NEET Practice Q & A

Hint

(a) When an energetic \(\gamma\)-ray photon falls on a heavy substance, it is absorbed by some nucleus of the substance and an electron and positron are emitted. This phenomenon is called pair production.

Hint

(a) A nuclear reactor is in a critical state when the neutron population remains constant over time.
This means the rate of neutron production equals the rate of neutron loss.
The operation of a nuclear reactor is said to be critical if the multiplication factor \((k)\) has a value of 1.

Hint

(b) \(\mathbf{C o}^{\mathbf{6 0}}\). Cobalt-60 is a radioactive isotope used in radiation therapy for cancer treatment. It emits gamma rays that are directed at tumors to damage the DNA of cancer cells and prevent their growth and division. This technique is also known as cobalt therapy.

Hint

(d) first increases and then decreases with increase of mass number.

Explanation:
Initial Increase:
As the number of nucleons (protons and neutrons) increases in a nucleus, the binding energy per nucleon generally increases initially. This is because the strong nuclear force, which holds the nucleus together, is more effective in smaller nuclei. The strong force is short-range, meaning each nucleon primarily interacts with its immediate neighbors.

Reaching a Maximum:
As the nucleus grows larger, the attractive forces from neighboring nucleons become less effective, and the repulsive forces between protons start to play a more significant role. The binding energy per nucleon reaches a maximum around iron56.

Decreasing for Larger Nuclei:
For even larger nuclei (mass number significantly higher than iron), the binding energy per nucleon starts to decrease. This is because the long-range repulsive forces between protons start to dominate, and the nucleus becomes less stable.

Hint

(b) Binding energy per nucleon increases with atomic number and is maximum for iron. After that it decrease. \({ }_{26}^{56} \mathrm{Fe}\) (Iron-56) has a BE/A of approximately 8.79 MeV. It is known to have the highest binding energy per nucleon among all nuclides.

Hint

(b) Understanding Fission: Fission is a nuclear reaction in which the nucleus of an atom splits into two or more smaller nuclei, along with the release of energy and neutrons. For fission to occur, the nucleus must absorb a neutron and become unstable.
Uranium Isotopes:
U-235 is a fissile isotope, meaning it can sustain a fission chain reaction when it absorbs a neutron.
U-238 is not fissile but can undergo fission under certain conditions, typically when bombarded with fast neutrons.
Neutron Bombardment:
When U-235 is bombarded with fast-moving neutrons, it can absorb the neutron and form U-236, which is unstable and can undergo fission. However, fast neutrons are less likely to cause fission in U-235 compared to thermal neutrons.
When U-235 is bombarded with thermal (slow-moving) neutrons, it is more likely to absorb the neutron and become U236, leading to fission. This is because thermal neutrons have a higher probability of being captured by the nucleus.
U-238 Interaction:
U-238 can also undergo fission when bombarded with fast neutrons, but it is not as efficient as U-235 with thermal neutrons. \(\mathrm{U}-238\) can capture a thermal neutron to become \(\mathrm{U}-239\), which is not a fission reaction but can lead to the production of plutonium-239, which is fissile.
Conclusion: The process that results in a fission reaction is when U-235 is bombarded with thermal neutrons. This leads to the formation of U-236, which is unstable and undergoes fission.
Final Answer: The process that will result in a fission reaction is when U-235 is bombarded with thermal neutrons.

Hint

(c) The initial number of atoms is \(N_0=10000\).
After one half-life, the average number of atoms remaining is \(\frac{N_0}{2}\).
\(
N_{a v g}=\frac{10000}{2}=5000
\)
Radioactive decay is a statistical process.
While the average number of atoms after one half-life is 5000 , individual containers will show variations.
The actual number of atoms in each container will fluctuate around this average.
Option (a) is incorrect because not all containers will have exactly 5000 atoms due to randomness.
Option (b) is incorrect because the number of atoms will not be the same in all containers.
Option (d) is incorrect as some containers might statistically have slightly more than 5000 atoms.
Option (c) correctly states that containers will have different numbers of atoms, but their average will be close to 5000.

Hint

(a) The principle of controlled chain reaction is used in atomic energy reactor.

Explanation: A controlled chain reaction means that the rate of fission reactions is carefully managed to produce a steady amount of energy, which is the key function of a nuclear reactor.

Key points about the options:
(b) Atom bomb: An atom bomb uses an uncontrolled chain reaction, where the fission process rapidly escalates, causing a large explosion.
(c) The core of the sun: The energy production in the sun is primarily through nuclear fusion, not fission, and the process is not considered a controlled chain reaction.
(d) Artificial radioactivity: Artificial radioactivity involves creating unstable isotopes by bombarding stable nuclei with particles, not a chain reaction.

Hint

(a) To sustain a chain reaction in atom bomb one of the neutrons emitted in the fission of \({ }^{235} \mathrm{U}\) must be captured by another \({ }^{235} \mathrm{U}\) nucleus and causes its fission.

Expplanation: When uranium -235 is bombarded by neutrons, each uranium nucleus is broken into two nearly equal fragments and along with it huge energy and two ot three fresh neutron are liberated. Under favourable conditions. These neutrons fission other uranium nuclei in the same way. Thus, a chain of nuclear fissions is established which continues until the whole of the uranium is consumed.

Hint

(a) Boron rods are used in nuclear reactors because boron can absorb neutrons.

Explanation: Boron rods are designed to control the nuclear chain reaction by absorbing excess neutrons, preventing them from causing further fission.
Why other options are incorrect:
(b) strength is given to the plant: Boron rods do not provide structural strength to the reactor.
(c) boron speeds up neutrons: Boron actually slows down the chain reaction by absorbing neutrons, not speeding them up.
(d) the reactors look good: Boron rods have no aesthetic function; their sole purpose is to control the nuclear reaction.

