Atoms and Nuclei

The nucleus is assumed as a sphere where the volume is proportional to the total number of nucleons (mass number, A).

Because the density of nuclear matter is approximately constant, the radius R of a nucleus is related to its mass number A by the following formula :

$ \mathrm{R}=\mathrm{R}_0 \mathrm{~A}^{1 / 3} $

Where R is the nuclear radius.

$R_0$ is a constant (approximately $1.2 \times 10^{-15} \mathrm{~m}$ or 1.2 fm ).

A is the mass number.

For the first nucleus : $\mathrm{R}_\alpha=\mathrm{R}_0(\alpha)^{1 / 3}$

For the second nucleus : $R_\beta=R_0(\beta)^{1 / 3}$

Taking the ratio $\frac{\mathrm{R}_\alpha}{\mathrm{R}_\beta}$ :

$ \frac{\mathrm{R}_\alpha}{\mathrm{R}_\beta}=\frac{\mathrm{R}_0 \alpha^{1 / 3}}{\mathrm{R}_0 \beta^{1 / 3}}=\left(\frac{\alpha}{\beta}\right)^{1 / 3} $

It is given that $\beta=8 \alpha$.

$ \frac{\mathrm{R}_\alpha}{\mathrm{R}_\beta}=\left(\frac{\alpha}{8 \alpha}\right)^{1 / 3}=\frac{1}{2}=0.5 $

Therefore, the ratio of the radius of the first nucleus to the second is 0.5 . Hence, the correct option is (D).

The mass number of initial nucleus is $\mathrm{A}_1=8$

The radius of a nucleus is proportional to $A^{1 / 3}\left(R=R_0 A^{1 / 3}\right)$.

So, the nuclear radius of initial nucleus is $R_1=R_0(8)^{1 / 3}=2 R_0$

Hence, initial surface area

$ \mathrm{S}_1=4 \pi \mathrm{R}_1^2 $

$\Rightarrow $ $ S_1=4 \pi\left(2 R_0\right)^2=16 \pi R_0^2 $

The atom absorbs 10 protons and 9 neutrons.

Total nucleons in final state of nucleus is 10 protons +9 neutrons $=19$ nucleons.

So, new mass number is $\mathrm{A}_2=\mathrm{A}_1+19=8+19=27$

The final radius,

$ \mathrm{R}_2=\mathrm{R}_0(27)^{1 / 3}=3 \mathrm{R}_0 $

So, final surface area is

$ S_2=4 \pi\left(3 R_0\right)^2=36 \pi R_0^2 $

The question has asked about the percentage growth recorded, that means final surface area is what percentage

So, the percentage growth is,

So, the percentage growth is,

$ \frac{\Delta S}{\mathrm{~S}_1} \times 100 \%=\frac{36 \pi \mathrm{R}_0^2-16 \pi \mathrm{R}_0^2}{16 \pi \mathrm{R}_0^2} \times 100 \%=125 \% $

A nucleus is more stable if it has a higher binding energy i.e., is in a lower energy state.

In exothermic reaction energy is released, $\Delta Q>0$. The products are more stable than the reactants.

In endothermic reaction energy is absorbed, $\Delta Q<0$. The products are less stable than the reactants.

The given two nuclear reactions are,

$ { }_2^3 \mathrm{He}+{ }_0^1 \mathrm{n} \rightarrow{ }_2^4 \mathrm{He}+20 \mathrm{MeV} \ldots . (1)$

$ { }_2^4 \mathrm{He}+{ }_0^1 \mathrm{n} \rightarrow{ }_2^5 \mathrm{He}-0.9 \mathrm{MeV} $......(2)

In ${ }_2^3 \mathrm{He}+{ }_0^1 \mathrm{n} \rightarrow{ }_2^4 \mathrm{He}+20 \mathrm{MeV}$ the released energy is $20 \mathrm{MeV}(\mathrm{Q}=+20 \mathrm{MeV})$.

Since energy is released, the product ${ }_2^4 \mathrm{He}$ is significantly more stable than the reactant ${ }_2^3 \mathrm{He}$.

Therefore, $\mathrm{X}_4>\mathrm{X}_3$.

In ${ }_2^4 \mathrm{He}+{ }_0^1 \mathrm{n} \rightarrow{ }_2^5 \mathrm{He}-0.9 \mathrm{MeV}$ the absorbed energy is $0.9 \mathrm{MeV}(\mathrm{Q}=-0.9 \mathrm{MeV})$.

Since energy must be supplied to this reaction, the product ${ }_2^5 \mathrm{He}$ is less stable than the reactant ${ }_2^4 \mathrm{He}$.

Therefore, $\mathrm{X}_4>\mathrm{X}_5$.

From reaction 1 , the jump in stability from $\mathrm{X}_3$ to $\mathrm{X}_4$ is very large ( 20 MeV ).

From reaction 2, the drop in stability from $\mathrm{X}_4$ to $\mathrm{X}_5$ is very small ( 0.9 MeV ).

Because $\mathrm{X}_5$ is only slighty less stable than $\mathrm{X}_4$, but $\mathrm{X}_3$ is significantly less stable than $\mathrm{X}_4$, we can conclude that $\mathrm{X}_5$ is still more stable than $\mathrm{X}_3$.

Therefore, the order of stability is $X_4>X_5>X_3$.

Hence, the correct option is (A).

The velocity of an electron in the $\mathrm{n}^{\text {th }}$ orbit is proportional to the atomic number Z and inversely proportional to the principal quantum number n .

$ \mathrm{v} \propto \frac{\mathrm{Z}}{\mathrm{n}} $

$\Rightarrow $ $ \mathrm{v}=\mathrm{k}_{\mathrm{v}} \cdot \frac{\mathrm{Z}}{\mathrm{n}} \ldots (i)$

Where $\mathrm{k}_{\mathrm{v}}$ is constant for the velocity.

The radius of the $\mathrm{n}^{\text {th }}$ orbit is proportional to the square of the principal quantum number n and inversely proportional to the atomic number Z.

$ r \propto \frac{n^2}{Z} $

$\Rightarrow $ $ \mathrm{r}=\mathrm{k}_{\mathrm{r}} \cdot \frac{\mathrm{n}^2}{\mathrm{Z}} \ldots(\mathrm{ii}) $

Let $\mathrm{r}_1$ be the radius for the first atom (with state $\mathrm{n}_1, \mathrm{Z}_1$ ) and $\mathrm{r}_2$ for the second (with state $\mathrm{n}_2, \mathrm{Z}_2$ ).

$ \mathrm{r}_1=\mathrm{r}_2 $

$\Rightarrow $ $\mathrm{k}_{\mathrm{r}} \cdot \frac{\mathrm{n}_1^2}{\mathrm{Z}_1}=\mathrm{k}_{\mathrm{r}} \cdot \frac{\mathrm{n}_2^2}{\mathrm{Z}_2}$

$\Rightarrow $ $\frac{\mathrm{n}_1^2}{\mathrm{Z}_1}=\frac{\mathrm{n}_2^2}{\mathrm{Z}_2}$

$ \frac{\mathrm{Z}_1}{\mathrm{Z}_2}=\frac{\mathrm{n}_1^2}{\mathrm{n}_2^2}=\left(\frac{\mathrm{n}_1}{\mathrm{n}_2}\right)^2 \ldots (iii)$

Let $\mathrm{v}_1$ and $\mathrm{v}_2$ be the speeds.

$ \mathrm{v}_1=\mathrm{k}_{\mathrm{v}} \frac{\mathrm{Z}_1}{\mathrm{n}_1} \ldots (iv)$

$\mathrm{v}_2=\mathrm{k}_{\mathrm{v}} \frac{\mathrm{Z}_2}{\mathrm{n}_2} \ldots(\mathrm{v})$

$\therefore $ $\frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{\mathrm{k}_{\mathrm{v}}\left(\frac{\mathrm{Z}_1}{\mathrm{n}_1}\right)}{\mathrm{k}_{\mathrm{v}}\left(\frac{\mathrm{Z}_2}{\mathrm{n}_2}\right)}$

$\Rightarrow $ $\frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{\mathrm{Z}_1}{\mathrm{n}_1} \times \frac{\mathrm{n}_2}{\mathrm{Z}_2}$ .....(vi)

$\Rightarrow $ $ \frac{\mathrm{v}_1}{\mathrm{v}_2}=\left[\left(\frac{\mathrm{n}_1}{\mathrm{n}_2}\right)^2\right] \cdot\left(\frac{\mathrm{n}_2}{\mathrm{n}_1}\right)=\frac{\mathrm{n}_1}{\mathrm{n}_2} $

It is given that the speed of first electron is $\mathrm{v}_1=3 \times 10^5 \mathrm{~m} / \mathrm{s}$ and that of electron in second hydrogen like atom is $\mathrm{v}_2=2.5 \times 10^5 \mathrm{~m} / \mathrm{s}$

$ \frac{n_1}{n_2}=\frac{3 \times 10^5}{2.5 \times 10^5} $

$\Rightarrow $ $\frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{3}{2.5}$

$\Rightarrow $ $\frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{6}{5}$

So, the smallest possible energy states in two atoms are 6 and 5.

