron to proton ratio ( \(\mathrm{N} / \mathrm{Z}\) ).

Hint

(c) The neutron has a larger rest mass than the proton. This mass difference allows for the decay of a neutron into a proton, an electron, and an antineutrino, releasing energy.
While option (a) is incorrect because neutrons are not made of protons and electrons, and
option (d) is not the primary reason, the mass difference is the key factor.
Option (b) is also incorrect, as the charged nature of the proton is not relevant to its decay potential.

Hint

(d) In a beta decay, either a neutron is converted to a proton or a proton is converted to a neutron such that the mass number does not change. Also, the number of the nucleons present in the nucleus remains the same. Thus, the active nucleus gets converted to one of its isobars after beta decay.

Hint

(d) \(\gamma\)-decay.

Explanation: Gamma decay only involves the release of energy in the form of gamma rays, not a change in the number of protons or neutrons in the nucleus, meaning the element remains the same.

Why other options are incorrect:
(a) \(\alpha\)-decay:

Alpha decay involves the emission of an alpha particle (a helium nucleus), which changes the number of protons and neutrons in the nucleus, resulting in a new element.
(b) \(\beta^{+}\)-decay:

Beta positive decay involves the emission of a positron (a positively charged electron), which also changes the number of protons and neutrons, leading to a new element.
(c) \(\beta^{-}\)-decay:

Beta negative decay involves the emission of an electron (a negatively charged particle), which also changes the number of protons and neutrons, resulting in a new element.

Hint

(a, b) In alpha particle decay, the unstable nucleus emits an alpha particle reducing its proton number (atomic number) \(Z\) as well as neutron number \(N\) by 2.
\(
{ }_Z^A X \rightarrow{ }_{Z-2}^{A-4} Y+{ }_2^4 \mathrm{He}
\)
During \(\beta^{-}\)-decay, a neutron is converted to a proton, an electron and an antineutrino. Thus, there is an increase in the atomic number.
\(
{ }_Z^A X \rightarrow{ }_{Z+1}^A Y+e^{-}+\bar{\nu}
\)
During \(\beta^{+}\)-decay, a proton in the nucleus is converted to a neutron, a positron and a neutrino in order to maintain the stability of the nucleus. Thus, there is a decrease in the atomic number.
\(
{ }_Z^A X \rightarrow{ }_{Z-1}^A Y+\beta^{+}+\nu
\)
When a nucleus is in higher excited state or has excess of energy, it comes to the lower state in order to become stable and release energy in the form of electromagnetic radiation called gamma ray. The element in the gamma decay doesn’t change. Therefore, alpha and beta plus decay suffer decrease in atomic number.

Hint

(d) gamma rays.
Explanation: Alpha and beta rays (both plus and minus) are charged particles, and therefore will be deflected by a magnetic field. Gamma rays, however, are electromagnetic radiation (photons) and have no charge. Since they lack charge, they will not interact with a magnetic field and are not deflected.

Hint

(d) Gamma rays.

Explanation: Gamma rays are considered electromagnetic waves because they are energy transmitted in the form of waves, unlike alpha and beta rays which are particles.

Why other options are incorrect:
(a) $\alpha$-rays:

Alpha rays are particles consisting of two protons and two neutrons, not electromagnetic waves.
(b) Beta-plus rays:

Beta-plus rays are high-energy positrons (positively charged electrons), which are particles, not electromagnetic waves.
(c) Beta-minus rays:

Beta-minus rays are high-energy electrons, particles, not electromagnetic waves.

Hint

(d) Coulomb repulsion does not allow the nuclei to come very close is the most accurate description of why two lithium nuclei do not combine to form a carbon nucleus at room temperature. While other factors like the instability of the resulting carbon nucleus and unfavorable energy conditions play a role, it’s the electrostatic repulsion between the positively charged lithium nuclei that prevents them from getting close enough for nuclear fusion to occur.

Here’s why:
Coulomb Repulsion:
Lithium nuclei, like all nuclei, are positively charged. Like charges repel each other, and at room temperature, the kinetic energy of the lithium nuclei is not high enough to overcome this repulsion and allow them to get close enough for the strong nuclear force to bind them together.

Hint

(b) the binding energy per nucleon decreases on an average as A increases
(c) if the nucleus breaks into two roughly equal parts, energy is released

Binding energy per nucleon varies in a way that it depends on the actual value of mass number (A). As the mass number (A) increases, the binding energy also increases and reaches its maximum value of 8.7 MeV for \(\mathrm{A}(50-80)\) and for \(\mathrm{A}>100\). The binding energy per nucleon decreases as A increases and the nucleus breaks into two or more atoms of roughly equal parts so as to attain stability and binding energy of mass number between 50-80.

Hint

(a) Radius of the nucleus, \(R=R_0 A^{1 / 3}\)
\(
\frac{R_{\mathrm{Fe}}}{R_{\mathrm{Al}}}=\left(\frac{A_{\mathrm{Fe}}}{A_{\mathrm{Al}}}\right)^{1 / 3}=\left(\frac{125}{27}\right)^{1 / 3}=\frac{5}{3}
\)
Radius, \(R_{\mathrm{Fe}}=\frac{5}{3} R_{\mathrm{Al}}=\frac{5}{3} \times 6.4 \simeq 10 \mathrm{fm}\)

Hint

(b) Density of nucleus (of water),
\(
\begin{aligned}
\rho & =\frac{3 m}{4 \pi R_0^3}=\frac{3 \times 1.67 \times 10^{-27}}{4 \times \frac{22}{7} \times\left(1.2 \times 10^{-15}\right)^3} \\
& =\frac{7 \times 3 \times 1.67 \times 10^{18}}{88 \times 1.2 \times 1.2 \times 1.2}=2.307 \times 10^{17} \mathrm{~kg} / \mathrm{m}^3
\end{aligned}
\)
Density of water, \(\rho^{\prime}=10^3 \mathrm{~kg} / \mathrm{m}^3\)
\(
\therefore \quad \frac{\rho}{\rho^{\prime}}=\frac{2.307 \times 10^{17}}{10^3}=2.307 \times 10^{14}
\)

Hint

(d)

