Step 1: Theoretical Analysis of Thomson’s Atomic Model

According to J.J. Thomson’s "Plum Pudding" model, the atom is envisioned as a sphere of uniform positive charge density with electrons embedded within it. In this model, the positive charge occupies the entire volume of the atom, which is approximately $10^{-10} \, \text{m}$ in radius. [Per Thomson's historical model]

Step 2: Theoretical Analysis of Rutherford’s Atomic Model

Ernest Rutherford’s alpha-particle scattering experiment led to the discovery of the atomic nucleus. Rutherford concluded that the positive charge and the bulk of the mass of the atom are concentrated in a tiny central region called the nucleus, with a radius of the order of $10^{-15} \, \text{m}$. The electrons revolve around this nucleus in a vast empty space. Thus, the actual "atomic size" remains the same order of magnitude ($10^{-10} \, \text{m}$), but the distribution of matter differs significantly.

Step 3: Comparative Assessment of Atomic Dimensions

The question asks to compare the "size of the atom" in both models. Based on the experimental outcomes of the gold foil experiment, both models conceptually define the total radius of an atom as approximately $10^{-10} \, \text{m}$.

Step 4: Logical Deduction of the Alternative

Since both models attribute the outer boundary of the atom to the extent of the electron cloud, the spatial extent (radius) of the atom in both models is identical. Mathematically, $R_{\text{Thomson}} = R_{\text{Rutherford}} \approx 10^{-10} \, \text{m}$.

Final Solution: The size of the atom in Thomson’s model is the same as the atomic size in Rutherford’s model.

Step 1: Establishing the Theoretical Foundation

To analyze the equilibrium of electrons in atomic models, we must evaluate the force vectors acting upon them. According to Classical Electromagnetic Theory (specifically Maxwell’s equations), an accelerated charge must radiate energy. If an electron were in a "stable equilibrium," it would imply a state where the net force is zero and no energy radiation occurs, preventing the collapse of the atom. In contrast, any orbit characterized by constant change in velocity (acceleration) results in a non-zero net force.

Step 2: Analyzing J.J. Thomson's Plum Pudding Model

In the Thomson model, the positive charge is distributed uniformly throughout the sphere, and the electron is embedded within it. When the electron is displaced from its equilibrium position at the center, the positive charge distribution exerts a restoring force proportional to the displacement ($F = -kx$). [By Hooke's Law and Electrostatic Force Analysis]. At the center, the net force is zero, representing a position of stable equilibrium.

Step 3: Analyzing the Rutherford/Bohr Planetary Model

In the planetary model, electrons revolve around the nucleus in circular orbits. The force acting on the electron is the centripetal force provided by the electrostatic Coulomb force ($F_c = \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r^2}$). Since the electron is undergoing circular motion, its velocity vector is constantly changing direction. [By Newton’s Second Law of Motion ($F = ma$)], a change in velocity implies a net force directed toward the center of the orbit. Therefore, in the planetary model, the electron is never in equilibrium.

Step 4: Comparative Data Table

Step 5: Visual Representation of Force Vectors

Step 6: Conclusion based on Clues

The first blank refers to the state where the electron is held by a restoring force in a static potential, characteristic of the Thomson model. The second blank refers to the state of constant acceleration in an orbital path, characteristic of the Rutherford model.

Final Solution: In the ground state of Thomson model electrons are in stable equilibrium, while in Rutherford model electrons always experience a net force.

Step 1: Given Variables and Fundamental Constants

• Mass of Earth ($M$): $6.0 \times 10^{24}\ \text{kg}$
• Radius of orbit ($r$): $1.5 \times 10^{11}\ \text{m}$
• Orbital speed ($v$): $3.0 \times 10^{4}\ \text{m/s}$
• Planck's constant ($h$): $6.63 \times 10^{-34}\ \text{J}\cdot\text{s}$

Step 2: Theoretical Framework

According to the Bohr quantization postulate, the angular momentum ($L$) of a revolving body is quantized and must be an integral multiple of the reduced Planck constant ($\hbar = \frac{h}{2\pi}$). The relationship is expressed as:

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

[Per Bohr’s Postulate of Angular Momentum Quantization]

Step 3: Derivation of the Quantum Number ($n$)

To find the quantum number $n$, we rearrange the equation to solve for $n$:

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

Step 4: Substitution and Numerical Computation

Substitute the given values into the equation:

$n = \frac{2 \times 3.14159 \times (6.0 \times 10^{24}\ \text{kg}) \times (3.0 \times 10^{4}\ \text{m/s}) \times (1.5 \times 10^{11}\ \text{m})}{6.63 \times 10^{-34}\ \text{J}\cdot\text{s}}$

Calculating the numerator:

$\text{Numerator} = 2 \times 3.14159 \times 6.0 \times 3.0 \times 1.5 \times 10^{(24 + 4 + 11)}$

$\text{Numerator} \approx 169.646 \times 10^{39} = 1.696 \times 10^{41}$

Calculating $n$:

$n = \frac{1.696 \times 10^{41}}{6.63 \times 10^{-34}} \approx 0.2558 \times 10^{75} \approx 2.56 \times 10^{74}$

Step 5: Visualization of Orbital Dynamics

Final Solution: The quantum number characterizing the Earth's revolution is $n \approx 2.56 \times 10^{74}$. This extremely large value indicates that for macroscopic systems, the quantum nature of angular momentum is effectively continuous, aligning with the correspondence principle.