Hence, the correct option is (D).

For light nuclei, the binding energy per nucleon $\left(\frac{B E}{A}\right)$ increases rapidly as mass increases. The curve reaches a maximum around Mass Number $\mathrm{A} \approx 56$ (Iron, Fe ). This is the most stable region. For heavy nuclei the $\frac{\mathrm{BE}}{\mathrm{A}}$ actually decreases as mass increases.

So, the binding energy per nucleon does not constantly increase with mass. It increases initially, peaks, and then decreases.

Hence, statement I is false.

Nature always tends towards the state of lowest potential energy (highest stability). Higher Binding Energy per Nucleon means the nucleus is more tightly bound and therefore more stable.

So, in nuclear fission a heavy nucleus (low $\frac{B E}{A}$ ) splits into two lighter nuclei (higher $\frac{B E}{A}$ ). And in nuclear fusion the light nuclei (low BE/A) combine to form a heavier nucleus (higher BE/A).

In both fundamental nuclear processes (fission and fusion), the system evolves from a state of lower binding energy per nucleon to a state of higher binding energy per nucleon. This is the driving force behind nuclear energy release.

Hence, statement II is true.

Therefore, statement I is false, statement II is true. Hence, the correct option is (A).

Atoms are represented by the notation $\mathrm{X}_{\mathrm{Z}}^{\mathrm{A}}$, where :

• 1. A is the Mass Number (total number of protons and neutrons).

• 2. $ \mathrm{Z}$ is the Atomic Number (number of protons).

Isobars are atoms of different chemical elements that have the same mass number (A) but different atomic numbers (Z).

Option A : $\mathrm{H}_1^3$ and $\mathrm{He}_2^3$

• Mass numbers: 3 and 3 (Same)

• Atomic numbers: 1 and 2 (Different)

Since they have the same mass number ( $A=3$ ) but represent different elements ( $Z=1$ vs $Z=2$ ), they are Isobars. Hence, option (A) is correct.

Option B : $\mathrm{Hg}_{80}^{198}$ and $\mathrm{Au}_{79}^{197}$

• Mass numbers: 198 and 197 (Different)

• Atomic numbers: 80 and 79 (Different)

So, $\mathrm{Hg}_{80}^{198}$ and $\mathrm{Au}_{79}^{197}$ are neither isobars nor isotopes. Hence, option (B) is incorrect.

Option C : $\mathrm{H}_1^2$ and $\mathrm{H}_1^3$

• Mass numbers: 2 and 3 (Different)

• Atomic numbers: 1 and 1 (Same)

So, $\mathrm{H}_1^2$ and $\mathrm{H}_1^3$ are Isotopes of Hydrogen (Deuterium and Tritium). Hence, option (C) is incorrect.

Option D : $\mathrm{U}_{92}^{236}$ and $\mathrm{U}_{92}^{238}$

• Mass numbers: 236 and 238 (Different)

• Atomic numbers: 92 and 92 (Same)

So, $\mathrm{U}_{92}^{236}$ and $\mathrm{U}_{92}^{238}$ are Isotopes (same element, different mass) not isobars. Hence, option (D) is incorrect.

Using the Rydberg formula :

$ \frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right) $

Where, $\lambda$ is the wavelength of radiated photon when electron makes a transition from initial energy state $n_2$ to final energy state $\mathrm{n}_1$, and R is the Rydberg's constant.

For statement A

The given series is Lyman ( $\mathrm{n}_1=1$ ).

Maximum wavelength implies minimum energy. The transition corresponds to the smallest energy gap, which is between the first excited state ( $\mathrm{n}_2=2$ ) and the ground state ( $\mathrm{n}_1=1$ ).

$ \frac{1}{\lambda_{\max }}=\mathrm{R}\left(\frac{1}{1^2}-\frac{1}{2^2}\right) $

$\Rightarrow $ $ \frac{1}{\lambda_{\max }}=\mathrm{R}\left(1-\frac{1}{4}\right)=\frac{3 \mathrm{R}}{4} \Rightarrow \lambda_{\max }=\frac{4}{3 \mathrm{R}} $

Hence, the statement A is true.

For statement B

The given series is Balmer ( $\mathrm{n}_1=2$ ).

The Balmer series corresponds to electron transitions from higher shells ( $n=3,4,5 \ldots$ ) to the second shell $(\mathrm{n}=2)$. The wavelengths produced (e.g., $656 \mathrm{~nm}, 486 \mathrm{~nm})$ fall within the range of approximately 400 nm to 700 nm , which is the visible spectrum

So, Balmer series lines are indeed visible. Hence, the statement B is true.

For Statement C

The given series is Paschen ( $\mathrm{n}_1=3$ ).

Minimum wavelength implies maximum energy. The transition corresponds to the largest energy gap, which is from infinity $\left(\mathrm{n}_2=\infty\right)$ to the level $\mathrm{n}_1=3$.

$ \frac{1}{\lambda_{\min }}=R\left(\frac{1}{3^2}-\frac{1}{\infty^2}\right) $

$\Rightarrow $ $ \frac{1}{\lambda_{\min }}=\mathrm{R}\left(\frac{1}{9}\right)=\frac{\mathrm{R}}{9} \Rightarrow \lambda_{\min }=\frac{9}{\mathrm{R}} $

Hence, the statement C is also true.

For statement D

The given series is Lyman ( $\mathrm{n}_1=1$ ).

Minimum wavelength means ( $\mathrm{n}_2=\infty \rightarrow \mathrm{n}_1=1$ ).

$ \frac{1}{\lambda_{\min }}=\mathrm{R}\left(\frac{1}{1^2}-\frac{1}{\infty^2}\right) $

$\Rightarrow $ $\frac{1}{\lambda_{\min }}=\mathrm{R}(1-0)=\mathrm{R}$

$\Rightarrow $ $ \lambda_{\min }=1 \mathrm{R} $

Hence, the statement D is false.

• Statement A is Correct.

• Statement B is Correct.

• Statement C is Correct.

• Statement D is Incorrect.

Therefore, the correct option is (D).

The energy of $n^{\text {th }}$ energy state of hydrogen atom is $E_n=-\frac{13.6}{n^2}$

The wavelengths of the spectral lines of hydrogen-like atoms are governed by the Rydberg formula

$ \frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right) $

Where :

• $\lambda$ is the wavelength of the emitted photon.

• R is the Rydberg constant.

• $\mathrm{n}_1$ is the principal quantum number of the lower energy level.

• $\mathrm{n}_2$ is the principal quantum number of the higher energy level ( $\mathrm{n}_2>\mathrm{n}_1$ ).

The smallest wavelength corresponds to the highest energy transition (from $\mathrm{n}_2=\infty$ to $\mathrm{n}_1$ ).

The largest wavelength corresponds to the lowest energy transition (from the immediate next energy level, $\mathrm{n}_2= \mathrm{n}_1+1$, down to $\mathrm{n}_1$ ).

For the Lyman series, the electron transitions fall to the ground state, so $\mathrm{n}_1=1$. The smallest wavelength $\left(\lambda_{\mathrm{L}(\min )}\right)$ occurs when $\mathrm{n}_2=\infty$.

It is given that $\lambda_{\mathrm{L}, \min }=91 \mathrm{~nm}$.

Using the Rydberg formula :

$ \frac{1}{\lambda_{\mathrm{L}, \min }}=\mathrm{R}\left(\frac{1}{1^2}-\frac{1}{\infty^2}\right) $

$\Rightarrow $ $ \frac{1}{91}=\mathrm{R}(1-0) \Rightarrow \mathrm{R}=\frac{1}{91} \mathrm{~nm}^{-1} $

For the Balmer series, the lower energy level is $\mathrm{n}_1=2$.

The largest wavelength ( $\lambda_{B, \max }$ corresponds to the smallest energy jump, which is from $\mathrm{n}_2=3$ to $\mathrm{n}_1=2$.

