NCERT Solutions Class 12 Physics Chapter 12 Atoms

NCERT Solutions Class 12 Physics Chapter 12 Atoms offers comprehensive, step-by-step guidance designed to help CBSE Class 12 students master atomic models, spectral series, and quantum energy transitions. Chapter 12 is a foundational unit in Modern Physics, carrying vital weightage in the CBSE Class 12 Board Examination and competitive entrance tests like JEE Main and NEET. This detailed guide covers every theoretical concept, mathematical derivation, NCERT textbook exercise (Exercises 12.1 to 12.10), and high-frequency CBSE board exam questions with complete working.

Chapter Overview: Class 12 Physics Chapter 12 Atoms

In NCERT Class 12 Physics Chapter 12, the structure of the atom is unraveled chronologically, beginning from early classical concepts to modern quantum mechanics. The chapter systematically examines why classical electrodynamics failed to explain atomic stability and how Niels Bohr bridged classical mechanics with Planck's quantum theory.

Key Topics Covered in Chapter 12

• Alpha-Particle Scattering Experiment (Geiger-Marsden Experiment): Experimental setup, trajectory of alpha particles, impact parameter ($b$), and distance of closest approach ($r_0$).
• Rutherford's Planetary Model of the Atom: Nuclear hypothesis and its two critical shortcomings—orbital instability due to continuous electromagnetic radiation and the inability to explain discrete line spectra.
• Bohr's Model of the Hydrogen Atom: Three fundamental postulates introducing quantized non-radiating stationary orbits and photon emission/absorption conditions.
• Mathematical Formulations: Radii of permissible orbits ($r_n \propto n^2$), orbital velocity ($v_n \propto 1/n$), kinetic energy, electrostatic potential energy, and total quantized energy ($E_n = -13.6/n^2\text{ eV}$).
• Hydrogen Spectral Series: Derivation of the Rydberg formula and spectral transitions categorized into Lyman (UV), Balmer (Visible), Paschen (Infrared), Brackett (Infrared), and Pfund (Infrared) series.
• De Broglie's Explanation of Bohr's Quantization Rule: Standing matter wave condition ($2\pi r = n\lambda$) providing physical justification for angular momentum quantization.

Real-World Applications of Atomic Physics

• Astronomical Spectroscopy: Analyzing emission and absorption lines from distant stars allows astrophysicists to determine stellar temperature, density, and chemical composition without physical sampling.
• Laser Technology: Population inversion and stimulated emission between discrete atomic energy levels form the operational basis of medical surgical lasers, optical communications, and barcode scanners.
• Atomic Clocks: Transitions between hyperfine energy levels of Cesium-133 atoms provide the primary international standard of time (1 second), vital for GPS navigation systems.
• Flame Photometry and Chemical Analysis: Characteristic spectral emission lines serve as unique atomic fingerprints to detect trace elements in clinical biochemistry and environmental forensics.

Detailed Concepts, Key Formulas, and Derivations

Understanding the mathematical derivations in this chapter is essential for scoring full marks in 3-mark and 5-mark board exam questions.

1. Distance of Closest Approach ($r_0$)

When an $\alpha$-particle of mass $m$ and initial velocity $v$ (kinetic energy $K = \frac{1}{2}mv^2$) travels head-on toward a nucleus of atomic number $Z$, its kinetic energy is completely converted into electrostatic potential energy at the turning point ($r_0$):

$$\frac{1}{2}mv^2 = \frac{1}{4\pi\varepsilon_0} \frac{(2e)(Ze)}{r_0} \implies r_0 = \frac{2Ze^2}{4\pi\varepsilon_0 K}$$

2. Bohr's Three Postulates

1. Stationary Orbits: Electrons revolve around the nucleus only in certain non-radiating stable orbits where classical electromagnetic radiation does not occur.
2. Quantization of Angular Momentum: The orbital angular momentum ($L$) of an electron is an integral multiple of $h / 2\pi$: $$L = m v r = \frac{n h}{2\pi}, \quad \text{where } n = 1, 2, 3, \dots$$
3. Frequency Condition: Radiation is emitted or absorbed only when an electron jumps from one stationary orbit ($E_i$) to another ($E_f$): $$h\nu = E_i - E_f \implies \nu = \frac{E_i - E_f}{h}$$

3. Derivation of Orbital Radius, Velocity, and Energy

For an electron of mass $m$ and charge $-e$ revolving at speed $v_n$ in an orbit of radius $r_n$ around a nucleus of charge $+Ze$:

