Past NEET Entrance Papers

Hint

(c) Step 1: Effect of Alpha ( \(\alpha\) ) Decay
An alpha particle \((\alpha)\) has a mass number of 4 and an atomic number of 2 .
When element \(X\) undergoes alpha decay:
\(
{ }_{82}^{290} X \xrightarrow_{80}^{286} Y+{ }_2^4 \alpha
\)
Thus, the new nucleus \(Y\) has:
Mass number \(=290-4=286\)
Atomic number \(=82-2=80\)
Step 2: Effect of Positron \(\left(e^{+}\right)\)Emission
Positron emission ( \(\beta^{+}\)) decreases the atomic number by 1 without changing the mass number:
\(
{ }_{80}^{286} Y \xrightarrow_{79}^{286} Z+e^{+}
\)
Thus, \(Z\) has:
Mass number \(=286\) (unchanged)
Atomic number \(=80-1=79\)
Step 3: Effect of Beta-Minus ( \(\beta^{-}\)) Decay
Beta-minus ( \(\beta^{-}\)) emission increases the atomic number by 1 without affecting the mass number:
\(
{ }_{79}^{286} Z\xrightarro_{80}^{286} P+\beta^{-}
\)
Thus, \(P\) has:
Mass number \(=286\) (unchanged)
Atomic number \(=79+1=80\)
Step 4: Effect of Electron Capture ( \(e^{-}\))
Electron capture decreases the atomic number by 1, keeping the mass number unchanged:
\(
{ }_{80}^{286} P+e^{-}\xrightarrow_{81}^{286} Q
\)
Thus, \(Q\) has:
Mass number \(=286\) (unchanged)
Atomic number \(=80+1=81\)
The final product \(Q\) has mass number 286 and atomic number 81.

Hint

(c) \(1: 1[latex] and [latex]14: 1\).

Explanation:
Nuclear Density:
Nuclear density is roughly constant for all nuclei, meaning it’s independent of the mass number. Therefore, the ratio of densities between two nuclei is always 1:1.

Nuclear Volume:
The volume of a nucleus is proportional to the cube of its radius, and the radius is proportional to the cube root of the mass number (A). Therefore, if the ratio of mass numbers \(A 1 / A 2=56 / 4=14\), then the ratio of the nuclear volumes is \(14: 1\).

Hint

(c) Alpha particle emission reduces the atomic number by 2 and the mass number by 4.
Beta minus ( \(e^{-}\)) emission increases the atomic number by 1 and does not change the mass number.
Beta plus ( \(e^{+}\)) emission decreases the atomic number by 1 and does not change the mass number.
Gamma radiation emission does not change the atomic number or the mass number.
Step 1: Identify the first emitted particle
The atomic number decreases by 2 and the mass number decreases by 4 . This corresponds to the emission of an alpha particle ( \(\alpha\) ).

Step 2: The atomic number increases by 1 and the mass number remains the same. This corresponds to the emission of a beta minus particle ( \(e^{-}\)).

Step 3: The atomic number decreases by 2 and the mass number decreases by 4 . This corresponds to the emission of an alpha particle ( \(\alpha\) ).

Step 4: The atomic number decreases by 1 and the mass number remains the same. This corresponds to the emission of a beta plus particle \(\left(e^{+}\right)\).

Step 5: The atomic number and mass number remain the same.
This corresponds to the emission of gamma radiation ( \(\gamma\) ).

The sequence of emitted particles or radiations is \(\alpha, e^{-}, \alpha, e^{+}, \gamma\).

Hint

(b) High specific heat capacity

Explanation:
Water is used as a coolant in nuclear reactors primarily due to its high specific heat capacity. This property allows water to absorb a significant amount of heat without a large increase in temperature. making it effective in regulating the reactor’s temperature and preventing overheating.

Hint

(a) (A) Slow neutron can cause fission in \({ }_{92}^{235} \mathrm{U}\) than fast neutrons because fast neutrons are too quick so they scatter of atoms instead of being captured by them
(B) \(\alpha\)-rays are Helium nuclei, is a true statement
(C) \(\beta\)-rays are generated when neutron is converted into proton by releasing electron or proton is converted into neutron by releasing positron.
(D) \(\gamma\)-rays have higher energies as compared to \(X\)-rays. So they have smaller wavelength as compared to X-rays.

Hint

(d) To solve the nuclear reaction given as \(\mathrm{Na}_{11}^{22} \rightarrow X+e^{+}+\nu\), we need to determine the element \(X\).
Identify the Initial Element:
The initial element is Sodium ( Na ) with atomic number 11 and mass number 22, represented as \({ }_{11}^{22} \mathrm{Ne}\)[/latex].
2. Understand the Reaction:
The reaction indicates that Sodium is undergoing a beta-plus decay ( \(\beta+\) decay), where a positron ( \(e^{+}\)) and a neutrino ( \(\nu\) ) are emitted.
3. Apply the Conservation Laws:
Mass Number Conservation: The mass number before the reaction must equal the mass number after the reaction. Since the positron and neutrino have negligible mass, the mass number of \(X\) must remain 22 .
Atomic Number Conservation: The atomic number decreases by 1 due to the emission of the positron. Therefore, the atomic number of \(X\) will be \(11-1=10\).
4. Determine the Element:
The element with atomic number 10 is \(\mathrm{Neon}(\mathrm{Ne})\). Therefore, we can write:
\(
X=\mathrm{Ne}_{10}^{22}
\)
5. Final Answer: The final answer for the element \(X\) is \({ }_{10}^{22} \mathrm{Ne}\).
\({ }_{10}^{22} \mathrm{Ne}\)

Hint

(d) The distance from the center to the point where the density has reduced to half its initial value is known as the nuclear radius, or \(R\). These radii can be measured experimentally by fast neutron deflection tests.
For this nuclear radii are given as \(R=R_0 A^{1 / 3}\)
where \(R=\) nuclear radii, \(A=\) atomic mass number, \(R_0=1.3 \mathrm{FM}\) which is constant for every nuclei.
Calculation:
Given:
Atomic mass number of nuclei \(1\left(\mathrm{~A}_1\right)=125\)
atomic mass number of nuclei \(2\left(\mathrm{~A}_2\right)=64\)
the ratio of the radius of two nuclei is \(=\frac{R(125)}{R(64)}=\frac{R_0 A_1^{1 / 3}}{R_0 A_2^{1 / 3}}\)
\(
\Rightarrow \frac{R(125)}{R(64)}=(125 / 64)^{1 / 3}=5 / 4
\)

Hint

(b) Mass number of the original nucleus: 240
Mass number of each fragment: 120
Binding energy per nucleon of the original nucleus: 7.6 MeV
Binding energy per nucleon of each fragment: 8.5 MeV
The initial total binding energy is the product of the mass number and the binding energy per nucleon of the original nucleus.
\(
E_{\text {initial }}=240 \times 7.6 \mathrm{MeV}=1824 \mathrm{MeV}
\)
Since the nucleus breaks into two fragments, each with a mass number of 120 , the final total binding energy is the sum of the binding energies of the two fragments.
\(
E_{\text {final }}=2 \times(120 \times 8.5 \mathrm{MeV})=2 \times 1020 \mathrm{MeV}=2040 \mathrm{MeV}
\)
The gain in binding energy is the difference between the final and initial total binding energies.
\(
\Delta E=E_{\text {final }}-E_{\text {initial }}=2040 \mathrm{MeV}-1824 \mathrm{MeV}=216 \mathrm{MeV}
\)
The total gain in binding energy is 216 MeV.