Applying the formula :

$ \frac{1}{\lambda_{\mathrm{B}, \max }}=\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\mathrm{R}\left(\frac{5}{36}\right) $

$\Rightarrow $ $\frac{1}{\lambda_{\mathrm{B}, \max }}=\frac{1}{91} \times \frac{5}{36} \mathrm{~nm}^{-1}$

$\Rightarrow $ $ \lambda_{\mathrm{B}, \max }=\frac{91 \times 36}{5} \mathrm{~nm}=655.2 \mathrm{~nm} $

For the Paschen series, the lower energy level is $\mathrm{n}_1=3$.

The largest wavelength ( $\lambda_{P, \max }$ ) corresponds to the smallest energy jump, which is from $\mathrm{n}_2=4$ to $\mathrm{n}_1=3$.

Applying the formula :

$ \frac{1}{\lambda_{\mathrm{P}, \max }}=\mathrm{R}\left(\frac{1}{3^2}-\frac{1}{4^2}\right)=\mathrm{R}\left(\frac{1}{9}-\frac{1}{16}\right) $

$\Rightarrow $ $\frac{1}{\lambda_{\mathrm{P}, \max }}=\frac{1}{91} \times \frac{7}{144} \mathrm{~nm}^{-1}$

$ \Rightarrow \lambda_{P, \max }=\frac{91 \times 144}{7} \mathrm{~nm}=1871 \mathrm{~nm} $

So, the difference between these two largest wavelengths :

$ \Delta \lambda=\lambda_{\mathrm{P}, \max }-\lambda_{\mathrm{B}, \max } $

$\Rightarrow $ $\Delta \lambda=1872 \mathrm{~nm}-655.2 \mathrm{~nm}$

$\Rightarrow $ $ \Delta \lambda=1216.8 \mathrm{~nm} \approx 1217 \mathrm{~nm} $

Therefore, the difference between the largest wavelengths of Paschen and Balmer series is nearly 1217 nm.

Hence, the correct option is (C).

Einstein's equation tells us that mass and energy are interchangeable.

$ \mathrm{E}=\mathrm{mc}^2 $

When a nuclear reaction occurs, the total mass of the products is often slightly different from the initial mass of the reactants.

• If the products are lighter, the missing mass is released as energy.

• If the products are heavier, energy must be supplied to create that extra mass.

It is given that a particle breaks into $4 \alpha$ particles.

Initial mass of the single particle. $\mathrm{M}_{\text {initial }}=15.348 \mathrm{amu}$

Final mass of $4 \alpha$ particles $=\mathrm{M}_{\text {final }}=4 \times$ mass of He nucleus

$ \mathrm{M}_{\mathrm{final}}=4 \times 4.002 \mathrm{amu}=16.008 \mathrm{amu} $

The mass defect ( $\Delta \mathrm{m}$ ) during this nuclear reaction is,

$ \Delta \mathrm{m}=\mathrm{M}_{\text {final }}-\mathrm{M}_{\text {initial }} $

$\Rightarrow $ $\Delta \mathrm{m}=16.008 \mathrm{amu}-15.348 \mathrm{amu}$

$\Rightarrow $ $ \Delta \mathrm{m}=0.66 \mathrm{amu} $

It is given that $1 \mathrm{amu}=1.66 \times 10^{-27} \mathrm{~kg}$.

$ \Delta \mathrm{m}=0.66 \times\left(1.66 \times 10^{-27}\right) \mathrm{kg} $

$\Rightarrow $ $ \Delta \mathrm{m}=1.0956 \times 10^{-27} \mathrm{~kg} $

Now using Einstein's equation,

$\mathrm{E}=\Delta \mathrm{m} \cdot \mathrm{c}^2$

$\Rightarrow $ $E=\left(1.0956 \times 10^{-27}\right) \times\left(3 \times 10^8\right)^2$

$\Rightarrow $ $E=\left(1.0956 \times 10^{-27}\right) \times\left(9 \times 10^{16}\right)$

$\Rightarrow $ $ E=9.8604 \times 10^{-11} \text { Joules } $

This energy is provided by a single photon. The energy of a photon is directly proportional to its frequency ( $v$ ), given by Planck's equation:

$ \mathrm{E}=\mathrm{h} \nu $

Where h is Planck's constant $\left(6.6 \times 10^{-34} \mathrm{~J} \cdot \mathrm{~s}\right)$.

$ v=\frac{E}{h} $

$\Rightarrow $ $v=\frac{9.8604 \times 10^{-11}}{6.6 \times 10^{-34}}$

$\Rightarrow $$ v=1.494 \times 10^{23} \mathrm{~Hz} $

$\Rightarrow $$ v_{\mathrm{kHz}}=1.494 \times 10^{23} \times 10^{-3} \mathrm{kHz} $

$\Rightarrow $ $ v_{\mathrm{kHz}}=14.94 \times 10^{19} \mathrm{kHz} $

Therefore, the minimum frequency of photon required is $14.94 \times 10^{19} \mathrm{kHz}$.

Hence, the correct option is (D).

Let an $\alpha$-particle fired directly at the center of a target nucleus.

The $\alpha$-particle has maximum kinetic energy $(\mathrm{K})$ and practically zero electrostatic potential energy at very far away from target material.

As it gets closer, the positive charge of the $\alpha$-particle and the positive charge of the target nucleus repel each other. The $\alpha$-particle slows down, converting its kinetic energy into electrostatic potential energy.

At the closest approach ( $\mathrm{r}_0$ ) it momentarily stops before reversing direction. At this exact point, all its initial kinetic energy has been converted into potential energy.

By the Law of Conservation of Energy:

Initial Kinetic Energy $\left(\mathrm{K}_0\right)=$ Final Potential Energy $(\mathrm{U})$

The electrostatic potential energy between two point charges ( $\mathrm{q}_1$ and $\mathrm{q}_2$ ) separated by a distance r is :

$ U=\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r} $

An $\alpha$-particle is a Helium nucleus, so it has 2 protons. So, the charge of $\alpha$-particle $\left(\mathrm{q}_1\right)$ is $\mathrm{q}_1=+2 \mathrm{e}$

The atomic number of the target is $\mathrm{Z}=79$, so it has 79 protons, So, the charge of target nucleus $\left(\mathrm{q}_2\right)$ is $\mathrm{q}_2= +79 \mathrm{e}$

Substituting into our energy conservation equation :

$ \mathrm{K}_0=\frac{1}{4 \pi \epsilon_0} \frac{(2 \mathrm{e})(79 \mathrm{e})}{\mathrm{r}_0} $

$\Rightarrow $ $ \mathrm{r}_0=\frac{1}{4 \pi \epsilon_0} \frac{158 \mathrm{e}^2}{\mathrm{~K}_0} $

The kinetic energy of alpha particle is,

$ \mathrm{K}_0=7.9 \mathrm{MeV}=7.9 \times 10^6 \mathrm{eV} $

Since $1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}$, we get:

$ K_0=\left(7.9 \times 10^6\right) \times\left(1.6 \times 10^{-19}\right) \mathrm{J} $

$\Rightarrow $ $ \mathrm{K}_0=12.64 \times 10^{-13} \mathrm{~J} $

Putting in the closest approach formula,

$ r_0=\frac{\left(9 \times 10^9\right) \times 158 \times\left(1.6 \times 10^{-19}\right)^2}{12.64 \times 10^{-13}} $

$\Rightarrow $ $r_0=\frac{(9 \times 158 \times 2.56) \times 10^{-29}}{12.64 \times 10^{-13}}$

$\Rightarrow $ $\mathrm{r}_0=288 \times 10^{-16} \mathrm{~m}$

$\Rightarrow $ $ \mathrm{r}_0=2.88 \times 10^{-14} \mathrm{~m} $

Therefore, the distance of closest approach is $\mathrm{r}_0=2.88 \times 10^{-14} \mathrm{~m}$.

The distance from the centre of the nucleus to this stopping point is the distance of closest approach. Because the $\alpha$-particle cannot physically penetrate the nucleus using just the electrostatic force in this classical model, $r_0$ gives us an upper estimate for the radius of the target nucleus.

Diameter $=2 \times$ Radius

Diameter $=2 \times \mathrm{r}_0$

Diameter $=2 \times\left(2.88 \times 10^{-14} \mathrm{~m}\right)$

Diameter $=5.76 \times 10^{-14} \mathrm{~m}$

Therefore, the diameter of nuclei of the target material is approximately $5.76 \times 10^{-14} \mathrm{~m}$.

Hence, the correct option is (B).