\(
\begin{aligned}
\text { Atomic weight } & =\text { Average weight of the isotopes } \\
& =\frac{6.01512 \times 7.5+7.01600 \times 92.5}{7.5+92.5} \\
& =\frac{45.1134+648.98}{100}=6.941 \mathrm{u}
\end{aligned}
\)

Hint

(a) Isotones have same number of neutrons,
\(
N=A-Z
\)
(i) \({ }_1 \mathrm{H}^3\) and \({ }_2 \mathrm{He}^4\)
Number of neutrons \(=3-1=1\)
\(
4-2=2
\)
(ii) \({ }_{12} \mathrm{Mg}^{24}\) and \({ }_{11} \mathrm{Na}^{23}\)
Number of neutrons \(=24-12=12\)
\(
23-11=12
\)

Note: Nuclei having same mass number \(A\) but with proton number \((Z)\) and neutron number ( \(A-Z\) ) interchanged are called mirror nuclei. e.g. \({ }_1 H^3\) and \({ }_2 H^3,{ }_3 L^7\) and \({ }_4 B e^7\).

Isobars: The atoms of different elements having the same mass number \((A)\) but different atomic number \((Z)\) are called isobars. They have different places in periodic table. Their chemical as well as physical properties are different.
\({ }_1 \mathrm{H}^3[latex] and [latex]{ }_2 \mathrm{He}^3,{ }_8 \mathrm{O}^{17}\) and \({ }_9 \mathrm{~F}^{17}\) are examples of isobars.

Isotones: The atoms of different elements having equal number of neutrons ( \(A-Z\) ) but different number of protons are called isotones. \({ }_3 \mathrm{Li}^7 \&{ }_4 \mathrm{Be}^8\) and \({ }_1 \mathrm{H}^3 \&{ }_2 \mathrm{He}^4\) are examples of isotones.

Hint

(c) neutron is unstable. Outside a nucleus, a neutron will decay into a proton, an electron, and an antineutrino. Protons, on the other hand, are considered stable outside of nuclei.

Explanation:
Neutron:
When neutrons are not part of an atomic nucleus, they are unstable and undergo beta decay. This means they transform into other particles. In beta decay, a neutron converts into a proton, an electron, and an antineutrino.

Hint

(c) \(\text { Nucleus does not contain electron. }\)

Hint

(b) For stability, in case of lighter nuclei \(\frac{N}{Z}=1\) and for heavier nuclei \(\frac{N}{Z}>1\).

Hint

(c) Helium nucleus \(={ }_2 \mathrm{He}^4\)
Number of protons \(=Z=2\)
Number of neutrons \(=A-Z=4-2=2\)

Hint

(b) \({ }_Z X^A={ }_{88} \mathrm{Ra}^{226}\)
Number of protons, \(Z=88\)
Number of neutrons, \(N=A-Z=226-88=138\)

Hint

(c) For \({ }_6 \mathrm{C}^{12}\)
Proton \(=6\), electron \(=6\) and neutron \(=6\)
For \({ }_6 \mathrm{C}^{14}\)
Proton \(=6\), electron \(=6\) and neutron \(=8\)

Hint

(d) \(R=R_0 A^{1 / 3} \Rightarrow R \propto A^{1 / 3}\)

Hint

(a) Radius of nucleus, \(R=R_0 A^{1 / 3}\).
where, \(R_0\) is a constant and \(A\) is the mass number.
\(
\begin{aligned}
R_{\mathrm{Cu}} & =\left(1.2 \times 10^{-15}\right)(64)^{1 / 3}=\left(1.2 \times 10^{-15}\right) \times\left(4^3\right)^{1 / 3} \\
& =4.8 \times 10^{-15} \mathrm{~m}=4.8 \mathrm{fm}
\end{aligned}
\)

Hint

(a) The masses of three isotopes are \(19.99 \mathrm{u}, 20.99 \mathrm{u}\) and 21.99 u.
Their relative abundances are \(90.51 \%, 0.27 \%[latex] and [latex]9.22 \%\).
\(\therefore\) Average atomic mass of neon,
\(
\begin{aligned}
m & =\frac{90.51 \times 19.99+0.27 \times 20.99+9.22 \times 21.99}{(90.51+0.27+9.22)} \\
& =\frac{2017.7}{100}=20.17 u
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
& \text { Mass defect }=\text { Mass of nucleons }- \text { Mass of nucleus } \\
& \begin{array}{l}
\text { [Here, } A(\text { mass number })=4 \\
\text { and } Z(\text { atomic number })=2 \\
\therefore \quad \text { Number of protons }=2 \\
\text { and number of neutrons }=A-Z=4-2=2] \\
\Delta m=\text { Mass of } 2 \text { protons }+ \text { Mass of } 2 \text { neutrons }- \text { Mass of nucleus } \\
\quad=[2 \times 1.007276+2 \times 1.008665-4.001506] \\
\quad=0.030376 \mathrm{u}
\end{array}
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
&\text { Here in order to remove a neutron, energy has to be supplied. }\\
&\begin{aligned}
& \text { Mass defect, } \begin{aligned}
\Delta m & =M\left({ }_{20}^{39} \mathrm{Ca}\right)+M\left({ }_0^1 n\right)-M\left({ }_{20}^{40} \mathrm{Ca}\right) \\
& =38.970691+1.008665-39.962589 \\
& =0.016767 \mathrm{amu}
\end{aligned} \\
& \text { Equivalent energy }=0.016767 \times 931 \quad(\because 1 \mathrm{amu}=931 \mathrm{MeV}) \\
&=15.6 \mathrm{MeV}
\end{aligned}
\end{aligned}
\)

Hint

(c) We have, \({ }_0^1 n \rightarrow{ }_1^1 p+{ }_{-1}^0 e\)
\(\begin{aligned} & \text { Mass defect, } \Delta m=m_n-m_p-m_e \\ & \quad=\left[1.6747 \times 10^{-27}-1.6725 \times 10^{-27}-0.0009 \times 10^{-27}\right] \\ & \quad=1.3 \times 10^{-30} \mathrm{~kg}\end{aligned}\)
\(
\begin{aligned}
\text { Energy produced } & =\Delta m c^2=1.3 \times 10^{-30} \times\left(3 \times 10^8\right)^2 \\
& =11.7 \times 10^{-14} \mathrm{~J} \\
& =\frac{11.7 \times 10^{-14}}{1.6 \times 10^{-13}} \\
& =0.73 \mathrm{MeV} \quad\left(\because 1 \mathrm{MeV}=1.6 \times 10^{-13} \mathrm{~J}\right)
\end{aligned}
\)

