Atoms and Nuclei

Let's analyze each of the given statements to determine which ones are correct:

(A) The angular momentum of an electron in $n^{\text{th}}$ orbit is an integral multiple of $\hbar$.

This statement reflects Bohr's quantization rule for angular momentum in the Bohr model of the hydrogen atom. According to this rule, the angular momentum of an electron in a stationary orbit is quantized and given by:

where $n$ is a principal quantum number (which can be any positive integer), and $\hbar$ is the reduced Planck's constant. Therefore, this statement (A) is correct.

(B) Nuclear forces do not obey inverse square law.

Nuclear forces, specifically strong nuclear forces, do act over short ranges within the nucleus but do not obey the inverse square law, which is characteristic of the electromagnetic and gravitational forces. Thus, statement (B) is correct.

(C) Nuclear forces are spin dependent.

The strength of the nuclear force can depend on the spin alignment of the nucleons. This is why some isotopes are more stable than others depending on spin-related factors. Thus, statement (C) is correct.

(D) Nuclear forces are central and charge independent.

The strong nuclear force is indeed charge independent, meaning it is the same regardless of the types of nucleons involved (neutrons or protons). However, nuclear forces are not always central; they can be tensor forces too, which involve more complex interactions that are not purely central. Therefore, the entirety of statement (D) is not correct; the statement should specify that nuclear forces are charge independent but may not always be central.

(E) Stability of nucleus is inversely proportional to the value of packing fraction.

This statement (E) is correct.

Based on the above explanations, the correct statements are (A), (B), (C) and (E). Hence, the correct answer is:

The energy $E_n$ of an electron in the $n$th energy level of a hydrogen atom is given by the formula:

$ E_n = -\frac{13.6 \, \text{eV}}{n^2} $

where:

• $E_n$ is the energy in electron volts (eV),


• $13.6 \, \text{eV}$ is the ionization energy of hydrogen (the energy required to remove the electron from the ground state, $n=1$),


• $n$ is the principal quantum number (an integer starting from 1).

The Balmer Series

The Balmer series is a specific set of spectral lines of hydrogen that are visible to the naked eye. It results from the electron making transitions between energy levels, specifically from levels with $n \geq 3$ (i.e., from higher energy states) down to $n = 2$. These transitions release energy in the form of electromagnetic radiation, which we observe as the Balmer series.

1. Ground State Energy ($E_1$): The energy of the electron in the ground state ($n=1$) of hydrogen is:

$ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} $

This represents the electron's energy level when it's closest to the nucleus.

1. Energy for the $n=3$ Level ($E_3$): For an electron in the $n=3$ level, its energy is:

$ E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -1.51 \, \text{eV} $

1. Minimum Energy for Balmer Series Transition: The minimum energy transition within the Balmer series is from $n=3$ to $n=2$, but to understand the total energy required for an electron to emit radiation in the Balmer series, starting from the ground state, we consider the energy needed to first excite the electron from $n=1$ to $n=3$.


2. Energy Difference ($\Delta E$): The energy emitted as radiation for the electron to transition from the ground state to a state where it can participate in the Balmer series is the difference between the ground state energy and the energy of the $n=3$ state:

$ \Delta E = E_3 - E_1 = -1.51 \, \text{eV} - (-13.6 \, \text{eV}) = 12.09 \, \text{eV} $

Note

This calculation shows that the minimum energy required by a hydrogen atom in the ground state to emit radiation in the Balmer series is approximately 12.1 eV. The energy difference essentially represents the energy that must be supplied to an electron to excite it from the ground state ($n=1$) to an excited state ($n=3$) from which it can then transition to $n=2$, emitting radiation observable as part of the Balmer series.