Hint

(a)

\(
{ }_Z^A X \xrightarrow{\beta^{+}} Z_{-1} B \xrightarrow{\alpha} Z_{-3} C \xrightarrow{\beta^{-}} Z_{-2} D
\)
\(\beta^{+}\)decreases atomic number by 1
\(\alpha\) decreases atomic number by 2
\(\beta^{-}\)increases atomic number by 1.

Hint

(a)

\(
\begin{aligned}
&\text { From mass-energy equivalence, }\\
&\begin{aligned}
& E=m c^2 \\
& =0.5 \times 10^{-3} \times 9 \times 10^{16} \\
& =4.5 \times 10^{13} \mathrm{~J}
\end{aligned}
\end{aligned}
\)

Hint

(d) Suppose after bombarding neutron we get \(X_Z^A\) along with \({ }_{36}^{89} \mathrm{Kr}\) and 3 neutrons. The reaction equation looks like;
\(
U_{92}^{235}+{ }_0^1 n \rightarrow K r_{36}^{89}+3 n_0^1+X_Z^A
\)
where \(\mathrm{A}=\) Atomic mass number and, Z is the Atomic number.
Now, \(\Sigma\) Atomic No in left-hand side \(=92+0=92\)
\(\mathbf{\Sigma A t o m i c}\) No in right-hand side \(=36+3 \times 0+Z=36+Z\)
\(\because\) Atomic No must be conserved.
\(\therefore \Sigma\) Atomic No on the left-hand side \(=\Sigma\) Atomic No on the right-hand side
\(
\begin{aligned}
& \Rightarrow 36+Z=92 \\
& \Rightarrow Z=56
\end{aligned}
\)
Also from the conservation of Mass-Number we have;
\(\Sigma\) Atomic Mass No on the left-hand side =\(\Sigma\) Atomic Mass No on the right-hand side
\(
\begin{aligned}
&\begin{aligned}
& \Rightarrow 235+1=89+3+A \\
& \Rightarrow A=144
\end{aligned}\\
&\text { So, }{ }_{56}^{144} X \text { is generated where } \mathrm{X}=\mathrm{Ba} \text {. }
\end{aligned}
\)

Hint

(b) When an element emits gamma radiation, neither its mass number nor its atomic number changes. Gamma rays are electromagnetic waves with no mass or charge, and their emission only involves the nucleus shifting from a higher en ergy state to a lower one, without altering the number of pro tons or neutrons.

Hint

(d) Since, \(\alpha\)-particle is equivalent to He -atom nucleus, hence it has two protons and two neutrons only.

Hint

(c) As we know, \(\mathrm{R}=\mathrm{R}_0(\mathrm{~A})^{1 / 3}\) where \(\mathrm{A}=\) mass number
\(
\begin{aligned}
& \mathrm{R}_{\mathrm{Al}}=\mathrm{R}_0(27)^{1 / 3}=3 \mathrm{R}_0 \\
& \mathrm{R}_{\mathrm{Te}}=\mathrm{R}_0(125)^{1 / 3}=5 \mathrm{R}_0=\frac{5}{3} \mathrm{R}_{\mathrm{Al}}
\end{aligned}
\)

Hint

(d) The binding energy per nucleon of \({ }_3^7 \mathrm{Li}\) is 5.60 MeV.
The binding energy per nucleon of \({ }_2^4 \mathrm{He}\) is 7.06 MeV.
The number of nucleons in \({ }_3^7 \mathrm{Li}\) is 7.
The number of nucleons in \({ }_2^4 \mathrm{He}\) is 4.
The total binding energy of the reactants (lithium and hydrogen) is calculated as follows:
For \({ }_3^7 \mathrm{Li}\) :
Total Binding Energy of Li \(=\) Binding Energy per Nucleon \(\times \text { Number of Nucleons }=5.60 \mathrm{MeV} \times 7=39.2 MeV\)
For \({ }_1^1 \mathrm{H}:\)
Total Binding Energy of \(\mathrm{H}=0 \mathrm{MeV}\) (since H has no binding energy)
Thus, the total binding energy of the reactants is:
Total Binding Energy of Reactants \(=39.2 \mathrm{MeV}+0 \mathrm{MeV}=39.2 \mathrm{MeV}\)
The total binding energy of the products (two \({ }_2^4 \mathrm{He}\) nuclei) is calculated as follows:
Total Binding Energy of Products \(=2 \times \text { Binding Energy per Nucleon of } \mathrm{He} \times \text { Number of Nucleons in } \mathrm{He}=2 \times 7.06 \mathrm{MeV} \times 4\)
\(
=56.48 \mathrm{MeV}
\)
The energy \(Q\) released in the reaction can be calculated using the difference in binding energies:
\(Q=\) Total Binding Energy of Products – Total Binding Energy of Reactants

Substituting the values:
\(
Q=56.48 \mathrm{MeV}-39.2 \mathrm{MeV}=17.28 \mathrm{MeV}
\)
Thus, the value of energy \(Q\) released is approximately 17.3 MeV.

Hint

(b) Given
Mass defect \(=0.2886 u\)
Total energy liberated \(=0.2886 \times 931 \mathrm{MeV}=26.86 \mathrm{MeV}\)
We have been asked the energy liberated per unit mass
atomic mass of helium \(=4 \mathrm{u}\)
So, Energy liberated per unit mass \(=26.86 \mathrm{MeV} / 4\)
\(
=6.675 \mathrm{MeV}
\)

Hint

(d) From the graph of BE/A versus mass number A it is clear that, BE/A first increases and then decreases with increase in mass number.