Note:
\(
\begin{aligned}
& \text { Mass defect, } \Delta m=(\text { Mass of protons }+ \text { Mass of } \\
&\text { neutrons })- \text { Mass of the nucleus }
\end{aligned}
\)
Let us consider an atom of element \({ }_Z X^A\).
Its atomic number is \(Z\) and mass number is \(A\). Thus, its nucleus has \(Z\) protons and ( \(A-Z\) ) neutrons. If \(m_p\) is the mass of a proton, \(m_n\) is the mass of a neutrons and \(m_N\) is the mass of the nucleus, then the mass defect,
\(
\Delta m=Z m_p+(A-Z) m_n-m_N
\)

Hint

(a) \(\alpha\)-particle consists of 2 protons and 2 neutrons
\(
\text { Mass defect, } \Delta m=\left[Z m_p+N m_n-m\left({ }_2^4 \mathrm{He}\right)\right]
\)
\(
\begin{aligned}
& =2 \times 1.007277+2 \times 1.008666-4.001265 \\
& =0.03062 \mathrm{u}
\end{aligned}
\)
\(
\begin{aligned}
\text { Binding energy } & =\Delta m \times 931.4813 \mathrm{MeV} \\
& =0.03062 \times 931.48=28.52 \mathrm{MeV}
\end{aligned}
\)

Note:

\(
\text { Binding energy }=\Delta m c^2=\left[\left[Z m_p+(A-Z) m_n\right\}-m_N\right] \cdot c^2
\)

Hint

(c) Binding energy per nucleon of \({ }_{17} \mathrm{Cl}^{35}=\frac{\text { Binding energy }}{\text { Mass number }}\) \(=\frac{287.67}{35}=8.22 \mathrm{MeV}\)
Binding energy per nucleon of \({ }_{15} \mathrm{P}^{31}=\frac{262.48}{31}=8.47 \mathrm{MeV}\) BE of \({ }_{15} \mathrm{P}^{31}\) is more than BE of \({ }_{17} \mathrm{Cl}^{35}\).
Hence, \({ }_{15} \mathrm{P}^{31}\) is more stable.

Explanation:

Binding energy per nucleon
\(
\begin{aligned}
& =\frac{\text { Total binding energy }}{\text { Mass number (i.e. total number of nucleons) }} \\
& =\frac{\Delta m \times 931 \mathrm{MeV}}{A} \\
\Rightarrow & E_{\mathrm{bn}}=\frac{E_b}{A}
\end{aligned}
\)
Binding energy per nucleon \(\propto\) stability of nucleus

Hint

(b) Binding energy per nucleon, \(E_{\mathrm{bn}}=\frac{E_b}{A}\)
For deuteron, atomic mass, \(A=2\)
Substituting the values in above equation, we get
\(
\therefore \quad 1.115 \mathrm{MeV}=\frac{E_b}{2} \Rightarrow E_b=2 \times 1.115 \mathrm{MeV}
\)
Since, \(E_b=931.5 \times \Delta m\)
\(
\Rightarrow \quad \Delta m=\frac{E_b}{931.5} \Rightarrow \Delta m=\frac{2 \times 1.115}{931.5}=0.0024 \mathrm{u}
\)

Hint

(b) The nuclear reaction as per the given data is given as
\(
{ }_7 \mathrm{~N}^{14}+{ }_1 \mathrm{H}^2 \longrightarrow{ }_2 X^A+{ }_2 \mathrm{He}^4 \text { ( } \alpha \text {-particle) }
\)
According to laws of conservation,
\(
\text { (i) For charge, } \Sigma Z_{\text {initial }}=\Sigma Z_{\text {final }}
\)
\(
\begin{aligned}
7+1 & =Z+2 \\
Z & =6
\end{aligned}
\)
\(
\text { (ii) For mass number, } \Sigma A_{\text {initial }}=\Sigma A_{\text {final }}
\)
\(
\begin{aligned}
14+2 & =A+4 \\
A & =12
\end{aligned}
\)
\(
{ }_6 \mathrm{X}^{12} \text {, hence the element is carbon }{ }_6 \mathrm{C}^{12} \text {. }
\)

Hint

(c) Here, atomic masses are given (not the nuclear masses), so we shall use them for calculating the mass defect because mass of electrons get cancelled both sides. Thus, mass defect, \(\Delta m=m\left({ }_{10} \mathrm{Ne}^{23}\right)-m\left({ }_{11} \mathrm{Na}^{23}\right)\)
\(
=(22.9945-22.9898)=0.0047 \mathrm{u}
\)
Kinetic energy is given by \(Q=\Delta m \cdot c^2\)
\(
\therefore \quad Q=(0.0047 \mathrm{u})(931.5 \mathrm{MeV} / \mathrm{u})=4.4 \mathrm{MeV}
\)
Hence, the energy of beta particles can range from 0 to 4.4 MeV.

Hint

(c) It is given that, \(0.1 \%[latex] of 1 g of [latex]{ }_{92} \mathrm{U}^{235}\) is being used. So, mass used, \(m=\frac{0.1}{100} \times 1=10^{-3} \mathrm{~g}\)
\(
\begin{aligned}
\text { Number of fissions } & =\frac{\text { Given mass }}{\text { Atomic mass }} \times N_A \\
& =\frac{10^{-3}}{235} \times 6.023 \times 10^{23}=\frac{6.023 \times 10^{20}}{235}
\end{aligned}
\)
Energy produced, \(E=\) Number of fissions
\(\times\) Energy released per fission
\(
\begin{array}{r}
=\frac{6.023 \times 10^{20}}{235} \times 200 \times 1.6 \times 10^{-13}=8.2 \times 10^7 \mathrm{~J} \\
\left(\because 1 \mathrm{MeV}=1.6 \times 10^{-13} \mathrm{~J}\right)
\end{array}
\)

Hint

(b)