In the Bohr model of the atom, the energy of an electron in the $n^{\text {th }}$ orbit $\left(E_n\right)$ is related to the ground state energy ( $\mathrm{E}_0$ ) by the formula :

$ \mathrm{E}_{\mathrm{n}}=-\frac{\mathrm{E}_0}{\mathrm{n}^2} $

It is given that the energy in the current orbit is $-0.04 \mathrm{E}_0 \mathrm{eV}$.

$ -0.04 \mathrm{E}_0=-\frac{\mathrm{E}_0}{\mathrm{n}^2} $

$\Rightarrow $ $0.04=\frac{1}{\mathrm{n}^2}$

$\Rightarrow $ $\mathrm{n}^2=\frac{100}{4}=25$

$\Rightarrow $ $ \mathrm{n}=\sqrt{25}=5 $

So, the electron is in the $5^{\text {th }}$ orbit of the Bohr's atom.

Bohr's second postulate states that the angular momentum of an electron is quantized and is an integral multiple of $\mathrm{h} / 2 \pi$ :

$ \mathrm{L}=\frac{\mathrm{nh}}{2 \pi} $

Where n is the principal quantum number (orbit number).

$ \frac{2 \pi L}{h}=n=5 $

Therefore, the value of $\frac{2 \pi L}{h}$ is 5 .

Hence, the correct option is (C).

Let an $\alpha$-particle fired directly at the centre of a target nucleus, gold foil.

The $\alpha$-particle has maximum kinetic energy $(\mathrm{K})$ and practically zero electrostatic potential energy at very far away from target material.

The initial Kinetic Energy ( $\mathrm{K}_0$ ) of the alpha particle is :

$ \mathrm{K}_0=7.7 \mathrm{MeV} $

As it gets closer, the positive charge of the $\alpha$-particle and the positive charge of the target nucleus repel each other. The $\alpha$-particle slows down, converting its kinetic energy into electrostatic potential energy.

At the closest approach ( $\mathrm{r}_0$ ) it momentarily stops before reversing direction. At this exact point, all its initial kinetic energy has been converted into potential energy.

The electrostatic potential energy $(U)$ between two point charges $q_1$ and $q_2$ separated by a distance $r_0$ is :

$ \mathrm{U}=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}_0} $

By the Law of Conservation of Energy :

Initial Kinetic Energy $\left(\mathrm{K}_0\right)=$ Final Potential Energy $(\mathrm{U})$

The electrostatic potential energy between two point charges ( $\mathrm{q}_1$ and $\mathrm{q}_2$ ) separated by a distance r is :

$ U=\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r} $

An $\alpha$-particle is a Helium nucleus, so it has 2 protons. So, the charge of $\alpha$-particle $\left(\mathrm{q}_1\right)$ is $\mathrm{q}_1=+2 \mathrm{e}$ The atomic number of the target is $\mathrm{Z}=79$, so it has 79 protons, So, the charge of target nucleus $\left(\mathrm{q}_2\right)$ is $\mathrm{q}_2= +79 \mathrm{e}$

Substituting into our energy conservation equation :

$ \mathrm{K}_0=\frac{1}{4 \pi \epsilon_0} \frac{(2 \mathrm{e})(79 \mathrm{e})}{\mathrm{r}_0} $

$\Rightarrow $ $ \mathrm{r}_0=\frac{1}{4 \pi \epsilon_0} \frac{158 \mathrm{e}^2}{\mathrm{~K}_0} $

The kinetic energy of alpha particle is,

$ \mathrm{K}_0=7.7 \mathrm{MeV}=7.7 \times 10^6 \mathrm{eV} $

Since $1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}$, we get :

$ K_0=\left(7.7 \times 10^6\right) \times\left(1.6 \times 10^{-19}\right) \mathrm{J} $

$\Rightarrow $ $ \mathrm{K}_0=12.32 \times 10^{-13} \mathrm{~J} $

Putting in the closest approach formula,

$ r_0=\frac{\left(9 \times 10^9\right) \times 158 \times\left(1.6 \times 10^{-19}\right)^2}{12.32 \times 10^{-13}} $

$\Rightarrow $ $r_0=\frac{(9 \times 158 \times 2.56) \times 10^{-29}}{12.32 \times 10^{-13}}$

$\Rightarrow $ $\mathrm{r}_0=295 \times 10^{-16} \mathrm{~m}$

$\Rightarrow $ $ \mathrm{r}_0=2.95 \times 10^{-14} \mathrm{~m} $

Therefore, the distance of closest approach is $\mathrm{r}_0=2.95 \times 10^{-14} \mathrm{~m}$.

Hence, the correct option is (C).

In $\alpha$-decay, a parent nucleus (X) of mass number A decays into a daughter nucleus (Y) and an $\alpha$-particle (Helium nucleus, mass number 4).

$ \mathrm{X}_{\mathrm{Z}}^{\mathrm{A}} \rightarrow \mathrm{Y}_{\mathrm{Z}-2}^{\mathrm{A}-4}+\mathrm{He}_2^4 $

Using conservation of momentum, the momentum of the $\alpha$-particle ( $\mathrm{p}_\alpha$ ) must be equal and opposite to the momentum of the daughter nucleus $\left(\mathrm{p}_{\mathrm{D}}\right)$.

And using energy balance ( Q -value) the total kinetic energy released ( Q ) is shared between the $\alpha$-particle and the daughter nucleus.

$ Q=K_\alpha+K_D $

Kinetic energy can be expressed in terms of momentum (p) and mass (m) as $\mathrm{K}=\frac{\mathrm{p}^2}{2 \mathrm{~m}}$.

Since $\mathrm{p}_\alpha=\mathrm{p}_{\mathrm{D}}$, we have :

$\mathrm{K}_{\mathrm{D}}=\frac{\mathrm{p}^2}{2 \mathrm{~m}_{\mathrm{D}}}$ and $\mathrm{K}_\alpha=\frac{\mathrm{p}^2}{2 \mathrm{~m}_\alpha}$

Therefore, $\frac{\mathrm{K}_{\mathrm{D}}}{\mathrm{K}_\alpha}=\frac{\mathrm{m}_\alpha}{\mathrm{m}_{\mathrm{D}}}$.

Substituting $\mathrm{K}_{\mathrm{D}}=\mathrm{K}_\alpha\left(\frac{\mathrm{m}_\alpha}{\mathrm{m}_{\mathrm{D}}}\right)$ into the Q -value equation:

$ \mathrm{Q}=\mathrm{K}_\alpha+\mathrm{K}_\alpha\left(\frac{\mathrm{m}_\alpha}{\mathrm{m}_{\mathrm{D}}}\right)=\mathrm{K}_\alpha\left(1+\frac{\mathrm{m}_\alpha}{\mathrm{m}_{\mathrm{D}}}\right)=\mathrm{K}_\alpha\left(\frac{\mathrm{m}_{\mathrm{D}}+\mathrm{m}_\alpha}{\mathrm{m}_{\mathrm{D}}}\right) $

Since $\mathrm{m}_{\mathrm{D}}+\mathrm{m}_\alpha \approx \mathrm{A}$ (mass number of parent) and $\mathrm{m}_{\mathrm{D}} \approx \mathrm{A}-4$ :

$ \mathrm{K}_\alpha=\mathrm{Q}\left(\frac{\mathrm{~A}-4}{\mathrm{~A}}\right) $

The Q-value is the same ( 1 MeV ) for both substances.

For Substance A ( $\mathrm{A}_{\mathrm{A}}=200$ ) :

$ K_{\alpha A}=Q\left(\frac{200-4}{200}\right)=Q\left(\frac{196}{200}\right) $

For Substance $B\left(A_B=212\right)$ :

$ K_{\alpha B}=Q\left(\frac{212-4}{212}\right)=Q\left(\frac{208}{212}\right) $

So, the ratio is;

$ \frac{K_{\alpha A}}{K_{\alpha B}}=\frac{Q\left(\frac{196}{200}\right)}{Q\left(\frac{208}{212}\right)}=\frac{196}{200} \times \frac{212}{208} $

$\Rightarrow $ $ \frac{\mathrm{K}_{\alpha \mathrm{A}}}{\mathrm{~K}_{\alpha \mathrm{B}}}=\frac{49}{50} \times \frac{53}{52}=\frac{49 \times 53}{50 \times 52}=\frac{2597}{2600} $

Therefore, the ratio of energies of $\alpha$-rays produced by A and B is $\frac{2597}{2600}$.

Hence, the correct option is (C).

The binding energy (B.E.) of a nucleus is the energy required to disassemble it into its constituent nucleons.