• Centripetal Force Balance: $\frac{m v_n^2}{r_n} = \frac{1}{4\pi\varepsilon_0} \frac{Z e^2}{r_n^2} \implies m v_n^2 r_n = \frac{Z e^2}{4\pi\varepsilon_0}$
• From Bohr's Quantization: $v_n = \frac{n h}{2\pi m r_n}$
• Radius of $n$-th Orbit: $$r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m Z e^2}$$ For Hydrogen ($Z=1, n=1$), the Bohr radius is: $$r_1 = a_0 = \frac{h^2 \varepsilon_0}{\pi m e^2} \approx 0.529 \times 10^{-10}\text{ m} = 0.529\text{ \AA}$$
• Orbital Velocity: $$v_n = \frac{Z e^2}{2\varepsilon_0 n h} = \left(\frac{c}{137}\right) \frac{Z}{n}$$ For Hydrogen in ground state ($n=1$), $v_1 \approx 2.18 \times 10^6\text{ m/s}$.
• Kinetic Energy ($K$): $K = \frac{1}{2}m v_n^2 = \frac{Z e^2}{8\pi\varepsilon_0 r_n}$
• Potential Energy ($U$): $U = -\frac{Z e^2}{4\pi\varepsilon_0 r_n}$
• Total Energy ($E_n$): $$E_n = K + U = -\frac{Z e^2}{8\pi\varepsilon_0 r_n} = -\frac{m Z^2 e^4}{8 \varepsilon_0^2 n^2 h^2}$$ For Hydrogen ($Z=1$): $$E_n = -\frac{13.6}{n^2}\text{ eV}$$ Note the fundamental relationships: $K = -E_n$ and $U = 2E_n$.

4. Spectral Series of Hydrogen Atom

The wavelength ($\lambda$) of radiation emitted during an electronic transition from $n_2$ to $n_1$ ($n_2 > n_1$) is given by the Rydberg formula:

$$\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$

Where $R = \frac{m e^4}{8 \varepsilon_0^2 c h^3} \approx 1.097 \times 10^7\text{ m}^{-1}$ is the Rydberg constant.

Spectral Series	Lower Level ($n_1$)	Upper Level ($n_2$)	Spectral Region	Shortest Wavelength ($\lambda_{\text{min}}$: $n_2=\infty$)	Longest Wavelength ($\lambda_{\text{max}}$: $n_2=n_1+1$)
Lyman Series	1	2, 3, 4, ...	Ultraviolet (UV)	$\lambda = 1/R \approx 91.2\text{ nm}$	$\lambda = 4/(3R) \approx 121.6\text{ nm}$
Balmer Series	2	3, 4, 5, ...	Visible & Near UV	$\lambda = 4/R \approx 364.6\text{ nm}$	$\lambda = 36/(5R) \approx 656.3\text{ nm}$
Paschen Series	3	4, 5, 6, ...	Infrared (IR)	$\lambda = 9/R \approx 820.4\text{ nm}$	$\lambda = 144/(7R) \approx 1875.1\text{ nm}$
Brackett Series	4	5, 6, 7, ...	Far Infrared	$\lambda = 16/R \approx 1458.5\text{ nm}$	$\lambda = 400/(9R) \approx 4051.6\text{ nm}$
Pfund Series	5	6, 7, 8, ...	Far Infrared	$\lambda = 25/R \approx 2278.9\text{ nm}$	$\lambda = 900/(11R) \approx 7457.8\text{ nm}$

NCERT Class 12 Physics Chapter 12 Exercise Solutions

Here are the step-by-step solutions for all textbook exercise questions from NCERT Class 12 Physics Chapter 12 (Atoms).