Hint

\(
\text { (c) Mean life, } \tau=\frac{1}{\lambda}=\frac{1}{15 \times 10^{-9}}=6.67 \times 10^8 \mathrm{~s}
\)

Hint

\(
\begin{aligned}
(d) Q & =(1.002+1.004-1.001-1.003)(931.5) \mathrm{MeV} \\
& =1.863 \mathrm{MeV}
\end{aligned}
\)

Hint

(b) Helium is very stable. A sharp peak in the curve of binding energy per nucleon indicates high stability for that nucleus. The elements high on the BE versus mass number plot are very tightly bound and hence are stable. And the elements those are lower on this plot, are less tightly bound and hence are unstable.
Since, helium nucleus shown a peak on this plot, so it is very stable.

Explanation:

Binding Energy per Nucleon: This value represents the average energy needed to remove a nucleon (proton or neutron) from the nucleus. A higher binding energy per nucleon implies a more stable nucleus. 

Peak for Helium: The curve of binding energy per nucleon has a prominent peak at helium (specifically, helium -4 , which has 2 protons and 2 neutrons). This peak signifies that helium nuclei are exceptionally stable.

Stability: Since the peak indicates high binding energy, it means a significant amount of energy is required to break apart a helium nucleus. This makes it very difficult to break apart and, therefore, very stable.

Hint

(a) \({ }_{92} \mathrm{U}^{238} \xrightarrow{\alpha}{ }_{90} \mathrm{Th}^{234} \xrightarrow{\beta}{ }_{91} \mathrm{~Pa}^{234} \xrightarrow{\beta}{ }_{92} \mathrm{U}^{234}\)
So, \(\quad A=234, B=90, C=234, D=91\) and \(E=\beta\)

Explanation: The initial reaction is:
\(
{ }_{92} \mathrm{U}^{238} \rightarrow \text { (alpha particle) }+{ }_B \mathrm{Th}^A
\)
Here, Uranium-238 (U-238) undergoes alpha decay.
An alpha particle consists of 2 protons and 2 neutrons, which means it has a mass number of 4 and an atomic number of 2. When U-238 emits an alpha particle, the new element formed will have:
Mass number: \(238-4=234\)
Atomic number: \(92-2=90\)
Thus, we have:
\(
A=234, \quad B=90
\)
The resulting element is Thorium (Th).
The first decay product is:
\(
{ }_{90} \mathrm{U}^{234}} \rightarrow \text { (beta particle) }+{ }_D \mathrm{~Pa}^C
\)
Here, Thorium-234 undergoes beta decay.
In beta decay, a neutron is converted into a proton, resulting in:
Mass number remains the same: 234
Atomic number increases by 1: \(90+1=91\)
Thus, we have:
\(
C=234, \quad D=91
\)
The resulting element is Protactinium (Pa).
\(
{ }_{91} \mathrm{Pa}^{234}} \rightarrow(\mathrm{E})+{ }_{92} \mathrm{U}^{234}}
\)
Here, Protactinium-234 undergoes decay to form Uranium-234.
In this case, since the atomic number increases from 91 to 92 while the mass number remains the same, this indicates that a beta particle is emitted.
Thus, we have:
\(
E=\text { beta particle }
\)
Final Values
Now, we can summarize the values:
\(A=234\)
\(B=90\)
\(C=234\)
\(D=91\)
\(E=\) beta particle

Hint

\(
(d){ }_7 X^{15}+\alpha \text {-particle } \longrightarrow{ }_9 Y^{19} \xrightarrow{{ }_1 \mathrm{H}^1}{ }_8 \mathrm{Z}^{18}
\)

Explanation: Initial mass number: 15
Mass number of \(\alpha\)-particle: 4
Mass number after capture: \(15+4=19\)
Calculate the atomic number after capturing an \(\alpha\)-particle.
Initial atomic number: 7
Atomic number of \(\alpha\)-particle: 2
Atomic number after capture: \(7+2=9\)
Calculate the final mass number after emitting a proton.
Mass number after capture: 19
Mass number of proton: 1
Final mass number: \(19-1=18\)
Calculate the final atomic number after emitting a proton.
Atomic number after capture: 9
Atomic number of proton: 1
Final atomic number: \(9-1=8\)
The mass number and atomic number of the resulting product are 18 and 8 respectively.

Hint

\(
\begin{aligned}
&\text { (a) Three atoms of } B \longrightarrow \text { one atom of } A\\
&\therefore \quad Q=E_a-3 E_b
\end{aligned}
\)

Hint

(c) \({ }_{92}^{238} \mathrm{U} \longrightarrow{ }_{82}^{214} \mathrm{~Pb}+6{ }_2^4 \mathrm{He}+2{ }_{-1}^0 e\)
The total number of \(\alpha\) and \(\beta\)-particles emitted in the nuclear reaction is 8.

Explanation: Let \(x\) be the number of alpha particles and \(y\) be the number of beta particles.
\(
{ }_{92}^{238} \mathrm{U} \rightarrow{ }_{82}^{214} \mathrm{~Pb}+x \cdot{ }_2^4 \mathrm{He}+y \cdot{ }_{-1}^0 \mathrm{e}
\)
The sum of mass numbers on both sides must be equal.
\(
\begin{aligned}
& 238=214+4 x+0 y \\
& 238=214+4 x \\
& 4 x=238-214 \\
& 4 x=24 \\
& x=\frac{24}{4}=6
\end{aligned}
\)
The sum of atomic numbers on both sides must be equal.
\(
\begin{aligned}
& 92=82+2 x+(-1) y \\
& 92=82+2 x-y
\end{aligned}
\)
Substitute the value of \(x=6\) :
\(
\begin{aligned}
& 92=82+2(6)-y \\
& 92=82+12-y \\
& 92=94-y \\
& y=94-92 \\
& y=2
\end{aligned}
\)
\(
\begin{aligned}
& \text { Total particles }=\text { number of alpha particles }+ \text { number of beta particles. } \\
& \text { Total particles }=x+y=6+2=8 \text {. }
\end{aligned}
\)