Total B.E. $=($ Binding Energy per nucleon $) \times($ Number of nucleons A$)$

Energy Released (Q) = (Total B.E. of Products) - (Total B.E. of Reactants)

The given reaction is $\mathrm{H}_1^2+\mathrm{H}_1^2 \rightarrow \mathrm{He}_2^4$

The reactants are two nuclei of $\mathrm{H}_1^2$.

B.E. per nucleon of $\mathrm{H}_1^2=1.1 \mathrm{MeV}$

Number of nucleons (A) for one $\mathrm{H}_1^2=2$

So, total B.E. of reactants $=2 \times(1.1 \times 2)=4.4 \mathrm{MeV}$

The product is one nucleus of $\mathrm{He}_2^4$.

B.E. per nucleon of $\mathrm{He}_2^4=7.2 \mathrm{MeV}$

Number of nucleons (A) for $\mathrm{He}_2^4=4$

So, total B.E. of product $=7.2 \times 4=28.8 \mathrm{MeV}$

Hence, energy released during reaction is,

$ \mathrm{Q}=28.8 \mathrm{MeV}-4.4 \mathrm{MeV} $

$\Rightarrow $ $ \mathrm{Q}=24.4 \mathrm{MeV} $

Therefore, the energy released if hydrogen atoms are combined to form $\mathrm{He}_2^4$ is 24.4 MeV .

Thus, the correct option is (D).

According to the de Broglie relation, the momentum (p) of a photon is inversely proportional to its wavelength ( $\lambda$ ):

$ \mathrm{p}=\frac{\mathrm{h}}{\lambda} $

Therefore, the ratio of momenta is the inverse ratio of their wavelengths:

$ \frac{\mathrm{p}_1}{\mathrm{p}_2}=\frac{\lambda_2}{\lambda_1} $

The Balmer series occurs when an electron transitions from a higher energy level $\left(\mathrm{n}_2\right)$ to the second energy level ( $\mathrm{n}_1=2$ ). The wavelength is given by Rydberg formula:

$ \frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{\mathrm{n}^2}\right) $

For $1^{\text {st }}$ line of Balmer transition from $\mathrm{n}=3$ to $\mathrm{n}=2$.

$ \frac{1}{\lambda_1}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{9}\right)=\mathrm{R}\left(\frac{9-4}{36}\right)=\frac{5 \mathrm{R}}{36} \Rightarrow \lambda_1=\frac{36}{5 \mathrm{R}} $

For $2^{\mathrm{nd}}$ line of Balmer transition from $\mathrm{n}=4$ to $\mathrm{n}=2$.

$ \frac{1}{\lambda_2}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{16}\right)=\mathrm{R}\left(\frac{4-1}{16}\right)=\frac{3 \mathrm{R}}{16} \Rightarrow \lambda_2=\frac{16}{3 \mathrm{R}} $

So, the ratio $(\alpha / \beta)$,

$ \frac{\mathrm{p}_1}{\mathrm{p}_2}=\frac{\lambda_2}{\lambda_1}=\frac{16 / 3 \mathrm{R}}{36 / 5 \mathrm{R}}=\frac{16}{3} \times \frac{5}{36} $

$\Rightarrow $ $ \frac{\mathrm{p}_1}{\mathrm{p}_2}=\frac{4 \times 5}{3 \times 9}=\frac{20}{27}=\frac{\alpha}{\beta} $

So, $\alpha=20$ and $\beta=27$.

Therefore, the correct option is (D).

For ${ }_6^{12}\mathrm{C}$,

• Number of protons $Z = 6$

• Number of neutrons $N = 12 - 6 = 6$

Mass of 6 protons:

$ 6 \times 1.00727 = 6.04362\,u $

Mass of 6 neutrons:

$ 6 \times 1.00866 = 6.05196\,u $

So, total mass of separate nucleons is

$ 6.04362 + 6.05196 = 12.09558\,u $

Given actual mass of ${ }_6^{12}\mathrm{C}$ atom is

$ 12\,u $

Hence, mass defect

$ \Delta m = 12.09558 - 12 = 0.09558\,u $

Now convert into $\mathrm{MeV}/c^2$:

$ \Delta m = 0.09558 \times 931.5 $

$ \Delta m \approx 89.03\,\mathrm{MeV}/c^2 $

So, the correct answer is:

$ \boxed{89.03\,\mathrm{MeV}/c^2} $

Hence, Option B is correct.

Step 1: Find the quantum numbers of the orbits

In the Bohr model of the hydrogen atom, the radius of the $n^{\text{th}}$ orbit is given by:

$r_n \propto n^2$

So, the ratio of radii is:

$\frac{r_i}{r_f} = \frac{n_i^2}{n_f^2} = \frac{16}{4}$

Taking the square root on both sides:

$\frac{n_i}{n_f} = \frac{4}{2} = \frac{2}{1}$

Therefore:

• $n_i = 4$ (higher orbit)

• $n_f = 2$ (lower orbit)

Step 2: Apply the Rydberg formula

The wavelength of the emitted photon is given by the Rydberg formula:

$\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$

Substituting the values:

$\frac{1}{\lambda} = 1.0973 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{4^2} \right)$

$\frac{1}{\lambda} = 1.0973 \times 10^7 \left( \frac{1}{4} - \frac{1}{16} \right)$

$\frac{1}{\lambda} = 1.0973 \times 10^7 \left( \frac{4-1}{16} \right) = 1.0973 \times 10^7 \times \frac{3}{16}$

Step 3: Calculate the wavelength

$\frac{1}{\lambda} = \frac{3 \times 1.0973 \times 10^7}{16} = \frac{3.2919 \times 10^7}{16}$

$\frac{1}{\lambda} = 2.0574 \times 10^6 \text{ m}^{-1}$

Therefore:

$\lambda = \frac{1}{2.0574 \times 10^6} \text{ m} \approx 4.86 \times 10^{-7} \text{ m}$

Converting to nanometres:

$\lambda \approx 486 \text{ nm}$

Answer

Option C: 486 nm

Note: This transition ($n=4 \to n=2$) belongs to the Balmer series and corresponds to the $H_\beta$ line of the hydrogen spectrum, which lies in the visible region.

In Rutherford’s $\alpha$-particle scattering experiment, a very small number of $\alpha$-particles rebound back, that is, they are deflected through angles close to $180^\circ$.

This happens only when an $\alpha$-particle comes very close to the nucleus, or almost makes a head-on collision.

Now let us check each statement:

1. A. The size of gold nucleus is very small as compared to the size of gold atom.

   This is correct.

   Since the nucleus occupies a very small region inside the atom, only very few $\alpha$-particles can come close enough to it to be scattered backward.

2. B. Alpha particle and gold nucleus have equal charge.

   This is incorrect.

   An $\alpha$-particle has charge $+2e$, while a gold nucleus has charge $+79e$.

3. C. The impact parameter is minimum for a few alpha particles.

   This is correct.

   Only those $\alpha$-particles which have very small impact parameter come very close to the nucleus and can be strongly deflected or rebound.

4. D. A few alpha particles have very high kinetic energy.

   This is incorrect.

   Rebounding is not because a few particles have very high kinetic energy. In fact, higher kinetic energy would make deflection less likely.

5. E. Only a few alpha particles undergo head-on collision with the nuclei.

   This is correct.

   Backward scattering is mainly due to nearly head-on collisions.

So the correct statements are:

$ A,\ C,\ E $

Hence, the correct option is:

$ \boxed{\text{Option D}} $

The radius $ R $ of a nucleus is given by the empirical relation:

$ R = r_0 A^{\frac{1}{3}}, $

where $ r_0 $ is a constant and $ A $ is the mass number.

The volume $ V $ of the nucleus is approximately that of a sphere:

$ V = \frac{4}{3}\pi R^3. $

Substituting the expression for $ R $ into the volume formula:

$ V = \frac{4}{3}\pi \left(r_0 A^{\frac{1}{3}}\right)^3 = \frac{4}{3}\pi r_0^3 A. $

The mass density $ \rho $ is defined as the mass (which is proportional to $ A $) divided by the volume:

$ \rho = \frac{A}{V} = \frac{A}{\frac{4}{3}\pi r_0^3 A} = \frac{1}{\frac{4}{3}\pi r_0^3}. $

Notice that $ A $ cancels out in the numerator and denominator, so the density is:

$ \rho = \text{constant}, $

meaning it is independent of the mass number $ A $.

Thus, the correct option is:

Option B: Independent of A.