\(
\text { Fission equation, }{ }_{92} \mathrm{U}^{235}+{ }_0 n^1 \rightarrow{ }_{56} \mathrm{Ba}^{141}+{ }_{36} \mathrm{Kr}^{92}+3\left({ }_0 n^1\right)
\)
\(
\begin{aligned}
&\text { Mass defect, }\\
&\Delta m=\left[m\left({ }_{92} \mathrm{U}^{235}\right)+m_n\right]-\left[m\left({ }_{56} \mathrm{Ba}^{141}\right)+m\left({ }_{36} \mathrm{Kr}^{92}\right)+3 m_n\right]
\end{aligned}
\)
\(
\begin{array}{r}
=[\{235.0439+1.0087\}-\{140.9139 \\
+91.8973+3 \times 1.0087\}]
\end{array}
\)
\(
\begin{aligned}
& =0.2135 \mathrm{amu} \\
\text { Equivalent energy, } E & =\Delta m c^2=\Delta m \times 931 \mathrm{MeV} \\
& =0.2153 \times 931=200.4 \mathrm{MeV}
\end{aligned}
\)
Number of atoms/fission in 1 g of atom is
\(
n=\frac{N_A}{A}=\frac{1}{235} \times 6.023 \times 10^{23}=2.56 \times 10^{21}
\)
Energy released in fission of 1 g of \({ }_{92}^{235} \mathrm{U}\)
\(
\begin{aligned}
\Rightarrow \quad n \times E & =2.56 \times 10^{21} \times 200.4 \mathrm{MeV} \\
& =2.56 \times 10^{21} \times 200.4 \times 1.6 \times 10^{-13} \mathrm{~J} \\
& \quad\left(\because 1 \mathrm{MeV}=1.6 \times 10^{-13} \mathrm{~J}\right) \\
& =\frac{2.56 \times 10^{21} \times 200.4 \times 1.6 \times 10^{-13}}{3.6 \times 10^6} \mathrm{kWh} \\
& =2.28 \times 10^4 \mathrm{kWh}
\end{aligned}
\)

Hint

(c) Total number of nuclei in 1 kg of uranium,
\(
\begin{aligned}
n & =(\text { Number of moles }) \times(\text { Avogadro’s number }) \\
& =\left(\frac{1000}{235}\right)\left(6.02 \times 10^{23}\right)=2.56 \times 10^{24}
\end{aligned}
\)
By the fission of one nucleus, 185 MeV energy is released. Hence total energy released by the fission of 1 kg of uranium is,
\(
E=\left(2.56 \times 10^{24}\right)\left(185 \times 1.6 \times 10^{-13}\right) \mathrm{J}=7.58 \times 10^{13} \mathrm{~J}
\)
Power of the power plant is given 100 MW or \(10^8 \mathrm{~J} / \mathrm{s}\). Therefore total time upto which the power plant can be run by 1 kg uranium is, \(t=\frac{7.58 \times 10^{13}}{10^8}=7.58 \times 10^5 \mathrm{~s}\)

Hint

(a)

\(
\text { Given, fusion reaction, }{ }_1 \mathrm{H}^2+{ }_1 \mathrm{H}^2 \longrightarrow{ }_2 \mathrm{He}^4+\text { energy }
\)
Binding energy of a \(\left({ }_1 \mathrm{H}^2\right)\) deuterium nuclei
\(
=2 \times 1.1=2.2 \mathrm{MeV}
\)
Total binding energy of two deuterium nuclei
\(
=2.2 \times 2=4.4 \mathrm{MeV}
\)
Binding energy of a ( \({ }_2 \mathrm{He}^4\) ) nuclei \(=4 \times 7=28 \mathrm{MeV}\)
\(
\begin{aligned}
\text { So, energy released in fusion } & =\text { Final } \mathrm{BE}-\text { Initial } \mathrm{BE} \\
& =28-4.4=23.6 \mathrm{MeV}
\end{aligned}
\)

Hint

(d) \(\text { Given, fusion reaction, }{ }_1 \mathrm{H}^2+{ }_1 \mathrm{H}^2 \rightarrow{ }_2 \mathrm{He}^3+{ }_0 n^1\)
\(
\text { Mass defect, } \Delta m=\left[2 \times m\left({ }_1 \mathrm{H}^2\right)-m\left({ }_2 \mathrm{He}^4\right)+m_n\right]
\)
\(
=[(2 \times 2.015)-(3.017+1.009)]=4 \times 10^{-3} \mathrm{amu}
\)
Equivalent energy, \(E=\Delta m c^2=\Delta m \times 931.5 \mathrm{MeV} / c^2\)
\(
=4 \times 10^{-3} \times 931.5=3.726 \mathrm{MeV}
\)
Number of atoms in 1 kg of \({ }_1 \mathrm{H}^2\)
\(
\begin{aligned}
& =\frac{\text { Given mass }}{\text { Atomic mass }} \times \text { Avogadro number } \\
& =\frac{1000}{2} \times 6.023 \times 10^{23}
\end{aligned}
\)
In one reaction, two atoms of \({ }_1^2 \mathrm{H}\) are used.
So, total energy released
\(
\begin{aligned}
& =\frac{1}{2}\left(\frac{1000}{2} \times 6.023 \times 10^{23}\right) \times 3.726 \times 1.6 \times 10^{-13} \\
& =8.98 \times 10^{13} \simeq 9 \times 10^{13} \mathrm{~J}
\end{aligned}
\)