Hint

(b) Let \(N(t)\) be the amount of radioactive material remaining after time \(t\). The radioactive decay law states that \(N(t)=N_0 e^{-\lambda t}\), where \(N_0\) is the initial amount and \(\lambda\) is the decay constant.
Given that \(90 \%\) of the sample is left undecayed after time \(t\), we have:
\(
\begin{aligned}
& 0.9 N_0=N_0 e^{-\lambda t} \\
& 0.9=e^{-\lambda t}
\end{aligned}
\)
We want to find the percentage of the initial sample that decays in time \(2 t\). Let \(N(2 t)\) be the amount remaining after time \(2 t\).
\(
\begin{aligned}
& N(2 t)=N_0 e^{-2 \lambda t} \\
& N(2 t)=N_0\left(e^{-\lambda t}\right)^2 \\
& N(2 t)=N_0(0.9)^2 \\
& N(2 t)=N_0(0.81)
\end{aligned}
\)
So, \(81 \%\) of the initial amount remains undecayed after time \(2 t\). The percentage of the initial sample that decayed is:
\(
100 \%-81 \%=19 \%
\)
Therefore, \(19 \%\) of the initial sample will decay in a total time \(2 t\).

Hint

(b) Here, half-life of radioactive substance, \(T_{1 / 2}=20 \mathrm{~min}\) We have, \(N=N_0\left(\frac{1}{2}\right)^{t / T_{1 / 2}}\)
So, for \(20 \%\) decay, \(\frac{N}{N_0}=\frac{80}{100}=\left(\frac{1}{2}\right)^{t_1 / 20} \dots(i)\)
For \(80 \%\) decay, \(\frac{N}{N_0}=\frac{20}{100}=\left(\frac{1}{2}\right)^{t_2 / 20} \dots(ii)\)
On dividing Eq. (ii) by Eq. (i), we get
\(
\left(\frac{1}{2}\right)^{\left(\frac{t_2-t_1}{20}\right)}=\frac{1}{4} \quad \text { or } \quad\left(\frac{1}{2}\right)^{\left(\frac{t_2-t_1}{20}\right)}=\left(\frac{1}{2}\right)^2
\)
On comparing the exponents,
\(
\frac{t_2-t_1}{20}=2 \Rightarrow t_2-t_1=40 \mathrm{~min}
\)
\(\therefore\) The time between \(20 \%\) and \(80 \%\) decay \(t_2-t_1=40 \mathrm{~min}\)

Hint

(d) After \(n\) number of half-life, \(N=N_0\left(\frac{1}{2}\right)^n\)
\(
1=64\left(\frac{1}{2}\right)^n \Rightarrow\left(\frac{1}{2}\right)^6=\left(\frac{1}{2}\right)^n
\)
\(n=6\), i.e. 6 half-lives
\(\therefore\) Minimum time, \(t=6 \times 2=12 \mathrm{~h}\)

Hint

(b) We have, \(\frac{1}{64}=\left(\frac{1}{2}\right)^n\)
\(
n=6
\)
6 half-lives are equal to 60 s.
\(
\therefore 1 \text { half-life }=10 \mathrm{~s}
\)

Hint

(b) The nuclear reaction is \({ }_1^2 \mathrm{H}+{ }_1^3 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+{ }_0 n^1\).
The binding energy of \({ }_1^2 \mathrm{H}\) is \(a \mathrm{MeV}\).
The binding energy of \({ }_1^3 \mathrm{H}\) is \(b \mathrm{MeV}\).
The binding energy of \({ }_2^4 \mathrm{He}\) is \(c \mathrm{MeV}\).
Energy released in a nuclear reaction is the difference between the total binding energy of products and reactants.
A free neutron has no binding energy.
Calculate the total binding energy of the reactants:
The reactants are \({ }_1^2 \mathrm{H}\) and \({ }_1^3 \mathrm{H}\).
Total binding energy of reactants: \(B E_{\text {reactants }}=a+b\).
Calculate the total binding energy of the products:
The products are \({ }_2^4 \mathrm{He}\) and \({ }_0 n^1\).
The binding energy of \({ }_2^4 \mathrm{He}\) is \(c\).
The binding energy of a free neutron \(0^n\) is 0.
Total binding energy of products: \(B E_{\text {products }}=c+0=c\).
Energy released is the difference between the total binding energy of products and reactants.
Energy released \(=\boldsymbol{B} E_{\text {products }}-\boldsymbol{B} E_{\text {reactants }}\).
Energy released \(=c-(a+b)\).
Energy released \(=c-a-b\).

Hint

(d) Rate, \(R=\lambda N\)
\(
\therefore \quad \frac{R}{N}=\lambda \text { (constant) }
\)
So, graph between \(R / N\) and \(t\) be a straight line parallel to the time axis.

Hint

(d) We have, \(N=N_0 e^{-\lambda t}\) and \(R=R_0 e^{-\lambda t}=\lambda N_0 e^{-\lambda t}\)
\(
\left(\because R_0=\lambda N_0\right)
\)
\(
\therefore \quad N_{\text {decayed }}=N_0-N=N_0-N_0 e^{-\lambda t}=N_0-\frac{R}{\lambda}
\)
This is the equation of straight line with negative slope.

Hint

(d) By using, \(N=N_0 e^{-\lambda t}\)
It shows that \(N\) decreases exponentially with time.