To evaluate the assertion and reason provided, we need to consider the formula for the density of a nucleus. The density $ \rho $ can be expressed as follows:

$ \rho = \frac{M}{V} = \frac{m_n \times A}{\frac{4}{3} \pi R^3} $

Here, $ m_n $ represents the nucleon mass, $ A $ is the mass number, and $ R $ is the radius of the nucleus. According to the formula for the radius:

$ R = R_0 A^{1/3} $

This indicates that the radius $ R $ is proportional to $ A^{1/3} $. By substituting the expression for $ R $ in terms of $ A $ back into the formula for density, we get:

$ \rho = \frac{m_n \times A}{\frac{4}{3} \pi (A^{1/3} R_0)^3} $

Simplifying the equation shows that the $ A $ terms cancel out:

$ \rho = \frac{m_n}{\frac{4}{3} \pi R_0^3} $

This reveals that the nuclear density $ \rho $ is approximately constant and independent of the mass number $ A $. Therefore, the densities of copper $(^{64}_{29} \text{Cu})$ and carbon $(^{12}_{6} \text{C})$ nuclei are effectively the same because the density formula results in a similar value regardless of the specific mass number.

To determine the atomic number (Z) of a hydrogen-like ion, given that the energy difference between its second excitation state and the ground state is 108.8 eV, we use the following formula for the energy difference between two energy levels in a hydrogen-like atom:

$ \Delta E = 13.6 \, \text{eV} \times Z^2 \left[\frac{1}{n_1^2} - \frac{1}{n_2^2}\right] $

Here, for the second excitation, $ n_2 = 3 $ (since excitation means moving to the second higher energy level beyond the ground state, which means $ n_1 = 1 $ for the ground state to $ n_2 = 3 $). Plug these into the energy difference formula:

$ \Delta E = 13.6 \, Z^2 \left[\frac{1}{1^2} - \frac{1}{3^2}\right] $

Simplifying the term in brackets:

$ \Delta E = 13.6 \, Z^2 \left[1 - \frac{1}{9}\right] = 13.6 \, Z^2 \times \frac{8}{9} $

Given that $\Delta E = 108.8 \, \text{eV}$, set the equations equal:

$ 13.6 \, \frac{8Z^2}{9} = 108.8 $

Solving for $ Z $, first simplify:

$ 13.6 \times \frac{8}{9} \times Z^2 = 108.8 $

$ \frac{108.8 \times 9}{13.6 \times 8} = Z^2 $

$ Z^2 = 9 $

Thus, the atomic number $ Z $ is:

$ Z = 3 $

The atomic number of the ion is 3.

Lyman

$\begin{aligned} & \frac{1}{\lambda_1}=\mathrm{R}\left[\frac{1}{1}-\frac{1}{4}\right]=\frac{3 \mathrm{R}}{4} \\ & \lambda_1=\frac{4}{3 \mathrm{R}}\quad\text{..... (1)} \end{aligned}$

and Balmer

$\begin{aligned} &\begin{aligned} & \frac{1}{\lambda_2}=\mathrm{R}\left[\frac{1}{4}-\frac{1}{9}\right]=\frac{5 \mathrm{R}}{36} \\ & \lambda_2=\frac{36}{5 \mathrm{R}} \end{aligned}\\ &\text { Then, } \frac{\lambda_1}{\lambda_2}=\frac{5}{27} \end{aligned}$

$\begin{aligned} &\begin{aligned} & \mathrm{E}=\mathrm{hf} \\ & \mathrm{ML}^2 \mathrm{~T}^{-2}=[\mathrm{h}] \times\left[\mathrm{T}^{-1}\right] \\ & {[\mathrm{h}]=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]} \\ & \mathrm{L}=[\mathrm{MVR}]=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right] \\ & \mathrm{L}=\frac{\mathrm{nh}}{2 \pi} \end{aligned}\\ &\mathrm{L} \text { is integral multiple of } \frac{\mathrm{h}}{2 \pi} \end{aligned}$

In the Bohr model for a hydrogen-like atom, the radius of the $ n^{\text{th}} $ orbit is given by

$ r_n = \frac{n^2 a_0}{Z} $

where:

$ n $ is the principal quantum number,

$ a_0 $ is the Bohr radius, and

$ Z $ is the atomic number (the effective nuclear charge).

For the $ 5^{\text{th}} $ orbit ($ n = 5 $):

For $ \mathrm{Li}^{2+} $ (where $ Z = 3 $):

$ r_5(\mathrm{Li}^{2+}) = \frac{5^2 a_0}{3} = \frac{25a_0}{3} $

For $ \mathrm{He}^{+} $ (where $ Z = 2 $):

$ r_5(\mathrm{He}^{+}) = \frac{5^2 a_0}{2} = \frac{25a_0}{2} $

Now, the ratio of the radius of the 5th orbit of $ \mathrm{Li}^{2+} $ to that of $ \mathrm{He}^{+} $ is:

$ \frac{r_5(\mathrm{Li}^{2+})}{r_5(\mathrm{He}^{+})} = \frac{\frac{25a_0}{3}}{\frac{25a_0}{2}} = \frac{25a_0}{3} \times \frac{2}{25a_0} = \frac{2}{3} $

Thus, the ratio is $ \frac{2}{3} $.

The correct answer is Option B.

Option B

Both statements are true, and the absence of electron–electron repulsion in Bohr’s formulation is exactly why it works only for one‐electron systems (hydrogen and hydrogen‐like ions).

• Assertion A is true: Bohr’s model applies only to atoms with a single electron.

• Reason R is true: Bohr treated only the Coulomb attraction between nucleus and electron, omitting any repulsive force between multiple electrons.

• Since ignoring e–e repulsion prevents extension to multi–electron atoms, R correctly explains A.

Here’s the correct matching:

A.

$_0^1\text{n}+_{92}^{235}\text{U}\to_{54}^{140}\text{Xe}+_{38}^{94}\text{Sr}+2\,_0^1\text{n}$

–– This is nuclear fission ⇒ III.

B.

$2\text{H}_2+\text{O}_2\to2\text{H}_2\text{O}$

–– This is a chemical reaction ⇒ I.

C.

$_1^2\text{H}+_1^2\text{H}\to_2^3\text{He}+_0^1\text{n}$

–– This D–D fusion releases about 3.3 MeV ⇒ fusion with +ve Q ⇒ II.

D.

$_1^1\text{H}+_1^3\text{H}\to_1^2\text{H}+_1^2\text{H}$

–– The mass of products exceeds reactants ⇒ requires energy ⇒ fusion with –ve Q ⇒ IV.

So the answer is

Option B: A–III, B–I, C–II, D–IV.

To calculate the energy released when two deuterons (${}_1 \mathrm{H}^2$) fuse to form a helium nucleus (${}_2 \mathrm{He}^4$), consider the following:

The reaction is represented as:

$ {}_1{H^2} + {}_1{H^2} \to {}_2{He^4} $

Given the binding energy per nucleon:

For ${}_1 \mathrm{H}^2$: 1.1 MeV

For ${}_2 \mathrm{He}^4$: 7.0 MeV

Calculating Binding Energy

Binding Energy for Reactants:

Each deuteron (${}_1 \mathrm{H}^2$) has a binding energy of $1.1 \, \text{MeV}$ per nucleon.

Since there are two nucleons in a deuteron, the total binding energy for one deuteron is $1.1 \times 2 = 2.2 \, \text{MeV}$.

Therefore, for two deuterons: $2.2 \times 2 = 4.4 \, \text{MeV}$.

Binding Energy for Product:

For ${}_2 \mathrm{He}^4$: Each of the four nucleons has a binding energy of $7.0 \, \text{MeV}$.

Total binding energy for the helium nucleus: $7.0 \times 4 = 28.0 \, \text{MeV}$.

Energy Released (Q)

The energy released, $ \mathrm{Q} $, is the difference between the binding energy of the products and the reactants:

$ \mathrm{Q} = \mathrm{BE}_{\text{product}} - \mathrm{BE}_{\text{reactants}} = 28.0 \, \text{MeV} - 4.4 \, \text{MeV} = 23.6 \, \text{MeV} $

Thus, the energy released when two deuterons fuse to form a helium nucleus is 23.6 MeV.

The radius of a hydrogen-like ion according to Bohr's atomic model is given by the formula:

$ r = r_0 \frac{n^2}{Z} $

Where:

$ r_0 $ is the first Bohr radius ($ a_0 $)

$ n $ represents the principal quantum number

$ Z $ is the atomic number of the ion

For the ion $ \text{Li}^{++} $, the atomic number $ Z = 3 $ and we are considering the ground state, so $ n = 1 $.