Hint

(c) The radius of the nuclears is directly proportional to cube root of atomic number i.e. \(R \propto A^{1 / 3}\)
\(R=R_0 A^{1 / 3}\), where \(R_0\) is a constant of proportionality
\(
\frac{R_2}{R_1}=\left(\frac{A_2}{A_1}\right)^{1 / 3} \Rightarrow\left(\frac{64}{27}\right)^{1 / 3}=\frac{4}{3}
\)
where \(R_1=\) the radius of \({ }^{27} \mathrm{Al}\), and \(A_1=\) Atomic mass number of \(A 1\) \(R_2=\) the radius of \({ }^{64} \mathrm{Cu}\) and \(A_2=\) Atomic mass number of Cu
\(
R_2=3.6 \times \frac{4}{3}=4.8 \mathrm{~m}
\)

Hint

(b) The energy released is the product of power and time:
\(E=P \times t\)
\(E=10^6 \mathrm{~W} \times 3600 \mathrm{~s}\)
\(E=3.6 \times 10^9 \mathrm{~J}\)

Step 2: Using \(E=m c^2\), solve for \(m\) :
\(m=\frac{E}{c^2}\)
\(m=\frac{3.6 \times 10^9 \mathrm{~J}}{\left(3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)^2}\)
\(m=\frac{3.6 \times 10^9}{9 \times 10^{16}} \mathrm{~kg}\)
\(m=4 \times 10^{-8} \mathrm{~kg}\)

Step 3: Convert kilograms to micrograms:
\(m=4 \times 10^{-8} \mathrm{~kg} \times \frac{10^3 \mathrm{~g}}{1 \mathrm{~kg}} \times \frac{10^6 \mu \mathrm{~g}}{1 \mathrm{~g}}\)
\(m=40 \mu \mathrm{~g}\)
The mass decay of \(\mathrm{U}^{235}\) per hour is approximately \(40 \mu \mathrm{~g}\).

Hint

(c) At high temperatures, the atoms have a lot of kinetic energy, which allows them to move fast enough to overcome the electrostatic repulsion between the positively charged nuclei, enabling them to get close enough to fuse together.

Hint

(b) Momentum
\(
\mathrm{Mu}=\frac{\mathrm{E}}{\mathrm{c}}=\frac{\mathrm{h} \nu}{\mathrm{c}}
\)
Recoil energy
\(
\frac{1}{2} \mathrm{Mu}^2=\frac{1}{2} \frac{\mathrm{M}^2 \mathrm{u}^2}{\mathrm{M}}=\frac{1}{2 \mathrm{M}}\left(\frac{\mathrm{~h} \nu}{\mathrm{c}}\right)^2=\frac{\mathrm{h}^2 \nu^2}{2 \mathrm{Mc}^2}
\)

Hint

(a) When an alpha particle \(\left({ }_2^4 \mathrm{He}\right)\) is emitted, the mass number and the mass number and the atomic number of the daughter nucleus decreases by four and two respectively. When a beta particle \(\left(\beta^{-}\right)\)is emitted, the atomic number of the daughter nucleus increases by one but the mass number remains the same.
\(
\therefore \quad{ }_n^m X \xrightarrow{\alpha}{ }_{n-2}^{m-4} Y \xrightarrow{2 \beta^{-}}{ }_n^{m-4} X
\)

Hint

(b) Here, \(\Delta m=0.042 a . m . u\)
\(
\begin{aligned}
& \mathrm{BE}=\Delta m \times 931.5 \mathrm{MeV} \\
& =0.042 \times 931.5 \mathrm{MeV}=39.123 \mathrm{MeV} \\
& \text { Number of nucleons in }{ }_3^7 L i \text { is } 7 \text {. } \\
& \mathrm{BE} / \text { nucleon }=\frac{39.123}{7}\simeq 5.6 \mathrm{MeV}
\end{aligned}
\)

Hint

(d) Binding energy of two \({ }_1 H^2\) nuclei
\(
=2(1.1 \times 2)=4.4 \mathrm{MeV}
\)
Binding energy of one \({ }_2 \mathrm{He}^4\) nucelus
\(
\begin{aligned}
& =4 \times 7.0=28 \mathrm{MeV} \\
& \therefore \text { Energy released in fusion }=28-4.4=23.6 \mathrm{MeV}
\end{aligned}
\)

Hint

(c) Given that:
Initially, the radioactive element has undergone \(\alpha\)-decay and then \(\beta\)-decay
No: of \(\beta\) particles \(=2 \times\) No: of \(\alpha\) particles
Radioactive decay is as follows:
\(1 \alpha\)-decay: \({ }_Z^A \mathrm{X} \rightarrow{ }_{Z-4}^{A-4} \mathrm{Y}+{ }_2^4 \alpha\)
\(2 \beta\)-decay: \({ }_{Z-2}^{A-4} \mathrm{Y} \rightarrow{ }_{Z-1}^{A-4} \mathrm{Y} \rightarrow{ }_Z^{A-4} \mathrm{Y}\)
The daughter nucleus after \(\alpha\)-decay has its mass number reduced by 4.
There will be no change in the reduced mass number after \(\beta\)-decay.
The obtained daughter nucleus has a different mass number \(=(A-4)\) and the same atomic number \(=\mathrm{Z}\) as the parent nucleus.
Therefore, the resulting daughter is an isotope of the parent.

Hint

(a)

\(
\text { Mass defect }={ZM}_{{p}}+({A}-{Z}) {M}_{{n}}-{M}({~A}, {Z})
\)
\(
\text { B. E. }=\left[Z M_p+(A-Z) M_n-M(A, Z)\right] c^2
\)
\(
\therefore \frac{B . E .}{c^2}=Z M_p+(A-Z) M_n-M(A, Z)
\)
\(
M(A, Z)=Z M_p+(A-Z) M_n-B . E . / c^2
\)

Hint

(d) The radius of a nucleus is given by \(r=r_0 A^{\frac{1}{3}}\), where \(r_0\) is a constant and \(A\) is the mass number.
The volume of a sphere is given by \(V=\frac{4}{3} \pi r^3\).
Density is defined as mass per unit volume: \(\rho=\frac{m}{V}\).
Nuclear density is approximately constant for all nuclei.
Let \(A_1\) and \(A_2\) be the mass numbers of the two nuclei. Given \(\frac{A_1}{A_2}=\frac{1}{3}\).
The radii of the nuclei are \(r_1=r_0 A_1^{\frac{1}{3}}\) and \(r_2=r_0 A_2^{\frac{1}{3}}\).
The ratio of the radii is \(\frac{r_1}{r_2}=\frac{r_0 A_1^{\frac{1}{3}}}{r_0 A_2^{\frac{1}{3}}}=\left(\frac{A_1}{A_2}\right)^{\frac{1}{3}}=\left(\frac{1}{3}\right)^{\frac{1}{3}}\).
The volumes of the nuclei are \(v_1=\frac{4}{3} \pi r_1^3\) and \(v_2=\frac{4}{3} \pi r_2^3\).
The ratio of the volumes is \(\frac{V_1}{V_2}=\frac{\frac{4}{3} \pi r_1^3}{\frac{4}{3} \pi r_2^3}=\left(\frac{r_1}{r_2}\right)^3=\left(\left(\frac{1}{3}\right)^{\frac{1}{3}}\right)^3=\frac{1}{3}\).
The densities of the nuclei are \(\rho_1=\frac{m_1}{V_1}\) and \(\rho_2=\frac{m_2}{V_2}\).
Since the mass of a nucleus is proportional to its mass number, we can write \(m_1=k A_1\) and \(m_2=k A_2\), where \(k\) is a constant.
The ratio of the densities is \(\frac{\rho_1}{\rho_2}=\frac{\frac{m_1}{V_1}}{\frac{m_2}{V_2}}=\frac{m_1}{m_2} \cdot \frac{V_2}{V_1}=\frac{A_1}{A_2} \cdot \frac{V_2}{V_1}=\frac{1}{3} \cdot \frac{3}{1}=1\).
The ratio of the nuclear densities is \(1: 1\).