Exercise 12.1

Question: Choose the correct alternative between the given options:

• (a) The size of the atom in Thomson's model is .......... the atomic size in Rutherford's model. (much greater than / no different from / much less than)
• (b) In the ground state of .........., electrons are in stable equilibrium, while in .......... electrons always experience a net force. (Thomson's model / Rutherford's model)
• (c) A classical atom based on .......... is doomed to collapse. (Thomson's model / Rutherford's model)
• (d) An atom has a nearly continuous mass distribution in .......... but a highly non-uniform mass distribution in .......... (Thomson's model / Rutherford's model)
• (e) The positively charged part of the atom possesses most of the mass in .......... (Rutherford's model / both the models)

Solution:

1. (a) no different from (both models predict an atomic diameter of the order of $10^{-10}\text{ m}$).
2. (b) Thomson's model; Rutherford's model (in Thomson's model, electrostatic forces balance in equilibrium; in Rutherford's model, orbiting electrons experience centripetal acceleration).
3. (c) Rutherford's model (accelerated charges radiate electromagnetic energy continuously according to Maxwell's theory, causing the electron to spiral into the nucleus).
4. (d) Thomson's model; Rutherford's model (Thomson assumed mass was uniformly distributed across the atomic sphere; Rutherford proved over 99.9% of mass resides in the central nucleus).
5. (e) both the models (both models attribute the dominant portion of atomic mass to the positive charge structure).

Exercise 12.2

Question: Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is solid at temperatures below 14 K.) What results do you expect?

Solution:

Given: Target nucleus = Hydrogen nucleus (proton, mass $m_p \approx 1\text{ u}$), Projectile = $\alpha$-particle (mass $m_\alpha \approx 4\text{ u}$).

Explanation: In an elastic collision where the incident particle ($\alpha$-particle) is much heavier than the target nucleus ($m_\alpha \approx 4 m_p$), the heavy $\alpha$-particle will knock the lighter proton forward with high velocity and continue moving in the forward direction with minimal deviation.

Answer: There will be virtually no large-angle scattering ($\theta > 90^\circ$) or backward deflection of $\alpha$-particles. The experiment will fail to show nuclear back-scattering.

Exercise 12.3

Question: What is the shortest wavelength present in the Paschen series of spectral lines?

Solution:

Given:

• Series: Paschen series $\implies n_1 = 3$
• Shortest wavelength (series limit) occurs when transition originates from $n_2 = \infty$
• Rydberg constant, $R = 1.097 \times 10^7\text{ m}^{-1}$

To Find: Shortest wavelength ($\lambda_{\text{min}}$)

Formula:

$$\frac{1}{\lambda_{\text{min}}} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = R \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right) = \frac{R}{9}$$

Substitution and Calculation:

$$\lambda_{\text{min}} = \frac{9}{R} = \frac{9}{1.097 \times 10^7\text{ m}^{-1}} = 8.204 \times 10^{-7}\text{ m} = 820.4\text{ nm}$$

Answer: The shortest wavelength in the Paschen series is $820.4\text{ nm}$ (in the Infrared region).

Exercise 12.4

Question: A difference of $2.3\text{ eV}$ separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?

Solution:

Given:

• Energy difference, $\Delta E = 2.3\text{ eV} = 2.3 \times 1.6 \times 10^{-19}\text{ J} = 3.68 \times 10^{-19}\text{ J}$
• Planck's constant, $h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}$

To Find: Frequency of emitted radiation ($\nu$)

Formula:

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

Substitution and Calculation:

$$\nu = \frac{3.68 \times 10^{-19}\text{ J}}{6.63 \times 10^{-34}\text{ J}\cdot\text{s}} \approx 5.55 \times 10^{14}\text{ Hz} \approx 5.6 \times 10^{14}\text{ Hz}$$

Answer: The frequency of emitted radiation is $5.55 \times 10^{14}\text{ Hz}$ (or $5.6 \times 10^{14}\text{ Hz}$).

Exercise 12.5

Question: The ground state energy of hydrogen atom is $-13.6\text{ eV}$. What are the kinetic and potential energies of the electron in this state?

Solution:

Given:

• Total energy in ground state, $E = -13.6\text{ eV}$

To Find: Kinetic Energy ($K$) and Potential Energy ($U$)

Formula:

1. Kinetic Energy: $K = -E$
2. Potential Energy: $U = 2E$

Substitution and Calculation:

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

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

Answer: Kinetic energy is $+13.6\text{ eV}$ and Potential energy is $-27.2\text{ eV}$.

Exercise 12.6

Question: A hydrogen atom initially in the ground level absorbs a photon, which excites it to the $n = 4$ level. Determine the wavelength and frequency of the photon.