Step 1: Theoretical Foundation

According to the Bohr model of the hydrogen atom, the radius of the $n^{th}$ orbit is determined by the quantization of angular momentum. The expression for the radius $r_n$ of the $n^{th}$ orbit is given by:

$r_n = n^2 r_1$

[Per Bohr’s Quantization Postulate: $L = mvr = \frac{nh}{2\pi}$]

Where:

• $n$ is the principal quantum number (integer).
• $r_1$ is the radius of the innermost (ground state) orbit, defined as the Bohr radius $a_0$.
• $r_1 = 5.3 \times 10^{-11} \, \text{m}$.

Step 2: Calculation for the $n=2$ Orbit

Substituting $n = 2$ into the derived expression:

$r_2 = (2)^2 \times r_1$

$r_2 = 4 \times (5.3 \times 10^{-11} \, \text{m})$

$r_2 = 21.2 \times 10^{-11} \, \text{m}$

$r_2 = 2.12 \times 10^{-10} \, \text{m}$

Step 3: Calculation for the $n=3$ Orbit

Substituting $n = 3$ into the derived expression:

$r_3 = (3)^2 \times r_1$

$r_3 = 9 \times (5.3 \times 10^{-11} \, \text{m})$

$r_3 = 47.7 \times 10^{-11} \, \text{m}$

$r_3 = 4.77 \times 10^{-10} \, \text{m}$

Visual Representation of Orbital Radii

Comparative Data Summary

Final Solution: The radius of the $n=2$ orbit is $2.12 \times 10^{-10} \, \text{m}$ and the radius of the $n=3$ orbit is $4.77 \times 10^{-10} \, \text{m}$.

Step 1: Theoretical Foundation for Orbital Period

According to the Bohr model of the atom, the angular velocity $\omega$ of an electron in a stable orbit is related to the principal quantum number $n$. The orbital period $T$ is defined as the time taken for one complete revolution: $T = \frac{2\pi}{\omega}$.

From the derivation of the Bohr radius ($r_n$) and the velocity ($v_n$), we know that $v_n = \frac{v_1}{n}$ and $r_n = n^2 r_1$. Since $\omega = \frac{v_n}{r_n}$, we substitute the expressions:

$\omega_n = \frac{v_1/n}{n^2 r_1} = \frac{v_1}{n^3 r_1} = \frac{\omega_1}{n^3}$

Thus, the orbital period $T_n$ is given by: $T_n = \frac{2\pi}{\omega_1/n^3} = T_1 \cdot n^3$, where $T_1$ is the period of the ground state ($n=1$). [Per Bohr’s Postulates of Quantization]

Step 2: Calculating Ground State Period ($T_1$)

For the hydrogen atom, the velocity in the first orbit is $v_1 \approx 2.18 \times 10^6 \text{ m/s}$ and the Bohr radius is $r_1 \approx 0.529 \times 10^{-10} \text{ m}$.

The ground state angular frequency $\omega_1$ is: $\omega_1 = \frac{v_1}{r_1} = \frac{2.18 \times 10^6}{0.529 \times 10^{-10}} \approx 4.12 \times 10^{16} \text{ rad/s}$

Therefore, the ground state period $T_1$ is: $T_1 = \frac{2\pi}{\omega_1} = \frac{2 \times 3.14159}{4.12 \times 10^{16}} \approx 1.52 \times 10^{-16} \text{ s}$

Step 3: Calculating Periods for Higher Energy Levels

Using the relation $T_n = T_1 \cdot n^3$, we calculate the periods for $n=1, 2, 3$:

Visualization of Orbital Expansion

Final Solution: The orbital periods for $n=1, 2,$ and $3$ are $1.52 \times 10^{-16} \text{ s}$, $1.22 \times 10^{-15} \text{ s}$, and $4.10 \times 10^{-15} \text{ s}$ respectively.

Given Variables & Initial Setup

To determine the frequency of the emitted radiation during an electronic transition, we identify the following parameters provided by the problem:

• Energy difference between the two levels: $\Delta E = 2.3 \text{ eV}$
• Planck's constant: $h \approx 6.626 \times 10^{-34} \text{ J}\cdot\text{s}$
• Elementary charge (conversion factor): $1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}$

Step 1: Conversion of Energy to SI Units

The energy difference must be expressed in Joules ($\text{J}$) to maintain consistency with the SI units of Planck’s constant. [Per the definition of the electronvolt as a unit of energy].