$\begin{aligned} & \mathrm{R}_1=\frac{\mathrm{R}_2}{2} \\ & \mathrm{R}_0\left(\mathrm{~A}_1\right)^{1 / 3}=\frac{\mathrm{R}_0}{2}\left(\mathrm{~A}_2\right)^{1 / 3} \\ & \mathrm{~A}_1=\frac{1}{8} \mathrm{~A}_2 \\ & \mathrm{~A}_1=\frac{192}{8}=24 \end{aligned}$

$\begin{aligned} & \frac{1}{\lambda}=\frac{13.6 \mathrm{z}^2}{\mathrm{hc}}\left[\frac{1}{1^2}-\frac{1}{2^2}\right] ...... \text{(i)}\\ & \frac{1}{\lambda^{\prime}}=\frac{13.6 \mathrm{z}^2}{\mathrm{hc}}\left[\frac{1}{1^2}-\frac{1}{3^2}\right] ...... \text{(ii)} \end{aligned}$

On dividing (i) & (ii)

$\lambda^{\prime}=\frac{27}{32} \lambda$

To determine the relationship between the magnetic moment and the principal quantum number $ n $, we need to understand the formula for the magnetic moment of an electron in a Bohr orbit.

The magnetic moment ($ \mu_n $) of an electron in the nth Bohr orbit is given by:

$ \mu_n = \frac{n \cdot e}{2m} \cdot \frac{e \cdot Z \cdot h}{2 \pi m} $

Where:

• $ e $ is the charge of the electron
• $ m $ is the mass of the electron
• $ Z $ is the atomic number (for a hydrogen atom, $ Z = 1 $)
• $ h $ is Planck's constant

Since the Bohr's magneton ($ \mu_B $) is defined as:

$ \mu_B = \frac{e \hbar}{2m} $

The magnetic moment for the nth orbit can be written as:

$ \mu_n = n \cdot \mu_B $

From the above formula, it is clear that the magnetic moment $ \mu_n $ is directly proportional to the principal quantum number $ n $. Mathematically, this relationship can be expressed as:

$ \mu_n \propto n^1 $

Therefore, the value of $ x $ is 1.

Answer: Option D (1)

$\begin{aligned} & \begin{aligned} & (\mathrm{X}) \rightarrow(\mathrm{Y})+(\mathrm{Z})+(\mathrm{P}) \\ & \begin{array}{llll} \mathrm{M} & \mathrm{M} / 3 & \mathrm{M} / 3 & \mathrm{M} / 3 \end{array} \\ \end{aligned} \\ & \Delta \mathrm{Mc}^2=\frac{1}{2} \frac{\mathrm{M}}{3} \mathrm{~V}^2+\frac{1}{2} \frac{\mathrm{M}}{3} \mathrm{~V}^2+\frac{1}{2} \frac{\mathrm{M}}{3} \mathrm{~V}^2 \\ & \mathrm{~V}=\mathrm{c} \sqrt{\frac{2 \Delta \mathrm{M}}{\mathrm{M}}} \end{aligned}$

$\frac{1}{2}|P E|=K E$ for each value of $\mathrm{n}$ (orbit)

$\therefore \frac{K E}{|P E|}=\frac{1}{2}$

According to Rutherford atomic model, most of mass of atom and all its positive charge is concentrated in tiny nucleus & electron revolve around it.

According to Thomson atomic model, atom is spherical cloud of positive charge with electron embedded in it.

Hence, Statement I is true but statement II false.

$\begin{aligned} & { }_3 \mathrm{Li}^6+{ }_0 \mathrm{n}^1 \rightarrow{ }_2 \mathrm{He}^4+{ }_1 \mathrm{H}^3 \\ & { }_1 \mathrm{H}^2+{ }_1 \mathrm{H}^3 \rightarrow{ }_2 \mathrm{He}^4+{ }_0 \mathrm{n}^1 \\ & \hline{ }_3 \mathrm{Li}^6+{ }_1 \mathrm{H}^2 \rightarrow 2\left({ }_2 \mathrm{He}^4\right) \\ & \hline \end{aligned}$