Hint

\(
\text { (d) We have, } \frac{1}{e}=\frac{N_0 e^{-10 \lambda t}}{N_0 e^{-\lambda t}}=e^{-9 \lambda t} \text { or } t=\frac{1}{9 \lambda}\left(\because \text { given, } \frac{N_1}{N_2}=\frac{1}{e}\right)
\)

Hint

\(
\begin{aligned}
& (c) \text { We have, } N_A e^{-\lambda_A t}=N_B e^{-\lambda_B t} \\
& \therefore \quad e^{\left(\lambda_B-\lambda_A\right) t}=\frac{N_B}{N_A} \\
& \therefore \text { Time, } t=\frac{1}{\lambda_B-\lambda_A} \ln \left(\frac{N_B}{N_A}\right)
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (d) Binding energy per nucleon }\\
&\begin{aligned}
& =\frac{\left[\left\{m_p Z+m_n(A-Z)\right\}-M\right] \times 9314}{\text { Mass number }(A)} \\
& =\left(\frac{6 \times 1.0078+6 \times 1.0087-12}{12}\right) \times 931.4 \\
& =7.68 \mathrm{MeV}
\end{aligned}
\end{aligned}
\)

Hint

(c) If in the rock there is no \(Y\) element, then the time taken by element \(X\) to reduce to \(\left(\frac{1}{8}\right)\) th the initial value will be equal to
\(
\frac{1}{8}=\left(\frac{1}{2}\right)^n \text { or } n=3
\)
Therefore, from the beginning three half-life time is spent. Hence, the age of the rock is
\(
=3 \times 1.37 \times 10^9=4.11 \times 10^9 \mathrm{yr}
\)

Hint

(d) Energy released in the fission of one nucleus \(=200 \mathrm{MeV}\)
\(
\begin{aligned}
& =200 \times 10^6 \times 1.6 \times 10^{-19}=3.2 \times 10^{-11} \mathrm{~J} \\
P & =16 \mathrm{~kW}=16 \times 10^3 \mathrm{~W}
\end{aligned}
\)
Now, number of nuclei required per second
\(
n=\frac{P}{E}=\frac{16 \times 10^3}{3.2 \times 10^{-11}}=5 \times 10^{14}
\)

Hint

(c) In \(\beta^{-}\)decay, a neutron in the nucleus is transformed into a proton, an electron and an anti-neutrino.
\(
n \rightarrow P+\bar{e}+\bar{\nu}
\)
So, \(x\) and \(y\) stand for electron and anti-neutrino.

Explanation:
The atomic number (subscript) increases from 15 to 16.
This indicates a \(\beta^{-}\)-decay, where a neutron converts to a proton.
In \(\beta^{-}\)-decay, an electron ( \(\mathrm{e}^{-}\) or \(\beta^{-}\)) and an electron antineutrino ( \(\bar{\nu}}\) ) are emitted.
Therefore, \(x\) and \(y\) represent an electron and an antineutrino.

Hint

(c) Alpha ( \(\alpha\) ) emission decreases \(Z\) by 2 .
Beta-minus ( \(\beta^{-}\)) emission increases \(Z\) by 1 .
Beta-plus \(\left(\beta^{+}\right)\) emission decreases \(Z\) by 1 .
Number of \(\alpha\) emissions: 8
Number of \(\beta^{-}\)emissions: 4
Number of \(\beta^{+}\)emissions: 2
Change due to \(\alpha\) emissions: \(\Delta Z_\alpha=8 \times(-2)=-16\)
Change due to \(\beta^{-}\)emissions: \(\Delta \boldsymbol{Z}_{\beta^{-}}=4 \times(1)=4\)
Change due to \(\beta^{+}\)emissions: \(\Delta Z_{\beta^{+}}=2 \times(-1)=-2\)
Total change: \(\Delta Z_{\text {total }}=\Delta Z_\alpha+\Delta Z_{\beta^{-}}+\Delta Z_{\beta^{+}}\) \(\Delta Z_{\text {total }}=-16+4-2=-14\)
Final \(Z=\) Initial \(Z+\Delta Z_{\text {total }}\)
Final \(Z=92+(-14)=78\)

Hint

(c) Energy is released in a process when total binding energy (BE) of the nucleus is increased or we can say when total BE of products is more than the reactants. By calculation, we can see that only in case of option (c), this happens.
Given, \(W \rightarrow 2 Y\)
BE of reactants \(=120 \times 7.5=900 \mathrm{MeV}\) and BE of products \(=2 \times 60 \times 8.5=1020 \mathrm{MeV}\)
i.e. \mathrm{BE}[/latex] of products \(>\mathrm{BE}\) of reactants.
Hence, energy is released in the process of option (c).

Hint

(b) Momentum remains constant.
\(
\begin{array}{ll}
\therefore & K=\frac{p^2}{2 m} \text { or } K \propto \frac{1}{m} \\
\Rightarrow & \frac{K_n}{K_\alpha}=\frac{m_\alpha}{m_n}=\frac{4}{206}=\frac{2}{103}
\end{array}
\)
\(\therefore\) Kinetic energy of the residual nucleus
\(
K_n=\left(\frac{2}{103}\right) K_\alpha=\left(\frac{2}{103}\right) E
\)

Hint

(c) As, \(n=\frac{m}{A} N_A \Rightarrow E=\left(\frac{m}{A} N_A \times 200 \times 1.6 \times 10^{-13}\right) \mathrm{J}\)
For 1 day \(=24 \times 3600\)
Power \(=\frac{E}{t}=\frac{m N_A \times 3.2 \times 10^{-11}}{235 \times 24 \times 3600}=10^6 \quad\left(\because\right.\) given, \(\left.P=10^6 \mathrm{~W}\right)\)
\(\therefore\) Mass, \(m=\frac{1 \times 10^6 \times 24 \times 3600 \times 235}{3.2 \times 10^{-11} \times 6.02 \times 10^{23}} \approx 1 \mathrm{~g}\)

Hint

(a) Kinetic energy of the molecules of a gas at a temperature \(T\) is \(\frac{3}{2} k T\).
\(\therefore\) To initiate the reaction \(\frac{3}{2} k T=7.7 \times 10^{-14} \mathrm{~J}\)
\(
\begin{array}{rlr}
\Rightarrow & \frac{3}{2} \times 1.38 \times 10^{-23} \mathrm{~T} & =7.7 \times 10^{-14} \\
\Rightarrow & T =3.7 \times 10^9 \mathrm{~K}
\end{array}
\)
Thus, the order of temperature is \(10^9 \mathrm{~K}\).