Hint

(d) It has been known that a nucleus of mass number \(A\) has radius
\(
R=R_0 A^{1 / 3},
\)
where \(R_0=1.2 \times 10^{-15} \mathrm{~m}\)
and \(A=\) mass number
In case of \({ }_{13}^{27} \mathrm{~A}\), let nuclear radius be \(R_1\)
and for \({ }_{32}^{125} \mathrm{Te}\), nuclear radius be \(R_2\)
For \({ }_{13}^{27} \mathrm{Al}, R_1=R_0(27)^{1 / 3}=3 R_0\)
For \({ }_{32}^{125} \mathrm{Te}, R_2=R_0(125)^{1 / 3}=5 R_0\)
\(
\frac{R_2}{R_1}=\frac{5 R_0}{3 R_0}=\frac{5}{3} R_1=\frac{5}{3} \times 3.6=6 \mathrm{fm}
\)

Hint

(b) Beta decay can involve the emission of either electrons or positrons. The electrons or positrons emitted in a \(\beta\)-decay do not exist inside the nucleus. They are only created at the time of emission, just as photons are created when an atom makes a transition from higher to a lower energy state.

In negative \(\beta\)-decay a neutron in the nucleus is transformed into a proton, an electron and an antineutrino.
Hence, in radioactive decay process. the negatively charged emitted \(\beta\)-particles are the electrons produced as a result of the decay of neutrons present inside the nucleus.

Hint

(b) The difference in mass of a nucleus and its constituents, \(\Delta M\), is called the mass defect and is given by
\(
\begin{aligned}
& \Delta M=\left[Z M_p+(A-Z) M_n\right]-M \\
& \text { and binding energy }=\Delta M c^2 \\
& =\left[\left\{Z M_p+(A-Z) M_n\right\}-M\right] c^2
\end{aligned}
\)

Hint

(b) Binding energy of deuteron: 2.2 MeV
Binding energy of \({ }_2 \mathrm{He}^4: 28 \mathrm{MeV}\)
The binding energy of one deuteron is 2.2 MeV.
The total binding energy of two deuterons is \(2 \times 2.2 \mathrm{MeV}=4.4 \mathrm{MeV}\).
The energy released is the difference between the binding energy of \({ }_2 \mathrm{He}^4\) and the total binding energy of the two deuterons.
Energy released \(=28 \mathrm{MeV}-4.4 \mathrm{MeV}=23.6 \mathrm{MeV}\).
The energy released is 23.6 MeV.

Hint

(c) We use the formula,
\(
R=R_0 A^{1 / 3}
\)
This represents relation between atomic mass and radius of the nucleus.
For berillium,
\(
R_1=R_0(9)^{1 / 3}
\)
For germanium,\(\quad R_2=R_0 A^{1 / 3}\)
\(
\begin{aligned}
& \frac{R_1}{R_2}=\frac{(9)^{1 / 3}}{(A)^{1 / 3}} \Rightarrow \frac{1}{2}=\frac{(9)^{1 / 3}}{(A)^{1 / 3}} \\
& \Rightarrow \frac{1}{8}=\frac{9}{A} \Rightarrow A=8 \times 9=72
\end{aligned}
\)

Hint

(d) The binding energy per nucleon for the middle nuclides (from \(A=20 \rightarrow A=56\) ) is maximum. Hence, these are more stable.

As the mass number increases, the binding energy per nucleon gradually decreases and ultimately binding energy per nucleon of heavy nuclides (such as uranium, etc) is comparatively low. Hence, these nuclides are relatively unstable. So, they can be fissioned easily.

Hint

(c) Binding energy per nucleon for fission products is higher relative to Binding energy per nucleon for parent nucleus, i.e., more masses are lost and are obtained as kinetic energy of fission products. So, the given ratio \(<1\).

Hint

(c) Binding energy of \({ }_1^2 \mathrm{H}\) is \(a~ \mathrm{MeV}\).
Binding energy of \({ }_1^3 \mathrm{H}\) is \(b~ \mathrm{MeV}\).
Binding energy of \({ }_2^4 \mathrm{He}\) is \(c~ \mathrm{MeV}\).
The total binding energy of the reactants is the sum of the binding energies of \({ }_1^2 \mathrm{H}\) and \({ }_1^3 \mathrm{H}\).
\(
E_{\text {reactants }}=a+b
\)
The total binding energy of the products is the binding energy of \({ }_2^4 \mathrm{He}\).
\(
E_{\text {products }}=c
\)
The energy released is the difference between the binding energies of the products and the reactants.
\(
\begin{aligned}
& Q=E_{\text {products }}-E_{\text {reactants }} \\
& Q=c-(a+b) \\
& Q=c-a-b
\end{aligned}
\)
The energy released in the reaction is \(c-a-b \mathrm{MeV}\).

Hint

(b) Step 1: Understand the Binding Energy
The binding energy \(B\) of a nucleus is the energy required to disassemble the nucleus into its constituent protons and neutrons. It can be expressed in terms of the mass of the nucleus and the masses of the individual nucleons (protons and neutrons).