Plugging these values into the formula, we get:

$ r = r_0 \frac{1^2}{3} = \frac{r_0}{3} $

This shows that the radius of $\text{Li}^{++}$ in its ground state is $\frac{1}{3} a_0$, meaning $ X = 3 $.

In Bohr's atomic model, the energy of an electron in an atom is given by the formula:

$ E \propto \frac{Z^2}{n^2} $

where $ Z $ is the atomic number and $ n $ is the principal quantum number of the electron's orbit. Let’s consider the implications:

For hydrogen ($ \mathrm{H} $), $ Z = 1 $.

For helium ion ($ \mathrm{He}^{+} $), $ Z = 2 $.

For lithium ion ($ \mathrm{Li}^{++} $), $ Z = 3 $.

Now, identifying the relevant states:

Ground State: This corresponds to $ n = 1 $.

First Excited State: This corresponds to $ n = 2 $.

Second Excited State: This corresponds to $ n = 3 $.

Let's evaluate the energy comparisons:

Hydrogen Atom in Ground State: $ E \propto \frac{1^2}{1^2} = 1 $.

Helium Ion ($ \mathrm{He}^{+} $) in the First Excited State:

$ E \propto \frac{2^2}{2^2} = 1 $

This matches the energy of hydrogen in its ground state.

Lithium Ion ($ \mathrm{Li}^{++} $) in the Second Excited State:

$ E \propto \frac{3^2}{3^2} = 1 $

This also matches the energy of hydrogen in its ground state.

From this analysis, we can conclude:

The energy of a hydrogen atom in its ground state is equal to the energy of a $ \mathrm{He}^{+} $ ion in its first excited state.

The energy of a hydrogen atom in its ground state also equals the energy of a $ \mathrm{Li}^{++} $ ion in its second excited state.

Hence, statements (A) and (B) are correct.

When an e transition from n$^{th}$ to ground state then $\sum {(n - 1)} $ spectral lines are produced.

Given, $n=4$

So, $\sum {(4 - 1) = \sum {3 = 3 + 2 + 1 = 6} } $

Hence, 6 spectral lines are emitted by atomic Hydrogen that is in the $4^{th}$ energy level

Frequency of revolution $\propto \frac{1}{\mathrm{n}^3}$

We know, theoretical equation for $\beta^-$ decay,

Hence, option 2 is correct.

$\begin{aligned} & {[E]=M^1 L^2 T^{-2}} \\ & {[P c]=M^1 L^1 T^{-1} \cdot L^1 T^{-1}=M^1 L^2 T^{-2}} \\ & {\left[\mathrm{mc}^2\right]=\mathrm{M}^1 L^2 T^{-2}} \\ & E^2=\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2} \\ & E^2=\mathrm{P}^2 \mathrm{c}^2+\mathrm{m}^2 \mathrm{c}^4 \end{aligned}$

$ \frac{1}{\lambda_{AC}} = \frac{E_A - E_C}{hc}, \quad \frac{1}{\lambda_{BC}} = \frac{E_B - E_C}{hc} $

Since the electron transitions from state A to state C in two steps (A to B and then B to C), the energy difference for the transition from A to B is given by

$ E_A - E_B = (E_A - E_C) - (E_B - E_C) $

Expressing these energy differences in terms of wavelengths,

$ \frac{hc}{\lambda_{AB}} = \frac{hc}{\lambda_{AC}} - \frac{hc}{\lambda_{BC}} $

Cancelling out $hc$, we obtain

$ \frac{1}{\lambda_{AB}} = \frac{1}{\lambda_{AC}} - \frac{1}{\lambda_{BC}} $

Substitute the given values:

$ \frac{1}{\lambda_{AB}} = \frac{1}{2000} - \frac{1}{6000} $

Finding a common denominator:

$ \frac{1}{\lambda_{AB}} = \frac{3 - 1}{6000} = \frac{2}{6000} = \frac{1}{3000} $

Thus, the wavelength for the A to B transition is

$ \lambda_{AB} = 3000 \, \mathring{A}. $

This matches Option D.

Let's analyze both statements:

Assertion (A): “The binding energy per nucleon is found to be practically independent of the atomic number A, for nuclei with mass numbers between 30 and 170.”

In fact, for nuclei in this mass range, the binding energy per nucleon is nearly constant (around 7–8 MeV per nucleon). This is because the attractive nuclear force saturates, leading to a relatively uniform binding energy per nucleon in medium to heavy nuclei.

Therefore, Assertion (A) is true.

Reason (R): “Nuclear force is long range.”

The nuclear force (or strong force) that holds the nucleons (protons and neutrons) together is actually very short-ranged. Its effective range is approximately 1 to 3 femtometers (fm), after which the force rapidly decreases.

Hence, Reason (R) is false.

Since the assertion is correct and the reason is incorrect, the correct answer is:

Option A: (A) is true but (R) is false.

Let's work through the problem step by step.

The decay of a radioactive nucleus is given by the formula:

$N(t) = N_0 \, e^{-\lambda t},$

where:

$N(t)$ is the number of nuclei remaining at time $t$,

$N_0$ is the initial number of nuclei,

$\lambda$ is the decay constant.

For nucleus $n_1$ with decay constant $\lambda_1$, its half-life $t_{1/2}$ is:

$t_{1/2} = \frac{\ln 2}{\lambda_1}.$

After one half-life of $n_1$, the number of $n_1$ nuclei remaining is:

$N_1 = N_0 \, e^{-\lambda_1 \, t_{1/2}} = N_0 \, e^{-\lambda_1 \, \frac{\ln2}{\lambda_1}} = N_0 \, e^{-\ln2} = \frac{N_0}{2}.$

Nucleus $n_2$ has a decay constant that is 3 times that of $n_1$:

$\lambda_2 = 3\lambda_1.$

For nucleus $n_2$, the number of nuclei remaining after one half-life of $n_1$ (which is $t = \frac{\ln2}{\lambda_1}$) is:

$ N_2 = N_0 \, e^{-\lambda_2 \, t} = N_0\, e^{-3\lambda_1 \, \frac{\ln2}{\lambda_1}} = N_0 \, e^{-3\ln2} = N_0 \cdot \frac{1}{2^3} = \frac{N_0}{8}. $

Now, we find the ratio of $n_2$ nuclei to $n_1$ nuclei after this time:

$ \frac{N_2}{N_1} = \frac{\frac{N_0}{8}}{\frac{N_0}{2}} = \frac{1/8}{1/2} = \frac{1}{4}. $

Therefore, the correct ratio is $\frac{1}{4}$, which corresponds to Option A.

(P) For a hydrogen - like atom (atomic. number $z$ ), the energy of an

electron in the n-th orbit (Bohr model) is, $E_n=\frac{-13.6 z^2}{n^2} \mathrm{eV}$

So $E \propto z^2$

So energy of radiation due to electronic transitions from hydrogen like atoms can be written as

$ \begin{aligned} & E=-13.6 z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \\ \Rightarrow & E \propto z^2 \end{aligned} $

$ \text { so }(p) \rightarrow(s) $

(Q) Characteristic $x$-rays are emitted when an electron from a higher shell fills a vacancy in an inner shell (eg. K-shell). The energy of the emitted $x$-ray corresponds to the energy difference between the shells.

For $k$-shell $x$-rays,

$ \text { L- shell }(n=2) \rightarrow k \text {-shell }(n=1) $ transition occurs.

The energy levels, are modified by screening effects as the inner electron sees an effective nuclear charge of approximately $(z-1)$ due to screening by the other $k$-shell electron.

So using Bohr-like model with effective charge :

$ E_1=\frac{-13.6(z-1)^2}{1^2}, E_2=\frac{-13.6(z-1)^2}{2^2} $

So $\Delta E=E_2-E_1$

$ \begin{aligned} & =13 \cdot 6(z-1)^2\left[\frac{1}{1^2}-\frac{1}{2^2}\right] \\ \Rightarrow & \Delta E \propto(z-1)^2 \\ \Rightarrow & E \propto(z-1)^2 \end{aligned} $

So $($ a $) \longrightarrow(1)$

(R) $E \propto z(z-1)$

Let's find the electrostatic energy between, protons in the nucleus: A nucleus with atomic number z has z protons.

The electrostatic (coulomb) energy arises from the repulsion between each pair of protons.

So number of paps $=z_{c_2}=\frac{z(z-1)}{2}$

Coulomb energy between two protons separated by a distance $r$.

$ E_1=\frac{k q_1 q_2}{r}=\frac{1}{4 \pi \varepsilon_0} \frac{e \cdot e}{r}=\frac{e^2}{4 \pi \varepsilon_0 r} $

So total electrostatic energy.