$\Delta E (\text{in Joules}) = 2.3 \text{ eV} \times (1.602 \times 10^{-19} \text{ J/eV})$

$\Delta E = 3.6846 \times 10^{-19} \text{ J}$

Step 2: Theoretical Foundation of Emission Frequency

According to the Bohr postulate for atomic transitions, the frequency ($\nu$) of the emitted photon is directly proportional to the energy difference between the two states. [Per the Bohr Frequency Condition: $\Delta E = h\nu$].

Rearranging for frequency ($\nu$):

$\nu = \frac{\Delta E}{h}$

Step 3: Numerical Calculation

Substituting the values calculated in Step 1 and the known constant for Planck's constant:

$\nu = \frac{3.6846 \times 10^{-19} \text{ J}}{6.626 \times 10^{-34} \text{ J}\cdot\text{s}}$

$\nu \approx 0.55608 \times 10^{15} \text{ Hz}$

$\nu \approx 5.56 \times 10^{14} \text{ Hz}$

Analytical Summary

The transition results in the release of a photon with frequency $\nu$. By applying the relationship $\nu = \frac{\Delta E}{h}$, we establish that the energy gap of $2.3 \text{ eV}$ corresponds to the visible spectrum region (specifically green-yellow light).

Final Solution: The frequency of the emitted radiation is $5.56 \times 10^{14} \text{ Hz}$.

Step 1: Theoretical Foundation of Hydrogen Atom Energetics

According to the Bohr model of the hydrogen atom, the total energy ($E$), kinetic energy ($K$), and potential energy ($U$) of an electron in a stationary orbit are related to the electrostatic interaction between the nucleus (proton) and the orbiting electron [Per the Coulombic Law and Virial Theorem].

The total energy is defined as the sum of kinetic and potential energies:

$E = K + U$

From the virial theorem for a system under an inverse-square law force, we establish the following relationships:

• $K = -E$
• $U = 2E$

Step 2: Calculation of Kinetic Energy ($K$)

Given the ground state energy $E = -13.6 \text{ eV}$.

Using the derived relation $K = -E$:

$K = -(-13.6 \text{ eV})$

$K = 13.6 \text{ eV}$

[Justification: The kinetic energy must be positive as it represents the work done to bring the electron from rest to its orbital velocity].

Step 3: Calculation of Potential Energy ($U$)

Using the derived relation $U = 2E$:

$U = 2 \times (-13.6 \text{ eV})$

$U = -27.2 \text{ eV}$

[Justification: The negative sign indicates that the electron is in a bound state within the electrostatic potential well of the nucleus].

Visual Representation of Energy States

Summary of Energetic Parameters

Final Solution: The kinetic energy of the electron in the ground state is 13.6 eV, and the potential energy is -27.2 eV.

Step 1: Identifying the Energy Levels of Hydrogen

The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the Bohr formula: $E_n = -\frac{13.6}{n^2} \, \text{eV}$
The transition energy required to excite an electron from the ground state ($n=1$) to an excited state ($n$) is $\Delta E = E_n - E_1$.

Step 2: Determining Maximum Excitation Level

The incident electron beam has an energy of $12.5 \, \text{eV}$.
Since $12.09 \, \text{eV} < 12.5 \, \text{eV} < 12.75 \, \text{eV}$, the hydrogen atoms can be excited up to the $n=3$ energy level. The electrons cannot reach the $n=4$ level because the incident energy is insufficient [Per the Principle of Conservation of Energy].

Step 3: Calculating Possible Transitions

When the atoms return to the ground state from $n=3$, the following transitions are possible:

• Transition A: From $n=3$ to $n=2$ (Balmer Series)
• Transition B: From $n=2$ to $n=1$ (Lyman Series)
• Transition C: From $n=3$ to $n=1$ (Lyman Series)

Step 4: Calculating Wavelengths ($\lambda$)

Using the relation $\lambda = \frac{hc}{\Delta E}$, where $hc \approx 1240 \, \text{eV}\cdot\text{nm}$:

• For $n=3 \rightarrow n=2$: $\Delta E = -1.51 - (-3.4) = 1.89 \, \text{eV}$ $\lambda = \frac{1240}{1.89} \approx 656.3 \, \text{nm}$
• For $n=2 \rightarrow n=1$: $\Delta E = -3.4 - (-13.6) = 10.2 \, \text{eV}$ $\lambda = \frac{1240}{10.2} \approx 121.6 \, \text{nm}$
• For $n=3 \rightarrow n=1$: $\Delta E = -1.51 - (-13.6) = 12.09 \, \text{eV}$ $\lambda = \frac{1240}{12.09} \approx 102.6 \, \text{nm}$

Final Solution: The series of wavelengths emitted correspond to the Lyman series (102.6 nm and 121.6 nm) and the Balmer series (656.3 nm).