Step 2: Write the Mass of Constituent Particles
The total mass of the individual nucleons (protons and neutrons) in the nucleus can be calculated as:
Total mass of nucleons \(=Z \cdot M_p+N \cdot M_n\)
where \(M_p\) is the mass of a proton and \(M_n\) is the mass of a neutron.
Step 3: Relate Binding Energy to Mass
The binding energy \(B\) can be related to the mass defect (the difference between the mass of the individual nucleons and the mass of the nucleus) by the equation:
\(
B=(\text { mass of nucleons })-(\text { mass of nucleus }) \cdot c^2
\)
Rearranging this gives us:
\(
B=\left(Z \cdot M_p+N \cdot M_n-M(N, Z)\right) \cdot c^2
\)
Step 4: Solve for the Mass of the Nucleus
Rearranging the equation for \(M(N, Z)\) :
\(
M(N, Z)=Z \cdot M_p+N \cdot M_n-\frac{B}{c^2}
\)
Final Expression
Thus, the mass of the nucleus can be expressed as:
\(
M(N, Z)=Z \cdot M_p+N \cdot M_n-\frac{B}{c^2}
\)

Hint

(a) \({ }_Z^A \mathrm{X}\) has number of protons \(=Z\) and
number of neutrons \(N=A-Z\) where \(A\) is the total number of protons and neutron,
i.e., \(A=Z+N\).

Hint

(d) \(m_1\) and \(m_2\) are the masses of the fusing nuclei.
\(m_3\) is the mass of the resultant nucleus.
In nuclear fusion, mass is converted into energy according to Einstein’s equation \(E=m c^2\).
The mass of the resultant nucleus is less than the sum of the masses of the fusing nuclei.
In nuclear fusion, a small amount of mass is converted into energy.
This means that the mass of the resultant nucleus is less than the sum of the masses of the fusing nuclei.
\(
m_3<m_1+m_2
\)

Hint

(a)

\(
\text { Nuclear reaction equation: }{ }_Z^A \mathrm{X} \rightarrow{ }_{Z+1}^A \mathrm{Y}+{ }_{-1}^0 \mathrm{e}+\nu
\)
\(\beta\)-decay involves the emission of an electron (or positron) and a neutrino (or antineutrino).
\(\gamma\)-decay involves the emission of a gamma ray photon.
Fusion involves the combining of two or more atomic nuclei to form a heavier nucleus.
Fission involves the splitting of an atomic nucleus into two or more lighter nuclei.
The equation shows a nucleus X transforming into a nucleus Y , an electron \({ }_{-1}^0 \mathrm{e}\), and a neutrino \(\nu\).
The atomic number \(Z\) increases by 1 , while the mass number \(A\) remains the same.
\(\beta\)-decay involves the emission of an electron or a positron. \(\gamma\)-decay involves the emission of a gamma ray photon. Fusion involves combining nuclei.
Fission involves splitting a nucleus.
The emission of an electron and a neutrino, along with the increase in atomic number by 1 while the mass number remains the same, is characteristic of \(\beta\)-decay.

Hint

(c) Mass number of an element is the total number of protons and neutrons present inside the atomic nucleus of the element. It is represented by A, A is different for different elements. Mass number of a nucleus is sometimes equal to its atomic number,for example in case of hydrogen, number of neutrons= 0. So, mass number = atomic number.

Hint

(c) Mass of proton: \(m_p=1.0073 \mathrm{u}\)
Mass of neutron: \(m_n=1.0087 \mathrm{u}\)
Mass of Helium-4 nucleus: \(m_{H e}=4.0015 \mathrm{u}\)
Helium- 4 has 2 protons and 2 neutrons.
\(
1 \mathrm{u}=931.5 \frac{\mathrm{MeV}}{c^2}
\)
Binding energy formula: \(E_b=\Delta m c^2\), where \(\Delta m\) is the mass defect.
Mass defect formula: \(\Delta m=\left(Z m_p+N m_n\right)-m_{\text {nucleus }}\), where \(Z\) is the number of protons, \(N\) is the number of neutrons, and \(m_{\text {nucleus }}\) is the mass of the nucleus.
Total mass of protons: \(2 \times m_p=2 \times 1.0073 \mathrm{u}=2.0146 \mathrm{u}\)
Total mass of neutrons: \(2 \times m_n=2 \times 1.0087 \mathrm{u}=2.0174 \mathrm{u}\)
Total mass of nucleons: \(2.0146 \mathrm{u}+2.0174 \mathrm{u}=4.032 \mathrm{u}\)
\(
\text { Mass defect: } \Delta m=\left(2 m_p+2 m_n\right)-m_{H e}=4.032 \mathrm{u}-4.0015 \mathrm{u}=0.0305 \mathrm{u}
\)
\(
\text { Binding energy: } E_b=\Delta m \times 931.5 \frac{\mathrm{MeV}}{c^2} \times c^2=0.0305 \mathrm{u} \times 931.5 \frac{\mathrm{MeV}}{\mathrm{u}}=28.40075 \mathrm{MeV}
\)
\(
\text { The binding energy of the Helium-4 nucleus is approximately } 28.4 \mathrm{MeV} \text {. }
\)

Hint

(c) Cosmic rays, \(\gamma\)-rays, and \(X\)-rays are part of electromagnetic spectrum, while \(\beta\)-rays are emitted by radioactive elements. Hence \(\beta\)-rays is not electromagnetic waves.

Hint

(d) Radius of atom \(r_a \approx 10^{-10} \mathrm{~m}\).
Radius of nucleus \(r_n \approx 10^{-15} \mathrm{~m}\).
Volume of a sphere is \(V=\frac{4}{3} \pi r^3\).
The volume of the atom is:
\(V_a=\frac{4}{3} \pi r_a^3\)
\(V_a=\frac{4}{3} \pi\left(10^{-10} \mathrm{~m}\right)^3\)
\(\quad V_a=\frac{4}{3} \pi \cdot 10^{-30} \mathrm{~m}^3\)
The volume of the nucleus is:
\(V_n=\frac{4}{3} \pi r_n^3\)
\(V_n=\frac{4}{3} \pi\left(10^{-15} \mathrm{~m}\right)^3\)
\(V_n=\frac{4}{3} \pi \cdot 10^{-45} \mathrm{~m}^3\)
The ratio of the volume of the atom to the volume of the nucleus is:
\(\frac{V_a}{V_n}=\frac{\frac{4}{3} \pi \cdot 10^{-30} \mathrm{~m}^3}{\frac{4}{3} \pi \cdot 10^{-45} \mathrm{~m}^3}\)
\(\frac{V_a}{V_n}=\frac{10^{-30}}{10^{-45}}\)
\(\frac{V_a}{V_n}=10^{-30-(-45)}\)
\(\frac{V_a}{V_n}=10^{15}\)

The volume occupied by an atom is greater than the volume of the nucleus by a factor of about \(10^{15}\).

Hint

(d) As a result of fusion, enormous amount of heat is liberated which is the main cause of source of solar energy.