Hint

(c) Net process is, \(3{ }_1 \mathrm{H}^2 \rightarrow{ }_2 \mathrm{He}^4+p+n\)
Mass defect, \(\Delta m=(3 \times 2.014)-4.001-1.007-1.008\)
\(
=0.026 \mathrm{amu}
\)
Energy released by the fusion of three deuterons
\(
\begin{aligned}
& =\Delta m \times 931.48=0.026 \times 931.48 \mathrm{MeV} \\
& =0.026 \times 931.48 \times 1.6 \times 10^{-13} \mathrm{~J} \\
& =3.87 \times 10^{-12} \mathrm{~J}
\end{aligned}
\)
\(
\begin{aligned}
&\begin{aligned}
& \begin{aligned}
\therefore \text { Total energy released } & =\frac{10^{40}}{3} \times 3.87 \times 10^{-12} \mathrm{~J} \\
& =1.29 \times 10^{28} \mathrm{~J}
\end{aligned} \\
& \text { Total time }=\frac{1.29 \times 10^{28}}{10^{16}}=1.29 \times 10^{12} \mathrm{~s}\left(\because \text { power }=\frac{\text { energy }}{\text { time }}\right)
\end{aligned}\\
&\therefore \text { The order of time is } 10^{12} \mathrm{~s}
\end{aligned}
\)

Hint

(c) For a \(\beta\)-decay, \({ }_0^1 n \longrightarrow{ }_1^1 p+{ }_{-1} e^0\) where, \({ }_{-1} e^0\) is an electron emitted in the transformation of a neutron into proton in the radioactive nucleus. But electron does not exist inside the nucleus.

Hint

\(
\begin{aligned}
& \text { (a) } \begin{aligned}
\Delta m & =2(2015)-(3.017+1.009) \\
& =0.004 \mathrm{amu}
\end{aligned} \\
& \begin{aligned}
\because \text { Energy released } & =(0.004 \times 931.5) \mathrm{MeV} \\
& =3.726 \mathrm{MeV}
\end{aligned}
\end{aligned}
\)
Energy released per deuterium
\(
=\frac{3.726}{2}=1.863 \mathrm{MeV}
\)
Number of deuterons in 1 kg
\(
=\frac{6.02 \times 10^{26}}{2}=3.01 \times 10^{26}
\)
\(\because\) Energy released per kg of deuterium fusion
\(
\begin{aligned}
& =\left(3.01 \times 10^{26} \times 1.863\right) \\
& =5.6 \times 10^{26} \mathrm{MeV} \\
& =9 \times 10^{13} \mathrm{~J}
\end{aligned}
\)

Hint

(d) We know that, \(\boldsymbol{\alpha}\)-particle carries two units of positive charge and four units of mass. In \(\alpha\)-decays, charge or atomic number of daughter nucleus decreased by two units. Also, mass number reduces by four units.

Hint

(d) Energy released per fission,
\(
\begin{aligned}
E & =200 \mathrm{MeV}=200 \times 1.6 \times 10^{-13} \\
& =3.2 \times 10^{-11} \mathrm{~J}
\end{aligned}
\)
Total energy required to run a 1 MW reaction for one year
\(
\begin{aligned}
& =10^6 \mathrm{~J} \mathrm{~s}^{-1} \times(365 \times 24 \times 60 \times 60) \\
& =3.15 \times 10^{13} \mathrm{~J}
\end{aligned}
\)
Since, 1 fission (atom of \(\mathrm{U}^{235}\) ) produces \(3.2 \times 10^{-11} \mathrm{~J}\) of energy.
So, total number of \(\mathrm{U}^{235}\) atoms required is
\(
\frac{3.15 \times 10^{13}}{3.2 \times 10^{-11}}=9.84 \times 10^{23}
\)
\(\because\) Mass of \(\mathrm{U}^{235}\) containing \(9.84 \times 10^{23}\) atom,
\(
\begin{aligned}
M & =\frac{235}{6.02 \times 10^{23}} \times 9.84 \times 10^{23} \\
& =384 \mathrm{~g}=0.384 \mathrm{~kg}
\end{aligned}
\)

Hint

(a) Radius of nuclei having mass number \(A\) is determined as where,
\(
\begin{aligned}
R & =R_0 A^{1 / 3} \\
R_0 & =1.2 \times 10^{-15} \mathrm{~m} \text { (where, } R_0=\text { constant) }
\end{aligned}
\)
This implies, \(\frac{R_1}{R_2}=\left(\frac{A_1}{A_2}\right)^{1 / 3}=\left(\frac{216}{64}\right)^{1 / 3}=\frac{6}{4} \Rightarrow R_1: R_2=3: 2\)

Hint

\(
\text { (d) The reaction should be }{ }_{88} \mathrm{Ra}^{226} \longrightarrow{ }_{86} \mathrm{Rn}^{222}+{ }_2 \mathrm{He}^4
\)
\(
Q \text {-value }=\left[\begin{array}{c}
\text { (mass number of Ra) } \\
-(\text { mass number of Rn }) \\
-m\left(\mathrm{He}^4\right)
\end{array}\right] \mathrm{u} \times 931.5 \mathrm{MeV}
\)
\(
\begin{aligned}
=0.005297 \times 9315 & =4.934 \mathrm{MeV} \\
\text { Energy corresponding mass defect } & =\frac{m(\mathrm{Rn})}{m(\mathrm{Ra})} \times Q \text {-value } \\
& \simeq 4.871 \mathrm{MeV}
\end{aligned}
\)

Hint

(b) The fission per second \(=4 \times 10^{20}\)
So, the energy released per second
\(
\begin{aligned}
& =\left(4 \times 10^{20} \times 250\right) \mathrm{MeV} \\
& =\left(4 \times 10^{20} \times 250 \times 10^6 \times 1.6 \times 10^{-19}\right) \mathrm{J}
\end{aligned}
\)
Therefore, energy released in \(10 \mathrm{~h}=36 \times 10^3 \mathrm{~s}\),
\(
\begin{aligned}
E & =\left(36 \times 10^3 \times 4 \times 10^{20} \times 250 \times 10^6 \times 1.6 \times 10^{-19}\right) \mathrm{J} \\
& =\left(36 \times 4 \times 250 \times 1.6 \times 10^{10}\right) \mathrm{J}=\left(57600 \times 10^{10}\right) \mathrm{J} \\
& =576 \times 10^{12} \mathrm{~J}
\end{aligned}
\)

Hint

(d) Nuclear fusion is not found in atom bomb because it works on the phenomenon of nuclear fission.