Solution:

Given:

• Initial level, $n_1 = 1$, Energy $E_1 = -13.6\text{ eV}$
• Final level, $n_2 = 4$, Energy $E_4 = -\frac{13.6}{4^2} = -\frac{13.6}{16} = -0.85\text{ eV}$
• Speed of light, $c = 3 \times 10^8\text{ m/s}$, $h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}$

To Find: Energy absorbed ($\Delta E$), Frequency ($\nu$), and Wavelength ($\lambda$)

Calculation:

$$\Delta E = E_4 - E_1 = -0.85\text{ eV} - (-13.6\text{ eV}) = 12.75\text{ eV}$$

$$\Delta E = 12.75 \times 1.6 \times 10^{-19}\text{ J} = 2.04 \times 10^{-18}\text{ J}$$

$$\nu = \frac{\Delta E}{h} = \frac{2.04 \times 10^{-18}\text{ J}}{6.63 \times 10^{-34}\text{ J}\cdot\text{s}} = 3.08 \times 10^{15}\text{ Hz} \approx 3.1 \times 10^{15}\text{ Hz}$$

$$\lambda = \frac{c}{\nu} = \frac{3 \times 10^8\text{ m/s}}{3.08 \times 10^{15}\text{ s}^{-1}} = 9.74 \times 10^{-8}\text{ m} = 97.4\text{ nm}$$

Answer: The absorbed photon has a frequency of $3.08 \times 10^{15}\text{ Hz}$ and a wavelength of $97.4\text{ nm}$ (Ultraviolet region).

Exercise 12.7

Question: (a) Using Bohr's model, calculate the speed of the electron in a hydrogen atom in the $n = 1, 2,$ and $3$ levels. (b) Calculate the orbital period in each of these levels.

Solution:

Given: $h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}$, $\varepsilon_0 = 8.854 \times 10^{-12}\text{ C}^2/\text{N}\cdot\text{m}^2$, $e = 1.6 \times 10^{-19}\text{ C}$, $r_1 = 5.29 \times 10^{-11}\text{ m}$.

Formula:

$$v_n = \frac{e^2}{2\varepsilon_0 n h} = \frac{v_1}{n}, \quad T_n = \frac{2\pi r_n}{v_n} = \frac{2\pi (n^2 r_1)}{(v_1 / n)} = n^3 T_1$$

Part (a) Calculation of Speeds:

• For $n = 1$: $v_1 = \frac{(1.6 \times 10^{-19})^2}{2 \times 8.854 \times 10^{-12} \times 1 \times 6.63 \times 10^{-34}} \approx 2.18 \times 10^6\text{ m/s}$
• For $n = 2$: $v_2 = \frac{v_1}{2} = \frac{2.18 \times 10^6}{2} = 1.09 \times 10^6\text{ m/s}$
• For $n = 3$: $v_3 = \frac{v_1}{3} = \frac{2.18 \times 10^6}{3} \approx 7.27 \times 10^5\text{ m/s}$

Part (b) Calculation of Orbital Periods:

• For $n = 1$: $T_1 = \frac{2\pi r_1}{v_1} = \frac{2 \times 3.1416 \times 5.29 \times 10^{-11}}{2.18 \times 10^6} \approx 1.52 \times 10^{-16}\text{ s}$
• For $n = 2$: $T_2 = 2^3 \times T_1 = 8 \times 1.52 \times 10^{-16}\text{ s} = 1.22 \times 10^{-15}\text{ s}$
• For $n = 3$: $T_3 = 3^3 \times T_1 = 27 \times 1.52 \times 10^{-16}\text{ s} = 4.10 \times 10^{-15}\text{ s}$

Answer:

• Speeds: $v_1 = 2.18 \times 10^6\text{ m/s}$, $v_2 = 1.09 \times 10^6\text{ m/s}$, $v_3 = 7.27 \times 10^5\text{ m/s}$
• Periods: $T_1 = 1.52 \times 10^{-16}\text{ s}$, $T_2 = 1.22 \times 10^{-15}\text{ s}$, $T_3 = 4.10 \times 10^{-15}\text{ s}$

Exercise 12.8

Question: The radius of the innermost electron orbit of a hydrogen atom is $5.3 \times 10^{-11}\text{ m}$. What are the radii of the $n = 2$ and $n = 3$ orbits?

Solution:

Given: $r_1 = 5.3 \times 10^{-11}\text{ m}$.

Formula: $r_n = n^2 r_1$

Calculation:

• For $n = 2$: $r_2 = 2^2 \times r_1 = 4 \times 5.3 \times 10^{-11}\text{ m} = 2.12 \times 10^{-10}\text{ m}$
• For $n = 3$: $r_3 = 3^2 \times r_1 = 9 \times 5.3 \times 10^{-11}\text{ m} = 4.77 \times 10^{-10}\text{ m}$

Answer: The radii are $r_2 = 2.12 \times 10^{-10}\text{ m}$ ($2.12\text{ \AA}$) and $r_3 = 4.77 \times 10^{-10}\text{ m}$ ($4.77\text{ \AA}$).