Hint

(d) Deuteron: \({ }_{\mathrm{1}} \mathrm{H}^2\)
Oxygen-16 nucleus: \({ }_8 \mathrm{O}^{16}\)
Alpha particle: \({ }_2 \mathrm{He}^4\)
The reaction is: \({ }_1 \mathrm{H}^2+{ }_8 \mathrm{O}^{16} \rightarrow{ }_2 \mathrm{He}^4+{ }_z \mathrm{X}^A\)
Where \({ }_Z \mathrm{X}^{\boldsymbol{A}}\) is the unknown product nucleus.

The total mass number on the left side is \(2+16=18\).
The total mass number on the right side is \(4+A\).
Therefore, \(18=4+A\), so \(A=18-4=14\).

The total atomic number on the left side is \(1+8=9\).
The total atomic number on the right side is \(2+Z\).
Therefore, \(9=2+Z\), so \(Z=9-2=7\).

With \(A=14\) and \(Z=7\), the product nucleus is \({ }_7 \mathrm{~N}^{14}\).
The product nucleus is nitrogen-14.

Hint

(a) For nuclear fusion process the nuclei with low mass are suitable.

Hint

(a) Initial nucleus \({ }_a^b X\)
Decays:
Positron emission
Two \(\alpha\) particles emission
Two \(\beta\) particles emission
One \(\alpha[l/atex] particle emission

Final nucleus [latex]{ }_d^c \mathrm{Y}\)
\(
\text { Positron emission: }{ }_1^1 \mathrm{e} \rightarrow{ }_{-1}^0 \mathrm{e}+{ }_{\mathrm{p}}^1 \mathrm{1} \text { (atomic number decreases by } 1 \text { ) }
\)
\(\alpha\) particle emission: \({ }_2^4 \mathrm{He}\) (atomic number decreases by 2 , mass number decreases by 4)
\(\beta\) particle emission: \({ }_0^{\mathrm{1}} \mathrm{n} \rightarrow{ }_1^1 \mathrm{p}+{ }_{-1}^0 \mathrm{e}\) (atomic number increases by 1 )
Positron emission:
Change in atomic number: \(a-1\)
Change in mass number: \(b-0\)
\(\alpha\) particle emission (2 times):
Change in atomic number: \(a-4\)
Change in mass number: \(b-8\)
\(\beta\) particle emission (2 times):
Change in atomic number: \(a+2\)
Change in mass number: \(b-0\)
\(\alpha\) particle emission (1 time):
Change in atomic number: \(a-2\)
Change in mass number: \(b-4\)
Total change in atomic number: \(a-1-4+2-2=a-5\)
Total change in mass number: \(b-8-4=b-12\)
The final nucleus is \({ }_{a-5}^{b-12} \mathrm{Y}\)
\(
\begin{aligned}
& c=b-12 \\
& d=a-5
\end{aligned}
\)
Compare the result with the options
The correct relation is \(c=b-12, d=a-5\).

Hint

(b) In nuclear fission, some mass is converted into energy, and the total binding energy \((\mathrm{BE})\) of the fragments is greater than the BE of the parental element. This difference in \(B E\) represents the energy released in the fission process.

Hint

(a) Given : Mass of neutron \(=M_n\)

Mass of proton \(=M_p\); Atomic mass of the element \(=M\); Number ofneutrons in the element \(=N\) and number of protons in the element \(=Z\).
We know that the atomic mass ( \(M\) ) of any stable nucleus is always less than the sum of the masses of the constituent particles.
Therefore, \(M<\left[N M_n+Z M_p\right]\).
\(X\) is a neutrino, when \(\beta\)-particle is emitted.

Hint

(a) Nuclear reaction: \(\mathrm{X}(n, \alpha) \rightarrow{ }_3^7 \mathrm{Li}\)
\(n\) represents a neutron: \({ }_0^1 n\)
\(\alpha\) represents an alpha particle (Helium nucleus): \({ }_2^4 \mathrm{He}\)
Lithium isotope: \({ }_3^7 \mathrm{Li}\)
Write the complete nuclear reaction equation.
\(
\mathrm{X}+{ }_0^1 n \rightarrow{ }_3^7 \mathrm{Li}+{ }_2^4 \mathrm{He}
\)
\(
\begin{aligned}
&\text { Let the mass number of } \mathrm{X} \text { be } A \text {. }\\
&\begin{aligned}
& A+1=7+4 \\
& A=7+4-1 \\
& A=10
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
&\text { Let the atomic number of } \mathrm{X} \text { be } \boldsymbol{Z} \text {. }\\
&\begin{aligned}
& Z+0=3+2 \\
& Z=3+2 \\
& Z=5
\end{aligned}
\end{aligned}
\)
The element with atomic number 5 is Boron (B). Therefore, X is \({ }_5^{10} B\).
The unknown element X is Boron- 10.

Hint

(b) \(\alpha \text {-rays are positively charged particles }\)

Hint

(a) Nuclear fission is best explained by the Liquid Droplet Model, while the Yukawa $\pi$-meson theory explains the nuclear force. The Independent Particle Model describes the nucleus as a collection of nucleons. The Proton-Proton cycle describes the fusion of hydrogen in the sun.

According to liquid drop model of nucleus, an excited nucleus breaks into lighter nuclei just like an excited drop breaks into tiny drops.

Hint

(c) The charge on the left side is +6.
The charge on the right side is +5 from Boron and +1 from positron, totaling +6 . Therefore, the charge of \(X\) is 0.
The lepton number on the left side is 0.
The lepton number of the positron ( \(\beta^{+}\)) is -1.
To balance the lepton number, X must have a lepton number of +1.
A particle with charge 0 and lepton number +1 is a neutrino \(\left(v_e\right)\).

The unknown particle X is a neutrino.

Hint

(a) Emission of \(1 \alpha\) particle led to decrease in atomic number by 2 while mass number by 4. On the other hand, emission of \(2 \beta\) particles increases atomic number by 2. Hence, overall emission of, \(1 \alpha\) and \(2 \beta\) particles led to decrease in mass number by 4.

Hint

(d) We know that alpha particles are the nucleus of ionized helium atoms which contain two protons and two neutrons. These are emitted by the nuclei of certain radioactive substances. Streams of alpha particles, called $\alpha$-rays, produce intense ionisation in gases through which they pass and are easily absorbed by matter.

Hint

(a) The explosion of an atomic bomb takes place due to nuclear fission.
Explanation:
Nuclear fission:
This is the process where a heavy atom’s nucleus splits into smaller fragments, releasing a large amount of energy. This is the primary mechanism behind the explosion of an atomic bomb.

Hint

(d) Nuclear reactions follow all law of Conservation principles.