Hint

(c)
\(
\begin{aligned}
&\text { Given, output power, } P_o=2000 \mathrm{MW}=2 \times 10^9 \mathrm{~W}\\
&\begin{aligned}
\text { Efficiency, } \eta & =\frac{\text { Output power } P_o}{\text { Input power } P_i} \\
\frac{20}{100} & =\frac{2 \times 10^9}{P_i} \Rightarrow P_i=10^{10} \mathrm{~W}
\end{aligned}
\end{aligned}
\)
Consider the mass of uranium be \(m\).
Number of fissions \(=\) Number of atoms
\(
\begin{aligned}
& =\frac{\text { Given mass }}{\text { Atomic mass }} \times \text { Avogadro number } \\
& =\frac{m}{A} \times N_A=\frac{m}{235} \times 6.023 \times 10^{23}
\end{aligned}
\)
\(
\begin{aligned}
&\text { Energy produced, } E=\text { Number of fissions } \times \text { Energy per fission }\\
&\begin{aligned}
& =\frac{m}{235} \times 6.023 \times 10^{23} \times 100 \mathrm{MeV} \\
& =\frac{m}{235} \times 6.023 \times 10^{23} \times 100 \times 1.6 \times 10^{-13} \mathrm{~J} \left(\because 1 \mathrm{MeV}=1.6 \times 10^{-13} \mathrm{~J}\right)
\end{aligned}
\end{aligned}
\)
\(
\begin{gathered}
=m \times 0.041 \times 10^{12} \mathrm{~J}=m \times 4.1 \times 10^{10} \mathrm{~J} \\
\text { Power }=\frac{\text { Energy }}{\text { Time }} \Rightarrow 10^{10}=\frac{m \times 4.1 \times 10^{10}}{5 \times 365 \times 24 \times 3600} \\
m=3.84 \times 10^7 \mathrm{~g}=38.4 \times 10^3 \mathrm{~kg}
\end{gathered}
\)

Hint

(b) Mass defect, \(\Delta m=1-0.993=0.007 \mathrm{~g}\)
\(\therefore\) Energy released, \(E=(\Delta m) c^2\)
\(
=\left(0.007 \times 10^{-3}\right) \times\left(3 \times 10^8\right)^2=63 \times 10^{10} \mathrm{~J}
\)

Hint

(b) Mass of electron \(=\) Mass of positron \(=9.1 \times 10^{-31} \mathrm{~kg}\)
Energy released,
\(
\begin{aligned}
E & =(2 \mathrm{~m}) \cdot c^2 \\
& =2 \times 9.1 \times 10^{-31} \times\left(3 \times 10^8\right)^2=1.6 \times 10^{-13} \mathrm{~J}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& (d) E=\Delta m c^2 ; \Delta m=0.3 \mathrm{~g}=0.3 \times 10^{-3} \mathrm{~kg}(\text { given }) \\
& \Rightarrow \text { Amount of energy } E=\frac{0.3}{1000} \times\left(3 \times 10^8\right)^2=2.7 \times 10^{13} \mathrm{~J} \\
& =\frac{2.7 \times 10^{13}}{3.6 \times 10^6}=7.5 \times 10^6 \mathrm{kWh} \\
& \quad\left(\because 1 \mathrm{kWh}=3.6 \times 10^6 \mathrm{~J}\right)
\end{aligned}
\)

Hint

(a) The mass of nucleus formed is always less than the sum of the masses of the constituent proton and neutrons.
i.e. \(m<(A-Z)_{m n}+Z_{m p}\)

Hint

(c)

\(
\begin{aligned}
\text { Energy released } & =\text { Products energy }- \text { Reactants energy } \\
& =8.5 \times 234-7.6 \times 236 \\
& =195.4 \mathrm{MeV} \\
& \approx 200 \mathrm{MeV}
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
\begin{aligned}
\mathrm{BE} & =\Delta m c^2=\Delta m \times 931 \mathrm{MeV} \\
\Rightarrow \mathrm{BE} & =\left[Z m_p+(A-Z) m_n-m_N\right] \times 931 \\
& =[2(1.0087+1.0073)-4.0015] \times 931 \\
& =28.4 \mathrm{MeV}
\end{aligned}
\end{aligned}
\)

Hint

(b) Decreases with mass number at high mass numbers.

Explanation:
Binding energy per nucleon and stability: A higher binding energy per nucleon indicates a more stable nucleus.
Trend with mass number: As the mass number of a nucleus increases, the binding energy per nucleon generally decreases, especially for high mass numbers.
Fission process: When a heavy nucleus undergoes fission, it splits into smaller, more stable nuclei with a higher binding energy per nucleon, releasing energy in the process.

Hint

(c) A chain reaction is continuous because of the production of more neutrons in fission.

Explanation: When a uranium atom undergoes fission, it releases neutrons. These neutrons can then collide with other uranium atoms, causing them to fission as well, releasing even more neutrons. This process continues, creating a chain reaction.

Hint

(a) In fusion reaction, two lighter nuclei combine to form a heavy nucleus.
This reaction shows two isotopes of hydrogen, tritium \({ }^3 \mathrm{H}_1\) and deuterium \({ }^2 \mathrm{H}_1\), combining.
They form a helium nucleus ( \({ }^4 \mathrm{He}_2\) ) and a neutron ( \({ }^1 n_0\) ).
This is a combination of lighter nuclei into a heavier one. option (a) is the correc option.

(b) This reaction shows a heavy uranium nucleus \({ }^{238} \mathrm{U}_{92}\) decaying. It forms lead \({ }^{206} \mathrm{~Pb}_{82}\), alpha particles \({ }^4 \mathrm{He}_2\), and beta particles \({ }^0 \mathrm{~B}_1\). This is a decay process, not a combination of nuclei.

(c) This reaction shows a carbon isotope ( \({ }^{12} \mathrm{C}_7\) ) transforming. It forms another carbon isotope \(\left({ }^{12} \mathrm{C}_6\right)\), a beta particle \((\beta)\), and a gamma ray \((\gamma)\). This is a beta decay process, not a fusion reaction.

Only reaction (a) involves the combination of two lighter nuclei to form a heavier one. The fusion reaction is (a) where two hydrogen isotopes combine to form helium.

Hint

(b) Fusion of protons during synthesis of heavier elements.

Explanation: In the Sun’s core, hydrogen atoms fuse together to form helium, releasing a vast amount of energy in the process. This fusion reaction is the primary source of solar energy.

Hint

(c) energy production in sun, hydrogen bomb.
Explanation: Nuclear fusion is the process where two atomic nuclei merge to form a single, heavier nucleus, releasing a large amount of energy. Both the energy production in the sun and the detonation of a hydrogen bomb involve this process of nuclear fusion.

Hint

(d) kinetic energy is high enough to overcome the Coulomb repulsion between nuclei.

Explanation: In a fusion reaction, two nuclei need to get close enough to combine, but due to their positive charges, they repel each other electrostatically. High temperatures provide the nuclei with enough kinetic energy to overcome this repulsion and collide, allowing them to fuse together.