Hint

(d) Nuclear mass density is independent of mass number, therefore ratio of nuclear mass densities of \(\mathrm{Au}^{197}\) and \(\mathrm{Ag}^{107}\) is \(1: 1\).

Hint

(d) In \(\beta^{-}\)decay, the atomic number increases by 1 and the mass number remains unchanged.
In \(\alpha\) decay, the atomic number decreases by 2 and the mass number decreases by 4.
Atomic number of \(X_1: Z_{X_1}=Z_X+1=72+1=73\).
Mass number of \(X_1: A_{X_1}=A_X=180\).
Atomic number of \(X_2: Z_{X_2}=Z_{X_1}-2=73-2=71\).
Mass number of \(X_2: A_{X_2}=A_{X_1}-4=180-4=176\).
Atomic number of \(X_3: Z_{X_3}=Z_{X_2}+1=71+1=72\).
Mass number of \(X_3: A_{X_3}=A_{X_2}=176\).
Atomic number of \(X_4: Z_{X_4}=Z_{X_3}-2=72-2=70\).
Mass number of \(X_4: A_{X_4}=A_{X_3}-4=176-4=172\).
The atomic number and atomic mass number of \(X_4\) are 70 and 172 respectively.

Hint

(a) Let there is \(n\) number of fission per second produces a power of 6.4 W , then
\(
\begin{gathered}
n \times 200 \times 10^6 \times 1.6 \times 10^{-19}=6.4 \\
\therefore \quad n=\frac{6.4}{200 \times 10^{-13} \times 1.6}=\frac{4}{2 \times 10^{-11}}=2 \times 10^{11}
\end{gathered}
\)

Hint

(b) The number of \(\alpha\)-particles, \(n_1\)
\(
=\frac{\text { change in mass number }}{4}=\frac{238-206}{4} \Rightarrow n_1=\frac{32}{4}=8
\)
Now, number of \(\beta\)-particles, \(n_2=82-\left(92-2 n_1\right)\)
\(
=82-(92-2 \times 8)=82-76=6
\)

Hint

(b) Given, \(\frac{R_1}{R_2}=\frac{1}{2}\)
According to conservation of linear momentum, we have
\(
m_1 v_1=m_2 v_2
\)
\(
\frac{v_1}{v_2}=\frac{m_2}{m_1}=\left(\frac{R_2}{R_1}\right)^3 \quad\left(\because m \propto v \propto R^3\right)
\)
\(
\frac{v_1}{v_2}=\left(\frac{2}{1}\right)^3=\frac{8}{1}
\)

Hint

(a) Energy released in one fission of \({ }_{92}^{235} \mathrm{U}\) nucleus \(=200 \mathrm{MeV}\) Mass of uranium \(=1 \mathrm{~g}\)
We know that, 235 g of \({ }^{235} \mathrm{U}\) has \(6.023 \times 10^{23}\) atoms or nuclei.
\(\therefore\) Energy released in fission of 1 g of \(\mathrm{U}^{235}\),
\(
\begin{aligned}
E & =\frac{6.023 \times 10^{23} \times 1 \times 200}{235}=5.1 \times 10^{23} \mathrm{MeV} \\
& =\left(5.1 \times 10^{23} \times 1.6 \times 10^{-13}\right) \mathrm{J} \\
& =\frac{5.1 \times 10^{23} \times 1.6 \times 10^{-13}}{3.6 \times 10^6} \mathrm{kWh}=2.26 \times 10^4 \mathrm{kWh}
\end{aligned}
\)

Hint

(a) We have, \(\frac{d N}{d t}=\lambda N\) and
\(
\lambda=\frac{0.6931}{T_{1 / 2}}=\frac{0.6931}{1620 \times 365 \times 24 \times 60 \times 60}
\)
Total number of atoms, \(N=\frac{6.023 \times 10^{23}}{226}\)
\(
\therefore \frac{d N}{d t}=\frac{0.6931 \times 6.023 \times 10^{23}}{1620 \times 365 \times 24 \times 60 \times 60 \times 226}=3.61 \times 10^{10}
\)

Hint

(b) The relation between decay constant \(\lambda\) and half-life \(T\) can be given as
\(
\begin{array}{rlrl}
& & T & =\frac{0.693}{\lambda} \\
\therefore & \lambda & =\frac{\log _e 2}{T}
\end{array}
\)

Hint

(a) The binding energy of a nucleus is the energy required to take its nucleons away from one another. It is generally expressed as binding energy per nucleon. It is a measure of the stability of the nucleus. Higher the binding energy per nucleon, more stable is the nucleus.

Hint

(d) An \(\alpha\)-particle is a helium nucleus, \({ }_2^4 \mathrm{He}\). A \(\beta\)-particle is an electron, \({ }_{-1}^0 \mathrm{e}\).
An \(\alpha\)-particle emission reduces the atomic number by 2.
An \(\alpha\)-particle emission reduces the mass number by 4.
New atomic number: \(Z^{\prime}=92-2=90\).
New mass number: \(A^{\prime}=238-4=234\).
A \(\beta\)-particle emission increases the atomic number by 1.
A \(\beta\)-particle emission does not change the mass number.
Final atomic number: \(Z^{\prime \prime}=90+1=91\).
Final mass number: \(A^{\prime \prime}=234+0=234\).
The atomic number of the final nucleus is 91 and the mass number is 234.

Hint

(b) Isotope of the original nucleus is produced: Not possible as atomic number always changes in both \(\alpha\) and \(\beta\) decay.

(a) Isobar of original nucleus is produced: Possible in \(\beta\) decay.
(c) Nuclei with higher atomic number than that of the original nucleus is produced: Possible in \(\beta^{-}\)decay.
(d) Nuclei with lower atomic number than that of the original nucleus is produced: Possible in \(\alpha\) and \(\beta^{+}\)decay.