Hint

(a) Suppose that,

The number of \({ }^{10} B\) type atoms \(=x\)
and the number of \({ }^{11} B\) type atoms \(=y\)
Weight of \({ }^{10} B\) type atoms \(=10 x\)
Weight of \({ }^{11} B\) type atoms \(=11 y\)
Total number of atoms \(=x+y\)
\(
\begin{aligned}
& \therefore \text { Atomic weight }=\frac{10 x+11 y}{x+y}=10.81 \\
& \Rightarrow 10 x+11 y=10.81 x+10.81 y \\
& \Rightarrow 0.81 x=0.19 y \Rightarrow \frac{x}{y}=\frac{19}{81}
\end{aligned}
\)

Note: If relative abundance of isotopes in an element has ratio \(n_1: n_2\) whose atomic massers are \(m_1\) and \(m_2\) then atomic masses of elements is \(M=\frac{n_1 m_1+n_2 m_2}{n_1+n_2}\).

Hint

(b)

\(
\text { Nuclear reaction: }{ }_{92}^{235} \mathrm{U}+{ }_0^1 \mathrm{n} \rightarrow{ }_{56}^{144} \mathrm{Ba}+\mathrm{X}+3{ }_0^1 \mathrm{n}
\)
The total mass number on the left side is \(235+1=236\).
The total mass number on the right side is \(144+3 \times 1=147\).
The mass number of the unknown element X is \(236-147=89\).
The total atomic number on the left side is \(92+0=92\).
The total atomic number on the right side is \(56+3 \times 0=56\).
The atomic number of the unknown element X is \(92-56=36\).

The element with atomic number 36 is Krypton (Kr). Therefore, the missing element X is \({ }_{36}^{89} \mathrm{Kr}\).
The missing element is Krypton-89.

Hint

(c) An alpha particle is a helium nucleus, which consists of 2 protons and 2 neutrons. Therefore, when an alpha particle is emitted from the nucleus \({ }_n^m X\) :
The mass number decreases by 4 (because 2 protons and 2 neutrons are lost).
The atomic number decreases by 2 (because 2 protons are lost).
So, after emitting one alpha particle, the new nucleus will be:
\(
{ }_n^m X \rightarrow_{n-2}^{m-4} Y
\)
A beta particle is essentially an electron, which is emitted when a neutron in the nucleus is converted into a proton. Each beta particle emission results in:
The mass number remaining the same (since a neutron is converted to a proton).
The atomic number increases by 1 (since a proton is added).
Since two beta particles are emitted, the changes will be:
The mass number remains \(m-4\).
The atomic number increases by 2 ( 1 for each beta particle).
So, after emitting two beta particles from the nucleus \({ }_{n-2}^{m-4} Y\) :
\(
{ }_{n-2}^{m-4} Y \rightarrow{ }_{(n-2)+2}^{m-4} Z
\)
This simplifies to:
\(
{ }_n^{m-4} Z
\)
After both emissions, the resulting nucleus is:
\(
{ }_n^{m-4} Z
\)
Final Result:
The resulting nucleus after the emission of one alpha particle and two beta particles from the original nucleus \({ }_n^m X[latex] is:
[latex]
{ }_n^{m-4} Z
\)

Hint

(d) From the given information, we know that after two beta emissions, the nucleus becomes \({ }_7^{14} N\) :
Atomic number \((Z)=7\)
Mass number \((A)=14\)
Since each beta emission increases the atomic number by 1 , we can deduce the atomic number of the parent nucleus \(X\) :
After 1 beta emission: Atomic number \(=7-1=6\)
After 2 beta emissions: Atomic number \(=6-1=5\)

Thus, the atomic number of the parent nucleus \(X\) is 5 .
The mass number remains unchanged during beta decay. Therefore, the mass number of the parent nucleus \(X\) is the same as that of the final nucleus:
Mass number of \(X=14\)
The number of neutrons can be calculated using the formula:
Number of Neutrons = Mass Number – Atomic Number

Substituting the values we have:
Number of Neutrons \(=14-5=9\)
The number of neutrons in the parent nucleus \(X\) is 9.

Hint

(a) The nuclear radius \(R=R_0 A^{1 / 3}\)
or \(R \propto A^{1 / 3}\)
\(
\begin{aligned}
& \text { or, } R^3 \propto A \Rightarrow A \propto R^3 \\
& \therefore \frac{A_2}{A_1}=\left(\frac{R_2}{R_1}\right)^3=\left(\frac{R_1 / 2}{R_1}\right)^3=\frac{1}{8} \\
& A_2=\frac{A_1}{8}=\frac{56}{8}=7
\end{aligned}
\)
Thus, stable nucleus will be \(L i^7\)

Hint

(b) The binding energy per nucleon of \({ }^{236} U\) is given as 7.6 MeV .
The binding energy per nucleon of \(X\) and \(Y\) is given as 8.5 MeV .
The total number of nucleons in \(X\) and \(Y\) is:
\(
117+117=234 \text { nucleons }
\)
\(
\begin{aligned}
& \text { The total binding energy for } X \text { and } Y \text { is: } \\
& \text { Total Binding Energy }{ }_{X+Y} \\
& =\text { Binding Energy per Nucleon } \times \text { Total Nucleons } \\
& =8.5 \mathrm{MeV} \times 234=1989 \mathrm{MeV}
\end{aligned}
\)
The total binding energy for \({ }^{236} U\) is:
Total Binding Energy \({ }_U=7.6 \mathrm{MeV} \times 236=1793.6 MeV\)
The energy liberated \((\Delta E)\) is the difference between the total binding energy of the products and the original nucleus:
\(
\Delta E=\text { Total Binding Energy }{ }_{X+Y}
\)
Total Binding Energy \({ }_U\)
\(
\Delta E=1989 \mathrm{MeV}-1793.6 \mathrm{MeV}=195.4 \mathrm{MeV}
\)
Thus, in per fission of Uranium nearly 200 MeV (we rounded 195.4 to 200) energy is liberated

Hint

(a) Balance the mass number

The total mass number on the left side is \(235+1=236\).
The mass number of Strontium is 90.
Let the mass number of the unknown element be \(A\) and the number of neutrons be \(x\).

The equation for mass number balance is \(236=90+A+x \cdot 1\).
The total atomic number on the left side is \(92+0=92\).
The atomic number of Strontium is 38.
Let the atomic number of the unknown element be \(\boldsymbol{Z}\).
The equation for atomic number balance is \(92=38+Z\).
From the atomic number balance equation: \(Z=92-38=54\). The element with atomic number 54 is Xenon ( \(X e\) ).
From the mass number balance equation: \(236=90+A+x\).
We know that fission of Uranium-235 typically releases 3 neutrons, so let \(x=3\).
Then, \(236=90+A+3\).
Solving for \(A: A=236-90-3=143\).
The complete equation is \({ }_{92} U^{235}+{ }_0 n^1 \rightarrow{ }_{38} S r^{90}+{ }_{54} X e^{143}+3_0 n^1\)

Hint

(c) A moderator in a nuclear reactor slows down neutrons, making them more likely to cause fission. Heavy water (D2O) is commonly used as a moderator because its deuterium atoms effectively slow down neutrons.