Hint

(d) The function of a moderator in a nuclear reactor is to slow down neutrons of thermal energies.

Explanation:
Moderator function:
A moderator is a material placed in a nuclear reactor to slow down the fast-moving neutrons produced during fission. This is crucial because slower neutrons are more likely to cause fission reactions with uranium-235 nuclei, sustaining the chain reaction.

Why heavy water is used:
Heavy water (\(\mathrm{D}_2 \mathrm{O}\)) is used as a moderator in some reactors because it effectively slows down neutrons with a relatively low probability of absorbing them.

Note: Heavy water is a form of water where the hydrogen atoms are replaced with a heavier isotope called deuterium. Specifically, it’s water ( \(\mathrm{H}_2 \mathrm{O}\) ) where the hydrogen atoms are replaced by deuterium (D), resulting in deuterium oxide ( \(\mathrm{D}_2 \mathrm{O}\) ). This makes it slightly denser than regular water and gives it some unique properties.

Ordinary Water vs. Heavy Water:
Regular water \(\left(\mathrm{H}_2 \mathrm{O}\right)\) contains two hydrogen atoms \(\left({ }^1 \mathrm{H}\right)\) and one oxygen atom. In heavy water ( \(\mathrm{D}_2 \mathrm{O}\) ), the hydrogen atoms are replaced with deuterium ( \({ }^2 \mathrm{H}\) ), which has an extra neutron in its nucleus, making it heavier than regular hydrogen.

Hint

(b) Packing fraction is defined as \(\frac{m-A}{A}\), where \(m\) is the nuclear mass and \(A\) is the mass number.
Stable nuclei generally have higher specific binding energy.

Fusion reactions release more energy per unit mass than fission reactions.
Therefore, statement (a) is incorrect.

Evaluate statement (b).
Packing fraction can be positive, negative, or zero.
Positive packing fraction indicates instability, negative indicates stability.
Therefore, statement (b) is correct.

Evaluate statement (c).
\(\)\mathrm{Pu}^{239}\(\) is a fissile material, suitable for fission reactions.
Therefore, statement (c) is incorrect.

Evaluate statement (d).
Stable nuclei have high specific binding energy, indicating strong nuclear forces.
Therefore, statement (d) is incorrect.

The correct statement is that packing fraction may be positive or may be negative.

Hint

(a) \({ }_{90} \mathrm{Th}^{232} \longrightarrow{ }_{83} \mathrm{Bi}^{212}\)
Decrease in mass number \(=232-212=20\)
Number of \(\alpha\)-particles emitted due to the above decrease in mass number \(=\frac{20}{4}=5\)
Expected decrease in atomic number due to emission of \(5 \alpha\)-particles \(=5 \times 2=10\)
Expected atomic number of the nucleus formed \(=90-10=80\)
But the atomic number of nucleus formed \(=83\)
Increase in atomic number \(=83-80=3\)
Number of \(\beta^{-}\)-particles emitted \(=3\)
Thus, \(5 \alpha\)-particles and \(3 \beta^{-}\)-particles are emitted.

Hint

(d) In 1s, \(90 \%\) of the nuclei have remained undecayed, so in another 1s, \(90 \%\) of 90, i.e., 81 nuclei will remain undecayed.

Hint

(b)

(i) From the relation,
\(
R=R_0 e^{-\lambda t} \text { (here, } R=\text { activity of sample) }
\)
Substituting the values, we have \(2700=4750 e^{-5 \lambda}\)
\(
\therefore \quad 5 \lambda=\ln \left(\frac{4750}{2700}\right)=0.56
\)
\(\therefore\) Decay constant, \(\lambda=0.113 \mathrm{~min}^{-1}\)
(ii) Half-life of the sample, \(t_{1 / 2}=\frac{\ln 2}{\lambda}=\frac{0.693}{0.113}=6.132 \mathrm{~min}\)

Hint

(c) Activity of the sample after \(n\) half-life, \(R=R_0\left(\frac{1}{2}\right)^n\)
\(
\text { In } 8 \mathrm{~s}, \frac{R}{R_0}=\frac{100}{1600}=\frac{1}{16} \Rightarrow R=\frac{R_0}{16}
\)
\(
\frac{R_0}{16}=R_0\left(\frac{1}{2}\right)^n \text { or }\left(\frac{1}{2}\right)^4=\left(\frac{1}{2}\right)^n
\)
or number of half-lives, \(n=4\)
Four half-lives are equivalent to 8 s. Hence, 2 s is equal to one half-life. So in one half-life, activity will remain half of 1600 Bq , i.e. 800 Bq.

Hint

(a)
\(
\begin{aligned}
&\text { Charge, } q=I t=10^{-4} \times 1 \times 3600=0.36 \mathrm{C}\\
&\begin{aligned}
& \text { Number of protons }=\frac{q}{e}=\frac{0.36}{1.6 \times 10^{-19}}=2.25 \times 10^{18} \\
& \text { Number of }{ }^7 \text { Be nuclei }=\frac{2.25 \times 10^{18}}{1000}=2.25 \times 10^{15} \\
& \text { Activity, } R=\frac{\ln 2}{T_{1 / 2}} N \text {, where } R=2.8 \times 10^8 \mathrm{dps} \text { (given) } \\
& \text { Half-life of substance, } T_{1 / 2}=\frac{N \ln 2}{R}=\frac{2.25 \times 10^{15} \times 0.693}{2.8 \times 10^8} \\
& \qquad \begin{aligned}
=0.556 \times 10^7 \mathrm{~s}
\end{aligned}
\end{aligned}
\end{aligned}
\)

Hint

(b) The number of undecayed nuclei reduces from \(25 \%\) to \(12.5 \%\) in 10 s. It means that, the half-life of the sample is 10 s.
i.e. \(\quad T_{1 / 2}=10 \mathrm{~s}\)
But, \(\quad T_{1 / 2}=0.6931 / \lambda\), where \(\lambda\) is decay constant.
Thus, \(\quad \lambda=\frac{0.6931}{T_{1 / 2}}=\frac{0.6931}{10}\)
The mean life of the sample, \(\tau=1 / \lambda=\frac{10}{0.6931}=14.428 \mathrm{~s}\)