Hint

(c) Suppose a given nucleus is \({ }_Z X^A\).
In \(\alpha\)-decay, the mass number of the product nucleus is four less than that of decaying nucleus, while the atomic number decreases by two.
\(
{ }_Z X^A \xrightarrow{\alpha-\text { decay }}{ }_{Z-2} X^{A-4}+{ }_2 \mathrm{He}^4
\)
In \(\beta\)-decay, the mass number of product nucleus remains same but atomic number increases or decreases by one.
\(
{ }_z X^A \xrightarrow{\beta^{-} \text {decay }} { }_{Z+1}Y^A+{ }_{-1}^0 e+\nu
\)
\(
{ }_Z X^A \xrightarrow{\beta^{+} \text {decay }}{ }_{Z-1} Y^A+{ }_{+1}^0 e+\nu
\)
Gamma decay is the phenomenon of emission of gamma rays photon from a radioactive nucleus.
\(
{ }_z X^A \longrightarrow{ }_z X^A+\gamma
\)
Hence, the given reaction will have sequence as follows
\(
{ }_Z X^A \xrightarrow{\beta^{-} \text {emission }}{ }_{Z+1} Y^A \xrightarrow{\alpha \text {-emission }}{ }_{Z-1} K^{A-4} \xrightarrow{\gamma \text {-emission }}{ }_{Z-1} K^{A-4}
\)

Hint

(b) The control rods used in a nuclear reactor can be made up of cadmium. As they absorb fast moving neutrons.

Hint

(d) The fusion reaction in the sun is a multi-step process in which the hydrogen is burned into helium. Because in fusion, lighter nuclei combine to form heavier nucleus.

Explanation: The fusion reaction in the Sun involves the fusion of hydrogen nuclei (protons) to form helium nuclei. This process is a multi-step reaction, meaning it occurs in several stages. Ultimately, four hydrogen atoms are combined to create one helium atom, releasing a large amount of energy. 

Hint

(b) The nuclear fusion is a very difficult process. Since, the nuclei to be fused are positively charged, they would repel one another strongly. Hence, they must be brought very close together not only by very high pressure but also with high kinetic energy ( \(\simeq 0.1 \mathrm{MeV}\) ). For this, a very high temperature of the order of \(10^8 \mathrm{~K}\) is required. Such high temperature are available in the sun and stars. On the earth, they may be produced by exploding a nuclear fission bomb.

Hint

(a) Power, \(P=\frac{n E}{t}\)
Given, \(n=2 \times 10^{18}\) fission per second and \(E=185 \mathrm{MeV}\)
Here, \(\quad P=\frac{2 \times 10^{18} \times 185 \times 10^6 \mathrm{eV}}{1}\)
\(\Rightarrow \quad=2 \times 10^{18} \times 185 \times 10^6 \times 16 \times 10^{-19}=59.2 \mathrm{MW}\)

Hint

(d) To decrease the energy of fast neutrons to thermal energy.

Explanation: Heavy water is used in a nuclear reactor as a moderator, which means its primary function is to slow down the fast-moving neutrons produced during fission. These slow neutrons are more likely to interact with uranium-235 nuclei and cause further fission, thus sustaining the chain reaction.

Hint

(c) Nucleus of mass number \(A\) has a radius, \(R=R_0 A^{1 / 3}\) where, \(\quad R_0=1.2 \times 10^{-15} \mathrm{~m}\)
\(
\therefore \quad \frac{R_8}{R_{64}}=\left[\frac{8}{64}\right]^{1 / 3}=\left[\frac{1}{8}\right]^{1 / 3}
\)
\(\therefore\) Ratio of volume \(=\frac{1}{8}=0.125\)

Hint

(b) Let \(R_1=N_1 \lambda\) and \(R_2=N_2 \lambda\)
Mean life, \(T=\frac{1}{\lambda}\)
\(
R_1-R_2=\left(N_1-N_2\right) \lambda=\left(N_1-N_2\right) \frac{1}{T}
\)
So, number of atoms disintegrated in ( \(T_2-T_1\) ) is
\(
=N_1-N_2=\left(R_1-R_2\right) T
\)

Hint

\(
\text { (d) Since, } N=13-7=6, C=12-6=6
\)

Isotones are nuclides with the same number of neutrons. 

Anlyzing options:

(a) \({ }_7 \mathrm{~N}^{13},{ }_7 \mathrm{~N}^{14}\) : Both have the same number of protons (7), but their neutron numbers differ ( 6 and 7 , respectively). Thus, they are not isotones.
(b) \({ }_6 \mathrm{C}^{14},{ }_6 \mathrm{C}^{12}\) : Both have the same number of protons (6), but their neutron numbers differ ( 8 and 6 , respectively). Thus, they are not isotones.
(c) \({ }_6 \mathrm{C}^{14},{ }_7 \mathrm{~N}^{14}\) : While both have a mass number of 14 , they have different numbers of protons ( 6 and 7 , respectively). Thus, they are not isotones.
(d) \({ }_7 \mathrm{~N}^{13},{ }_6 \mathrm{C}^{12}\) : Nitrogen-13 has 7 protons and 6 neutrons ( \(13-7=6\) ), and Carbon-12 has 6 protons and 6 neutrons \((12-6=6)\). They have the same number of neutrons (6), making them isotones.

Hint

(c) is a nuclear process which does not depend on external forces.

Explanation
Radioactivity is a nuclear process: Radioactivity originates from the unstable nucleus of an atom, which undergoes spontaneous decay to become more stable.
Independence from external forces: The rate of radioactive decay is generally unaffected by external factors like temperature and pressure. This is because the forces within the nucleus (the strong nuclear force and electrostatic repulsion) that govern stability are not significantly influenced by these external conditions.

Hint

\(
\begin{aligned}
&\text { (a) From, } E=(\Delta m) c^2\\
&\begin{array}{r}
\Delta m=\frac{E}{c^2}=\frac{9 \times 10^{13}}{\left(3 \times 10^8\right)^2}=10^{-3} \mathrm{~kg} \\
\therefore \frac{\Delta m}{m} \times 100=\frac{10^{-3}}{1} \times 100=0.1 \%
\end{array}
\end{aligned}
\)

Hint

(d) We know that, nuclear forces are charge independent and the chief source of stellar energy is fusion.
\(\beta\)-decay is accompanied by both proton and neutron.