Energy released in process

$\begin{aligned} & \mathrm{Q}=\Delta \mathrm{mc}^2 \\ & \mathrm{Q}=\left[\mathrm{M}(\mathrm{Li})+\mathrm{M}\left(\mathrm{H}^2\right)-2 \times \mathrm{M}\left({ }_2 \mathrm{He}^4\right)\right] \times 931.5 \mathrm{~MeV} \\ & \mathrm{Q}=[6.01690+2.01471-2 \times 4.00388] \times 931.5 \mathrm{~MeV} \\ & \mathrm{Q}=22.216 \mathrm{~MeV} \\ & \mathrm{Q}=22.22 \mathrm{~MeV} \end{aligned}$

$\begin{aligned} & { }_6 \mathrm{C}^{13}+\text { Energy } \rightarrow{ }_6 \mathrm{C}^{12}+{ }_0 \mathrm{n}^1 \\ & \Delta \mathrm{m}=(12.000000+1.008665)-13.003354 \\ & =-0.00531 \mathrm{u} \\ & \therefore \text { Energy required }=0.00531 \times 931.5 \mathrm{~MeV} \\ & =4.95 \mathrm{~MeV} \end{aligned}$

The radius of the nth stationary orbit in the Bohr model of an atom is directly proportional to the square of its principal quantum number n and inversely proportional to the atomic number Z. This relationship is represented by the formula $r_n = \frac{n^2h^2}{4\pi^2kme^2Z},$ where :

• $n$ is the principal quantum number,

• $h$ is Planck's constant,

• $k$ is the Coulomb constant,

• $m$ is the mass of the electron,

• $e$ is the charge of the electron, and

• $Z$ is the atomic number of the nucleus (for hydrogen, Z=1).

To find the radius of the fourth orbit ($r_4$), in comparison to the third orbit ($r_3$), we apply the formula with $n=4$ for the fourth orbit and $n=3$ for the third orbit, and simplify as follows:

$\begin{aligned} & \frac{r_4}{r_3} = \frac{(4^2)h^2}{4\pi^2kme^2Z} \div \frac{(3^2)h^2}{4\pi^2kme^2Z} = \frac{(4^2)}{(3^2)} \\\\ & = \frac{16}{9} \end{aligned}$

Thus, the radius of the fourth stationary orbit is $\frac{16}{9}$ times the radius of the third stationary orbit, $R$. Therefore, $r_4 = \frac{16}{9} R$.

The central force will provide necessary centripetal force

$\Rightarrow \mathrm{kr}=\frac{\mathrm{mv}^2}{\mathrm{r}}$

or, $\mathrm{kr}^2=\mathrm{mv}^2\quad \text{... (1)}$

By quantisation rule $\mathrm{n} \hbar=\mathrm{mvr}$

or, $\frac{\mathrm{n} \hbar}{\mathrm{r}}=\mathrm{mv}\quad \text{... (2)}$

$\begin{aligned} & \frac{(1)}{(2)^2} \Rightarrow \frac{\mathrm{kr}^2}{\frac{\mathrm{n}^2 \hbar^2}{\mathrm{r}^2}}=\frac{\mathrm{mv}^2}{\mathrm{~m}^2 \mathrm{v}^2} \\ & \Rightarrow \frac{\mathrm{k}}{\mathrm{n}^2 \hbar^2} \mathrm{r}^4=\frac{1}{\mathrm{~m}} \\ & \Rightarrow \mathrm{r}=\left(\frac{\mathrm{n}^2 \hbar^2}{\mathrm{~km}}\right)^{\frac{1}{4}} \Rightarrow \mathrm{r}^2=\frac{\mathrm{n} \hbar}{\sqrt{\mathrm{mk}}} \end{aligned}$

(B) Using (1), $\mathrm{K} \cdot \frac{\mathrm{n} \hbar}{\sqrt{\mathrm{mk}}}=\mathrm{mv}^2$

$\Rightarrow \mathrm{v}^2=\mathrm{n} \hbar \sqrt{\frac{\mathrm{k}}{\mathrm{m}^3}}$