Hint

(b) Applying law of conservation of momentum,
\(
\begin{aligned}
& m_1 v_1=m_2 v_2 \\
& \frac{v_1}{v_2}=\frac{m_2}{m_1}
\end{aligned}
\)
\(
\text { As } \mathrm{m}=\frac{4}{3} \pi r^3 \rho \Rightarrow m \propto r^3
\)
Hence, \(\frac{m_2}{m_1}=\frac{r_2^3}{r_1^3}\)
\(
\therefore \frac{v_1}{v_2}=\frac{r_2^3}{r_1^3} \Rightarrow \frac{r_2}{r_1}=\left(\frac{1}{2}\right)^{\frac{1}{3}}
\)

Hint

(b) Mass number (A) is the total number of protons \((Z)\) and neutrons \((N)\) in an atomic nucleus.
\(
A=Z+N
\)
Protons and neutrons are collectively called nucleons.
Therefore, the mass number is the number of nucleons.

The mass number of a nucleus is equal to the number of nucleons it contains.

Hint

\(
\text { (c) } \frac{R_S}{R_{H e}}=\left(\frac{A_S}{A_{H e}}\right)^{1 / 3}=\left(\frac{32}{4}\right)^{1 / 3}=2
\)

Hint

(d) Calculate the total binding energy of \({ }_3 \mathrm{Li}^7\):
The number of nucleons in \({ }_3 \mathrm{Li}^7\) is 7.
The total binding energy of \({ }_3 \mathrm{Li}^7\) is \(7 \times 5.60 \mathrm{MeV}=39.2 \mathrm{MeV}\).
The number of nucleons in \({ }_2 \mathrm{He}^4\) is 4.
The total binding energy of \(2_2 \mathrm{He}^4\) is \(2 \times 4 \times 7.06 \mathrm{MeV}=56.48 \mathrm{MeV}\).
The energy released is the difference between the total binding energy of the products and the reactants.
Energy released \(=56.48 \mathrm{MeV}-39.2 \mathrm{MeV}=17.28 \mathrm{MeV}\).
Rounding to one decimal place, the energy released is 17.3 MeV.

Hint

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

Explanation: A moderator like heavy water is used to reduce the speed of fastmoving neutrons produced during fission, allowing them to be more effectively captured by uranium-235 nuclei and sustain the chain reaction.

Hint

\(
\begin{aligned}
&\text { (b) Fission rate }\\
&\begin{aligned}
& =\frac{\text { total power }}{\frac{\text { energy }}{\text { fission }}}=\frac{5}{200 \times 1.6 \times 10^{-13}} \\
& =1.56 \times 10^{11} \mathrm{~s}^{-1}
\end{aligned}
\end{aligned}
\)

Hint

(b) \(\text { The radius of } \mathrm{Cs}^{189} \text { is } R_{C s}=R_0(189)^{\frac{1}{3}} \text {. }\)
The radius of the unknown nucleus is \(R_x=\frac{1}{3} R_{C s}\). \(R_x=\frac{1}{3} R_0(189)^{\frac{1}{3}}\).
\(
\begin{aligned}
&\text { Let the mass number of the unknown nucleus be } A_x \text {. }\\
&\begin{aligned}
& R_x=R_0\left(A_x\right)^{\frac{1}{3}} \\
& R_0\left(A_x\right)^{\frac{1}{3}}=\frac{1}{3} R_0(189)^{\frac{1}{3}} \\
& \left(A_x\right)^{\frac{1}{3}}=\frac{1}{3}(189)^{\frac{1}{3}}
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
A_x & =\frac{1}{27} \cdot 189 . \\
A_x & =7
\end{aligned}
\)
The nucleus with mass number 7 is \({ }_3^7 \mathrm{Li}\).

Hint

(c) The nuclear radius \(r\) varies with mass number A according to the relation
\(
\begin{aligned}
& r \propto A^{1 / 3} \text { or } A \propto r^3 \\
& \text { density }=\frac{\text { mass }}{\text { volume }}
\end{aligned}
\)
Further, mass \(\propto \mathrm{A}\) and volume \(\propto r^3\)
\(
\therefore \frac{\text { mass }}{\text { volume }}=\text { constant }
\)

Hint

(a) Solar energy is due to fusion reaction.
Explanation: In the Sun’s core, hydrogen atoms fuse together to form helium, releasing a vast amount of energy, which is what we perceive as solar energy.

Hint

(c) \(1 \text { a.m.u= } 931 \mathrm{MeV}\)

Hint

(c) \(\text { Relativistic mass formula: } m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\)
\(
\frac{m_0}{\sqrt{\frac{c^2-(0.8 c)^2}{c^2}}}=\frac{5 m_0}{3}
\)

Hint

(c) The nuclear force is charge-independent.
Therefore, the nuclear force between any two nucleons is the same.
\(
F_{p p}=F_{n n}=F_{p n}
\)
The nuclear forces between two protons, two neutrons, and a proton and neutron are approximately equal.

Hint

(a) Nucleus contains only neutrons and protons.

Hint

(a) \(Z=11\) i.e., number of protons \(=11, A=23\)
\(\therefore\) Number of neutrons \(=A-Z=12\)
Number of electron \(=0\) (No electron in nucleus)
Therefore \(11,12,0\).

Hint

(a) The number of neutrons is the mass number minus the atomic number.
\(
N=13-6=7
\)
The number of neutrons is the mass number minus the atomic number.
\(
N=14-7=7
\)
Compare the number of neutrons:
Both nuclei have 7 neutrons.
They have different numbers of protons ( 6 and 7). Therefore, they are isotones.

The nuclei \({ }_6 \mathrm{C}^{13}\) and \({ }_7 \mathrm{~N}^{14}\) are isotones

Hint

\(
\begin{aligned}
& \text { (a) } \quad R \propto(A)^{1 / 3} \\
& \therefore R_{A l} \propto(27)^{1 / 3} \text { and } R_{T e} \propto(125)^{1 / 3} \\
& \therefore \frac{R_{A l}}{R_{T e}}=\frac{3}{5}=\frac{6}{10}
\end{aligned}
\)

Hint

(c) Nuclear forces are short range attractive forces which balance the repulsive forces between the protons inside the nucleus.

Explanation: Nuclear forces are indeed short-range, attractive forces that act between nucleons (protons and neutrons) within the nucleus. These forces are strong enough to overcome the electrostatic repulsion between positively charged protons, holding the nucleus together.

Hint

(a) Average B.E./nucleon in nuclei is of the order of 8 MeV .