Hint

\(
\text { (b) According to the question, }{ }_Z X^A \longrightarrow{ }_{Z-2} Y^{A-4}+{ }_2 \mathrm{He}^4
\)
Let the recoil speed of daughter nucleus is \(v^{\prime}\). By the law of conservation of momentum, we have
Initial momentum \(=\) Final momentum \(\Rightarrow 0=(A-4) v^{\prime}+4 v\)
\(
(A-4) v^{\prime}=-4 v \text { or } v^{\prime}=-\frac{4 v}{(A-4)}
\)
Here, negative sign shows that the recoil speed is in opposite direction to that of \(\alpha\)-particles.

Hint

(c) Given, in 1st two seconds \(100 \beta\) particles and in next two seconds \(50 \beta\) particles are emitted.
\(
\therefore \quad T_{1 / 2}=2 \mathrm{~s}
\)
Hence, mean life, \(T=\frac{t_{1 / 2}}{0.693}=\frac{2}{0.693} \mathrm{~s}\)

Hint

(c) \(\alpha \text {-particle is He nucleus. Addition of an electron will change it into an ion. }\)

Explanation: An alpha particle is essentially a helium nucleus, consisting of two protons and two neutrons. It therefore carries a positive charge of +2 . When an alpha particle captures one electron, its charge changes from +2 to +1 . With a charge of +1 , but retaining its two protons, it becomes a singly ionized helium atom \(\left(\mathrm{He}^{+}\right)\)or, more simply, a helium ion.

Hint

(d) We know that, \(n=\frac{t}{T}\) where, \(T=\) half-life.
Given,
\(
\begin{aligned}
t & =3000 \mathrm{yr} \\
T & =600 \mathrm{yr} \\
n & =\frac{3000}{600}=5 \\
\frac{N}{N_0} & =\left(\frac{1}{2}\right)^n \\
\frac{N}{N_0} & =\left(\frac{1}{2}\right)^5=\frac{1}{32}
\end{aligned}
\)

Hint

(d) The given situation can be shown as
\(
{ }_Z X^A \xrightarrow{9 \alpha}{ }_{Z-18} X^{A-36} \xrightarrow{5 \beta}{ }_{Z-13} X^{A-36}
\)
\(\therefore \quad\) Number of protons, \(P=(Z-13)\) and number of neutrons,
\(
\begin{aligned}
& & N & =(A-36)-(Z-13)=(A-Z-23) \\
\therefore & & \frac{P}{N} & =\frac{(Z-13)}{(A-Z-23)}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (b) We know that, } N=N_0\left(\frac{1}{2}\right)^{\frac{t}{T_{1 / 2}}}\\
&\frac{1}{4}=\left(\frac{1}{2}\right) \Rightarrow\left(\frac{1}{2}\right)^{\frac{t}{10}}=\left(\frac{1}{2}\right)^2 \Rightarrow t=20 \mathrm{yr}
\end{aligned}
\)

Hint

(a) The total mass of the initial particles,
\(
m_i=1.007825+7.016004=8.023829 \mathrm{u}
\)
and the total mass of final particles,
\(
m_f=2 \times 4.002603=8.005206 \mathrm{u}
\)
Difference between initial and final mass of particles,
\(
\Delta m=m_i-m_f=8.023829-8.005206=0.018623 \mathrm{u}
\)
The \(Q\)-value is given by
\(
Q=(\Delta m) c^2=0.018623 \times 931.5=17.35 \mathrm{MeV}
\)

Hint

(b)

\(Q\)-value of the reaction is 5.5 MeV .
i.e. \(\quad K_1+K_2=5.5 \mathrm{MeV} \dots(i)\)
By conservation of linear momentum, \(p_1=p_2\)
\(
\begin{aligned}
\Rightarrow \sqrt{2(216) K_1} & =\sqrt{2(4) K_2} \\
K_2 & =54 K_1 \dots(ii)
\end{aligned}
\)
On solving Eqs. (i) and (ii), we get
The energy of emitted \(\boldsymbol{\alpha}\)-particle, \(K_2=5.4 \mathrm{MeV}\)

Hint

(c) We have, \(\lambda=\frac{0.693}{T_{1 / 2}}[latex] Mean life, [latex]\tau=\frac{1}{\lambda}=\frac{t_{1 / 2}}{0.693}=\frac{2}{2.303 \log 2}=\frac{2}{\ln 2} \mathrm{~s}\)

Hint

(c) Energy and mass are related by the relation, \(E=m c^2\)
Differentiating, we get
\(
\begin{gathered}
\frac{d E}{d t}=m \times[0]+c^2 \frac{d m}{d t} \\
\Rightarrow \quad 3.6 \times 10^{28}=9 \times 10^{16} \frac{d m}{d t} \Rightarrow \frac{d m}{d t}=4 \times 10^{11} \mathrm{~kg} / \mathrm{s}
\end{gathered}
\)

Hint

(c) From the relation, \(R=R_0 e^{-\lambda t}\) (where, \(R=\) activity of sample)
Given, \(R_0=5400 \mathrm{dis} / \mathrm{min}\)
\(
\begin{aligned}
R & =600 \mathrm{dis} / \mathrm{min} \text { and } t=5 \mathrm{~min} \\
\therefore \quad 600 & =5400 e^{-5 \lambda} \\
\frac{1}{9} & =e^{-5 \lambda} \Rightarrow \lambda=\frac{\ln 9}{5}
\end{aligned}
\)
\(\therefore\) Half-life of the sample, \(T_{1 / 2}=\frac{\ln 2}{\frac{\ln 9}{5}}=\frac{5 \ln 2}{\ln 9}=\frac{\ln 2^5}{\ln 9}=\frac{\ln 32}{\ln 9}\)