(C) $\frac{\mathrm{L}}{\mathrm{mr}^2}=\frac{\mathrm{mvr}}{\mathrm{mr}^2}=\frac{\mathrm{v}}{\mathrm{r}}=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$ from (1)

(D) $\mathrm{E}=\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{kr}^2=\frac{\mathrm{n} \hbar}{2} \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}+\frac{1}{2} \mathrm{k} \frac{\mathrm{n} \hbar}{\sqrt{\mathrm{mk}}}$

$\mathrm{E}=\mathrm{n} \hbar \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$

In a hydrogen atom, the energy levels are given by

$ E_{n}=\frac{-13.6}{n^{2}} \mathrm{eV} $

where, $ n= $ principal quantum number Energy difference between two energy levels,

$ \begin{array}{l} \Delta E=E_{i}-E_{f} \\\\ h \nu=E_{i}-E_{f} \end{array} $

where, $ E_{i} $ and $ E_{f} $ are the initial and final energy levels.

For the transition from second orbit to first orbit.

$ \begin{array}{l} \Delta E=E_{2}-E_{1} \\\\ h v=\frac{-13.6}{(2)^{2}}-\left(-\frac{13.6}{(1)^{2}}\right) \\\\ h v=-13.6\left[\frac{1}{1}-\frac{1}{4}\right]=-13.6 \times \frac{3}{4} \mathrm{eV} \\\\ v=\frac{-10.2}{h} \mathrm{eV} \end{array} $

Given, $ v=f $, then

$ f=\frac{-10.2}{h} \mathrm{eV} $

For the transition third excited states $ (n=4) $ to first excited state $ (n=2) $.

$ \begin{array}{l} \Delta E^{\prime}=E_{4}-E_{2} \\\\ h f^{\prime}=\frac{-13.6}{4^{2}}+\frac{13.6}{4}=-13.6\left[\frac{1}{4}-\frac{1}{16}\right] \\\\ f^{\prime}=\frac{-2.55}{h} \mathrm{eV} \end{array} $

On comparing Eq. (i) and Eq. (ii), we get

$ f^{\prime}=\frac{f}{4} $

Given,

$ \frac{R_{1}}{R_{2}}=\frac{5}{3} \text { and } A_{2}=27, A_{1}=A $

The radius $ R $ of a nucleus is related to its mass number $ A $ by the formula

$ R \propto A^{\frac{1}{3}} $

Using.Eq. (i),

$ \frac{R_{1}}{R_{2}}=\left(\frac{A_{1}}{A_{2}}\right)^{\frac{1}{3}} $

$ \frac{5}{3}=\left(\frac{A}{27}\right)^{\frac{1}{3}} \Rightarrow \frac{5}{3}=\frac{(A)^{\frac{1}{3}}}{3} $

$ A=5^{3} $

$ A=125 $

So, the mass number $ A $ of nucleus $ { }_{52} X^{125} $ is 125 .

Given that the nucleus $ { }_{52} X^{125} $ has 52 protons, the number of neutrons $ N $ in the nucleus is

$ \begin{array}{l} N=A-\text { number of protons } \\\\ N=125-52 \\\\ N=73 \end{array} $

Given,

Half-life of $ A, T_{A}=T $

Half-life of $ B, T_{B}=2 T $

Initial number of nuclei is same for $ A $ and $ B $, it means $ N_{A}=N_{B}=N_{0} $.

The number of remaining nuclei $ N(t) $ after time $ t $ is given by the exponential decay formula,

$ N(t)=N_{0}\left(\frac{1}{2}\right)^{\frac{t}{T_{1} / 2}} $

where, $ N_{0}= $ Initial number of nuclei, $ T_{1 / 2}= $ Half-life After a time $ t=4 T $

$ \begin{aligned} \frac{N_{A}(4 T)}{N_{B}(4 T)} & =\frac{N_{0}\left(\frac{1}{2}\right)^{\frac{4 T}{T}}}{N_{0}\left(\frac{1}{2}\right)^{\frac{4 T}{2 T}}}=\frac{\left(\frac{1}{2}\right)^{4}}{\left(\frac{1}{2}\right)^{2}} =\left(\frac{1}{2}\right)^{2}=\frac{1}{4} \end{aligned} $

Therefore, the ratio of the remaining number of nuclei of $ A $ and $ B $ is $ 1: 4 $.