Hint

(d) We know that, number of nuclei present, \(N=\frac{m}{M} N_A\)
where, \(m=\) given mass, \(M=\) atomic mass and \(\quad N_A=\) Avogadro’s number.
Activity, \(R=\lambda N=\frac{1}{\tau} \frac{m N_A}{M}\) where, \(\tau=\) mean life.
Mean life of substance, \(\tau=\frac{m N_A}{M R}=\frac{1 \times 6.023 \times 10^{23}}{226 \times 3.7 \times 10^{10}} \mathrm{~s}\) \(=7.2 \times 10^{10} \mathrm{~s}\)

Hint

(b) For \(\alpha\) – emission and \(\beta\)-emission respectively, according to parallel disintegration
\(
\begin{aligned}
& \lambda=\lambda_\alpha+\lambda_\beta \\
& =\frac{1}{T_\alpha}+\frac{1}{T_\beta}=\frac{1}{1600}+\frac{1}{400}=\frac{5}{1600} \mathrm{yr}^{-1} \\
& \quad\left(\because \text { mean life }=\frac{1}{\text { decay constant }}\right)
\end{aligned}
\)
If (7/8)th sample will decay, i.e. remaining (1/8)th.
After \(n\) half-life, \(N=N_0\left(\frac{1}{2}\right)^n \Rightarrow \frac{N_0}{8}=N_0\left(\frac{1}{2}\right)^n \Rightarrow n=8\)
Time during (7/8)th sample will decay, \(t=n T_{1 / 2}=n \frac{\ln 2}{\lambda}\)
\(
=3 \times \frac{0.693}{5 / 1600} \simeq 665 \mathrm{yr}
\)

Hint

(c) According to secular equilibrium,
\(
\lambda_{\mathrm{U}} N_{\mathrm{U}}=\lambda_{\mathrm{Ra}} N_{\mathrm{Ra}}
\)
where, \(\lambda_{\mathrm{U}}=\) decay constant of uranium,
\(\lambda_{\mathrm{Ra}}=\) decay constant of Ra,
\(N_{\mathrm{U}}=\) number of nuclei of U
and \(\quad N_{\mathrm{Ra}}=\) number of nuclei of Ra.
We know that,
\(
\text { Decay constant }(\lambda)=\frac{\ln 2}{\operatorname{Half-life}\left(t_{1 / 2}\right)}
\)
\(
\begin{aligned}
\frac{\ln 2}{\left(t_{1 / 2}\right)_{\mathrm{U}}} N_{\mathrm{U}} & =\frac{\ln 2}{\left(t_{1 / 2}\right)_{\mathrm{Ra}}} N_{\mathrm{Ra}} \Rightarrow \frac{3.6 \times 10^6}{\left(t_{1 / 2}\right)_{\mathrm{U}}}=\frac{1}{1500} \\
\left(t_{1 / 2}\right)_{\mathrm{U}} & =3.6 \times 10^6 \times 1500 \\
& =5400 \times 10^6=5.4 \times 10^9 \mathrm{yr}
\end{aligned}
\)

Hint

(c)

\(
\alpha \text {-rays are a bunch of particles consisting of } 2 \text { neutrons and } 2 \text { protons. Hence, a } \alpha \text {-particle is a doubly ionised helium-atom. }
\)

Explanation: An alpha particle is a helium nucleus, which consists of two protons and two neutrons, making it a doubly charged positive ion. 

Hint

(c) (c) \(\gamma\)-rays has the least ionising power than \(\alpha\) and \(\beta\)-particles.

(a) \(\boldsymbol{\alpha}\)-particles are not deflected by electric and magnetic fields. This statement is incorrect. Alpha particles are positively charged and are deflected by both electric and magnetic fields. For instance, in an electric field, alpha particles are attracted to the negative plate, and in a magnetic field, they deflect according to the right-hand rule for positively charged particles.
(b) \(\boldsymbol{\beta}\)-particle has less penetrating power than \(\boldsymbol{\alpha}\)-particle. This statement is incorrect. Beta particles are smaller and faster than alpha particles, giving them greater penetrating power. They can be stopped by thin aluminum foil or a layer of clothing, whereas alpha particles can be stopped by a sheet of paper or even human skin.
(c) \(\gamma\)-rays has the least ionising power than \(\boldsymbol{\alpha}\) and \(\boldsymbol{\beta}\)-particles. This statement is correct. Gamma rays are high-energy photons with no charge or mass. Because they don’t have a direct charge or mass to interact with, they are less likely to directly cause ionization compared to alpha and beta particles. However, when they do interact, they can cause ionization through secondary processes.
(d) \(\gamma\)-rays has the least penetrating power than \(\boldsymbol{\alpha}\) and \(\boldsymbol{\beta}\)-rays. This statement is incorrect. Gamma rays have the highest penetrating power among the three types of radiation. They require dense materials like lead or thick concrete for shielding. Alpha particles have the lowest penetrating power, followed by beta particles, and then gamma rays.

Hint

(a) The nuclear reaction is \({ }_7 \mathrm{~N}^{14}+\alpha \longrightarrow{ }_8 X^{17}+{ }_1 \mathrm{P}^1\). \(\alpha\) represents an alpha particle, which is a helium nucleus \({ }_2 \mathrm{He}^4\). \({ }_1 \mathrm{P}^1\) represents a proton.
The sum of atomic numbers on the left side must equal the sum on the right side.
\(
\begin{aligned}
& 7+2=8+1 \\
& 9=9
\end{aligned}
\)
The atomic number of \(X\) is given as 8 in the reaction.
An atomic number of 8 corresponds to Oxygen.
The mass number of \(x\) is given as 17 in the reaction.
\(x\) is an oxygen nucleus with a mass number of 17.
The unknown nucleus \(X\) is an oxygen nucleus with mass number 17.

Hint

(d) Radioactive nucleus. Beta decay is the emission of an electron (or a positron) from the nucleus of a radioactive atom.
Explanation:
Beta decay is a nuclear process: It involves changes within the nucleus of an atom, not the atom’s electron shells.
Unstable nuclei decay: Radioactive nuclei are unstable and undergo decay to become more stable.
Beta particles are emitted: In beta decay, either an electron (beta-minus decay) or a positron (beta-plus decay is emitted from the nucleus, along with other particles like neutrinos.