$ \begin{aligned} E & =\text { number of pairs } \times E_1 \\ \Rightarrow E & =\frac{z(z-1)}{2} \cdot \frac{e^2}{4 \pi \varepsilon_0 r} \\ \Rightarrow E & \propto z(z-1) \end{aligned} $

So $(R) \longrightarrow(2)$

(s) The binding energy per nucleon is a measure of the stability of a nucleus and is given by the semi-empirical mass formula. For most nuclei (except very light ones), the average binding energy per nucleon is approximately constant, around $8-9 \mathrm{Mev}$ due to the saturation of nuclear forces. i.e., it is independent of $z$.

so ( S ) $\rightarrow$ (4)

Hence, option (C) is correct.

To find the maximum wavelength, we use the ionisation energy as the energy of the photon.

$ \begin{aligned} \text { i.e., } \quad E & =\frac{h c}{\lambda} \\ \Rightarrow \quad \lambda & =\frac{h c}{E}=\frac{1240 \mathrm{eV} \cdot \mathrm{~nm}}{13.6 \mathrm{eV}} \\ {[h c=1240 \mathrm{eV} \cdot \mathrm{~nm}] } & \\ & =91.2 \mathrm{~nm}=912\mathop {\rm{A}}\limits^{\rm{o}}\end{aligned} $

Change in mass number

$ \begin{aligned} & =232-208 \\ & =24 \end{aligned} $

∴ Number of $\alpha$-particles $=\frac{24}{4}=6$

Each beta particle has no change in mass number but increases the atomic number by $1\left({ }_{-1}^0 e\right)$.

Initial atomic number of Thorium $(\mathrm{Th})=90$.

Atomic number after emitting 6 alpha particles $=90-(6 \times 2)=90-12=78$

Final atomic number of Lead $(\mathrm{Pb})=82$.

The increase in atomic number from 78 to 82 must be due to beta decay.

Number of beta particles $=$ Final atomic number - Atomic number after alpha decay

$ \begin{aligned} & =82-78 \\ & =4 \end{aligned} $

Momentum is conserved

$ \begin{aligned} & 200 \times V_2=8 V_1 \\ & V_2=\frac{8}{200} V_1=\frac{1}{25} V_1 \end{aligned} $

Energy of two $\alpha$-particles

$ \begin{aligned} & \quad=\frac{1}{2}\left(m_\alpha\right) V_1^2 \times 2=2 E \text { (Given) } \\ & \Rightarrow \quad \frac{1}{2} m_\alpha V_1^2=E \\ & \Rightarrow \quad \frac{1}{2} \times 4 \times V_1^2=E \quad\left(\therefore m_\alpha=4\right) \\ & \text { or } \quad V_1^2=\frac{E}{2} \end{aligned} $

Energy of $\alpha$-particles and daughter nucleus is,

$ \begin{aligned} \mathrm{KE}= & \frac{1}{2}(200) \cdot V_2^2+2\left(\frac{1}{2} 4 \times V_1^2\right) \\ & =\frac{1}{2} \times 200 \times \frac{1}{25 \times 25} \times V_1^2+4 V_1^2 \\ & =\frac{4}{25} V_1^2+4 V_1^2 \\ & =\frac{104}{25} \times V_1^2 \\ & =\frac{104}{25} \times \frac{E}{2}=\frac{52}{25} E . \end{aligned} $

In classical mechanics, the total energy (E) of a particle moving under an inverse square law force (like the electrostatic force between an electron and a nucleus) determines the nature of its orbit:

1. If E < 0 (Negative Total Energy): The particle is in a bound state. Its orbit is closed and elliptical (with a circular orbit being a special case of an ellipse). The particle is trapped by the central force.

2. If E = 0 (Zero Total Energy): The particle is at the boundary between bound and unbound states. Its orbit is open and parabolic. The particle just barely escapes the influence of the central force (it reaches infinity with zero kinetic energy).

3. If E > 0 (Positive Total Energy): The particle is in an unbound state. Its orbit is open and hyperbolic. The particle approaches the central force, is deflected, and then moves away to infinity, never to return. It has enough kinetic energy to overcome the attractive potential and escape.

Given that the total energy of the electron is positive (E > 0), the electron is in an unbound state. This means it will not be trapped in a closed orbit around the nucleus. Instead, it will follow an open trajectory (a hyperbola).

Let's evaluate the options:

• A. electron will revolve in a circular orbit. Circular orbits are closed orbits and correspond to E < 0. Incorrect.

• B. electron will revolve in an elliptical orbit. Elliptical orbits are closed orbits and correspond to E < 0. Incorrect.

• C. electron will not follow a closed orbit. This matches our understanding for E > 0, where the orbit is open (hyperbolic). Correct.

• D. electron will fall into the nucleus. Falling into the nucleus implies a highly negative energy or zero angular momentum, leading to a collision, or continuous energy loss through radiation. A positive energy means the electron has more than enough energy to escape, not fall in. Incorrect.

The final answer is $\boxed{C}$

Percentage of atoms left

$ \begin{aligned} & =100-87.5 \\ & =12.5 \% \\ \text { Using } \frac{N}{N_0} & =\left(\frac{1}{2}\right)^{\frac{t}{T_{1 / 2}}} \\ \frac{125}{100} & =\left(\frac{1}{2}\right)^{\frac{6}{T_{1 / 2}}} \\ \left(\frac{1}{2}\right)^3 & =\left(\frac{1}{2}\right)^{\frac{6}{T_{1 / 2}}} \Rightarrow 3=\frac{6}{T_{1 / 2}} \\ \Rightarrow \quad T_{1 / 2} & =2 \text { day } \end{aligned} $

Now, in 8 days, 4 half-life get completed

⇒ Fraction of atoms left

$ =\frac{100}{2^4 \times 100}=\frac{6.25}{100} $

⇒ Fraction of atoms decay $=\frac{100-6.25}{100}$

$ =\frac{93.75}{100}=\frac{15}{16} $

The mass number of a nucleus is shown by $A$.

The radius of a nucleus depends on the mass number as $R = R_0 A^{1/3}$, where $R_0$ is a constant.

To compare the radii of two nuclei, use the formula:

$\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1 / 3}$

Here, $\frac{A_1}{A_2} = \frac{27}{125}$.

So,

$\frac{R_1}{R_2} = \left(\frac{27}{125}\right)^{1/3} = \frac{3}{5}$

because $27^{1/3} = 3$ and $125^{1/3} = 5$.

The surface area of a sphere is $S = 4\pi R^2$.

The ratio of their surface areas is:

$\frac{S_1}{S_2} = \frac{4\pi R_1^2}{4\pi R_2^2} = \left(\frac{R_1}{R_2}\right)^2 = \left(\frac{3}{5}\right)^2 = \frac{9}{25}$

So, the ratio of their surface areas is $S_1 : S_2 = 9 : 25$.

The range of weak nuclear force is the order of $10^{-16} \mathrm{~m}$. This force is responsible for radioactive decay and acte over very short distances within the nucleus.

Step 1: Find the Total Energy (TE) from the Given Potential Energy (PE)

The potential energy of the electron is given as -6.8 eV.

We use the relation: $ \mathrm{PE} = 2(\mathrm{TE}) $

This means: $ \mathrm{TE} = \frac{\mathrm{PE}}{2} = \frac{-6.8}{2} = -3.4\,\mathrm{eV} $

Step 2: Find the Value of n (Principal Quantum Number)

For a hydrogen atom, the total energy formula is: $ \mathrm{TE} = \frac{-13.6}{n^2} $

Set this equal to the TE we found: $ -3.4 = \frac{-13.6}{n^2} $

Solve for $ n^2 $: $ n^2 = 4 $

So, $ n = 2 $

Step 3: The Electron is in the Second Orbit

Since $ n = 2 $, the electron is in the second Bohr orbit.

Step 4: Use Bohr's Model and de-Broglie Hypothesis

According to Bohr and de-Broglie:

$ 2\pi r_n = n\lambda $

This gives: $ \lambda = \frac{2\pi r_n}{n} \tag{i} $

Step 5: Use the Formula for the Radius of Orbit

Radius of the nth orbit: $ r_n = n^2 r_0 $

For $ n = 2 $ (second orbit): $ r_2 = 2^2 r_0 = 4r_0 $

Step 6: Calculate the de-Broglie Wavelength

Plug $ r_2 $ into the earlier formula:

$ \lambda = \frac{2\pi r_2}{2} = \frac{2\pi}{2} \times 4r_0 = 4\pi r_0 $