The ground state energy of a hydrogen atom is given as $-13.6 \, \text{eV}$. The total energy is equivalent to the kinetic energy, while the potential energy is twice the kinetic energy.

Therefore, the potential energy of the hydrogen atom in its ground state is:

$ \text{PE} = 2 \times (-13.6 \, \text{eV}) = -27.2 \, \text{eV} $

For the first excited state of the hydrogen atom, the energy can be calculated using the formula:

$ E_n = -\frac{13.6 \, \text{eV}}{n^2} $

When $n = 2$:

$ E_2 = \frac{-13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} $

Thus, the potential energy, when $n = 2$, is:

$ (\text{PE})_{n=2} = 2 \times (-3.4 \, \text{eV}) = -6.8 \, \text{eV} $

During a beta ($\beta$) decay process, the transformation of the parent nucleus can be represented as follows:

Let the parent nucleus be $ {}_B^A X $.

After a beta ($\beta$) decay, the transformation is:

$ {}_B^A X \longrightarrow {}_{B+1}^A Y + \beta^{-1} $

Here, $ {}_{B+1}^A Y $ represents the daughter nucleus.

In beta decay, the atomic number (denoted by $ B $) increases by 1, but the mass number (denoted by $ A $) remains unchanged. This means that the parent and daughter nuclei have the same mass number but different atomic numbers. Such nuclei are known as isobars.

The decay of a $\mathrm{A}_{92} \mathrm{U}^{238}$ nucleus into a ${ }_{82} \mathrm{Pb}^{214}$ nucleus involves emitting alpha and beta particles. Let's break down this process:

Decay Reaction:

$ { }_{92} \mathrm{U}^{238} \rightarrow { }_{82} \mathrm{~Pb}^{214} $

Mass Number Change:

Initial mass number: 238

Final mass number: 214

Difference in mass number: $238 - 214 = 24$

Each alpha particle reduces the mass number by 4 (since an alpha particle is $ \mathrm{He}^{4}_{2} $). Therefore, the number of alpha particles is:

$ \frac{24}{4} = 6 $

Atomic Number Change:

Initial atomic number: 92

Final atomic number: 82

Difference in atomic number: $92 - 82 = 10$

Alpha Particle Contribution:

Every alpha particle emission decreases the atomic number by 2.

With 6 alpha particles emitted, the atomic number reduces by:

$ 6 \times 2 = 12 $

This brings the atomic number to $92 - 12 = 80$.

Beta Particle Emission:

To adjust back from atomic number 80 to 82, beta particles are necessary.

Each beta particle ($\beta^-$) increases the atomic number by 1.

Therefore, 2 beta particles are emitted to increase the atomic number from 80 to 82.

In conclusion, in the decay process of ${ }_{92} \mathrm{U}^{238}$ to $ { }_{82} \mathrm{~Pb}^{214} $, 6 alpha particles and 2 beta particles are emitted:

$ { }_{92} \mathrm{U}^{238} \rightarrow { }_{82} \mathrm{~Pb}^{214} + 6 \, \alpha + 2 \, \beta^- $

Thus, the numbers of alpha and beta particles emitted are 6 and 2, respectively.

According to Bohr's postulate, the angular momentum is,

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

We have been asked for change in angular momentum in two consecutive orbit.

Let $ n_{1}=n $ and $ n_{2}=n+1 $

$ \begin{aligned} L_{1} & =\frac{n h}{2 \pi} \text { and } L_{2}=\frac{(n+1) h}{2 \pi} \\\\ \Delta L & =L_{2}-L_{1} \\\\ & =\frac{(n+1) h}{2 \pi}-\frac{n h}{2 \pi} \\\\ \Delta L & =\frac{h}{2 \pi} \end{aligned} $