Electron capture is related: In some cases, an electron from the atom’s inner shell can be captured by the nucleus, also considered a form of beta decay.

Note: \(\beta\) – emission takes place from a radioactive nucleus a
\(
\begin{aligned}
& { }_{15}^{32} P \xrightarrow{\beta}{ }_{16}^{32} S+{ }_{-1} e^0+\bar{\nu} \\
& { }_{15}^{32} P \xrightarrow{\beta}{ }_{16}^{32} S+{ }_{-1} e^0+\bar{\nu}
\end{aligned}
\)
where \(\bar{\nu}\) is the anti-neutrino.
In \(\beta^{+}\)decay a positron is emitted as
\(
{ }_{11}^{22} N a \rightarrow{ }_{11}^{22} N e++1 e^0+\nu
\)
where \(\nu\) is the neutrino.

Hint

(c) A neutron in the nucleus decays emitting an electron.

Explanation: In negative beta decay, a neutron within the nucleus transforms into a proton, emitting an electron and an antineutrino.
\(
\begin{aligned}
&\text { Negative } \beta \text {-decay is expressed by the equation, }\\
&n \longrightarrow p^{+}+e^{-}+\nu^{-}
\end{aligned}
\)

Hint

(a) \({ }_{92} X^{235} \xrightarrow{\alpha}{ }_{90} Z^{231} \xrightarrow{\beta^{-}}{ }_{91} Y^{231}\)
Therefore, one alpha and one electron are emitted.

Explanation:
\(
\begin{aligned}
&{ }_{92}^{235} \mathrm{X} \rightarrow{ }_{91}^{231} \mathrm{Y}+{ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{P}\\
&\text { Here, } \mathrm{A} \text { is the mass number and } \mathrm{Z} \text { is the atomic number of the emitted particle } \mathrm{P} \text {. }
\end{aligned}
\)
\(
\begin{aligned}
&\text { Balance the mass numbers. }\\
&\begin{aligned}
& 235=231+A \\
& A=235-231=4
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
&\text { Balance the atomic numbers. }\\
&\begin{aligned}
& 92=91+Z \\
& Z=92-91=1
\end{aligned}
\end{aligned}
\)
The emitted particle is \({ }_1^4 \mathrm{P}\).
This particle is a proton, \({ }_1 \mathrm{p}\).
However, the calculated mass number is 4 , not 1.
This indicates that a single particle is not sufficient to explain the decay.
If one alpha particle \(\left({ }_2^4 \alpha\right)\) is emitted:
Mass number change: \(235-4=231\). This matches.
Atomic number change: \(92-2=90\). This does not match the daughter nucleus’s atomic number of 91.
Let the emitted particles be \({ }_2^4 \alpha\) and \({ }_{-1}^0 \mathrm{e}\).
Total mass number change: \(4+0=4\).
Total atomic number change: \(2+(-1)=1\).
This matches the calculated \(\mathrm{A}=4\) and \(\mathrm{Z}=1\) for the combined emitted particles.
Hence, One alpha particle and one electron are emitted.

Hint

(d) Half-life of an element does not depend upon present amount of element, temperature and pressure. It depends upon the nature of the element.

The half-life of a radioactive element is solely determined by the specific isotope and is not affected by the amount of the element present, temperature, or pressure. 

Hint

(a) We know, after \(n\) half-life, undecayed nuclei left are
\(
N=N_0\left(\frac{1}{2}\right)^n ; t=n t_{1 / 2}
\)
We have, \(\frac{N_0}{32}=N_0\left(\frac{1}{2}\right)^{60 / T_{1 / 2}}\)
\(
\begin{array}{ll}
\Rightarrow & 5=\frac{60}{T_{1 / 2}} \\
\Rightarrow & T_{1 / 2}=12 \text { days }
\end{array}
\)
\(\therefore\) Half-life of a radioactive element, \(T_{1 / 2}=12\) days

Hint

(b) As, in 2 s, the substance is reduced to half of this previous value.
\(\therefore\) The half-life of radioactive substance, \(T_{1 / 2}=2 \mathrm{~s}\)

Hint

(b) We have, \(N=N_0 e^{-\lambda t}\)
\(
\frac{N_0}{2}=N_0 e^{-\lambda T_{1 / 2}} \Rightarrow 2=e^{\lambda T_{1 / 2}}
\)
By taking \(\log\) both sides, we get
\(
\begin{aligned}
\ln 2 & =\lambda T_{1 / 2} \\
\lambda T_{1 / 2} & =0.693
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (c) We have, } \frac{N}{N_0}=\left(\frac{1}{2}\right)^{t / 140} \quad\left[\because N=N_0\left(\frac{1}{2}\right)^n \text { and } t=n t_{1 / 2}\right] \\
& \quad \frac{1}{16}=\left(\frac{1}{2}\right)^4=\left(\frac{1}{2}\right)^{t / 140} \Rightarrow \frac{t}{140}=4 \\
& \Rightarrow \quad t=560 \text { days }
\end{aligned}
\)

Hint

\(
\text { (b) We have, } N=N_0\left(\frac{1}{2}\right)^{t / T} \Rightarrow \frac{N_0}{4}=N_0\left(\frac{1}{2}\right)^{16 / T} \Rightarrow T=8 \text { days }
\)

Hint

\(
\begin{aligned}
&\text { (b) Here, }\\
&\begin{aligned}
N & =N_0 \times\left(\frac{1}{2}\right)^{11400 / 5700} \quad\left(\because T_{1 / 2}=5700 \mathrm{yr} \text { and } T=11400 \mathrm{yr}\right) \\
& =N_0\left(\frac{1}{2}\right)^2=0.25 N_0
\end{aligned}
\end{aligned}
\)

Hint

\(
\text { (a) Mean life }=\frac{\text { Half-life }}{0.6931}=\frac{10}{0.6931}=14.4 \mathrm{~h}
\)

Hint

\(
\begin{aligned}
&\text { (d) Mean life, } \tau=1 / \lambda=100 \mathrm{~s}\\
&\text { Half-life }=\frac{0.693}{\lambda}=\frac{0.693 \times 100}{60}=1.155 \mathrm{~min}
\end{aligned}
\)