Exercise 12.9

Question: A $12.5\text{ eV}$ electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Solution:

Given: Incident electron energy = $12.5\text{ eV}$, Ground state energy $E_1 = -13.6\text{ eV}$.

Analysis: The maximum energy the hydrogen atom can acquire is:

$$E_{\text{max}} = -13.6\text{ eV} + 12.5\text{ eV} = -1.1\text{ eV}$$

Let us check the hydrogen energy levels:

• $n = 1: E_1 = -13.6\text{ eV}$
• $n = 2: E_2 = -3.4\text{ eV}$ (excitation energy required = $10.2\text{ eV}$)
• $n = 3: E_3 = -1.51\text{ eV}$ (excitation energy required = $12.09\text{ eV}$)
• $n = 4: E_4 = -0.85\text{ eV}$ (excitation energy required = $12.75\text{ eV}$)

Since $12.5\text{ eV} > 12.09\text{ eV}$ but $12.5\text{ eV} < 12.75\text{ eV}$, electrons can only be excited up to the $n = 3$ level.

Possible transitions and emitted wavelengths:

1. Transition $3 \to 1$ (Lyman series): $$\Delta E = -1.51 - (-13.6) = 12.09\text{ eV}$$ $$\lambda = \frac{hc}{\Delta E} = \frac{1242\text{ eV}\cdot\text{nm}}{12.09\text{ eV}} \approx 102.7\text{ nm} \quad (\text{UV})$$
2. Transition $2 \to 1$ (Lyman series): $$\Delta E = -3.4 - (-13.6) = 10.2\text{ eV}$$ $$\lambda = \frac{1242\text{ eV}\cdot\text{nm}}{10.2\text{ eV}} \approx 121.8\text{ nm} \quad (\text{UV})$$
3. Transition $3 \to 2$ (Balmer series): $$\Delta E = -1.51 - (-3.4) = 1.89\text{ eV}$$ $$\lambda = \frac{1242\text{ eV}\cdot\text{nm}}{1.89\text{ eV}} \approx 657.1\text{ nm} \quad (\text{Visible Red, } H_\alpha)$$

Answer: The emitted radiations belong to the Lyman series ($102.7\text{ nm}$ and $121.8\text{ nm}$) and the Balmer series ($657.1\text{ nm}$).

Exercise 12.10

Question: In accordance with Bohr's model, find the quantum number that characterizes the Earth's revolution around the Sun in an orbit of radius $1.5 \times 10^{11}\text{ m}$ with orbital speed $3 \times 10^4\text{ m/s}$. (Mass of Earth = $6.0 \times 10^{24}\text{ kg}$).

Solution:

Given:

• Radius of orbit, $r = 1.5 \times 10^{11}\text{ m}$
• Orbital speed, $v = 3 \times 10^4\text{ m/s}$
• Mass of Earth, $m = 6.0 \times 10^{24}\text{ kg}$
• Planck's constant, $h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}$

Formula:

$$m v r = \frac{n h}{2\pi} \implies n = \frac{2\pi m v r}{h}$$

Substitution and Calculation:

$$n = \frac{2 \times 3.1416 \times (6.0 \times 10^{24}\text{ kg}) \times (3 \times 10^4\text{ m/s}) \times (1.5 \times 10^{11}\text{ m})}{6.63 \times 10^{-34}\text{ J}\cdot\text{s}}$$

$$n = \frac{1.696 \times 10^{41}}{6.63 \times 10^{-34}} \approx 2.56 \times 10^{74} \approx 2.6 \times 10^{74}$$

Answer: The quantum number characterizing Earth's orbit is $n = 2.6 \times 10^{74}$. Significance: Because $n$ is extraordinarily large, quantum level spacing is undetectable, meaning classical physics perfectly describes macroscopic planetary motion.

Important CBSE Board Exam Questions and Official Keys

Practice these essential questions curated from official CBSE question banks and previous years' board papers:

1. Electric Flux Definition and Units

Question: Define electric flux. Write its SI unit.

Answer: Electric flux ($\Phi$) through a given surface placed in an electric field represents the total number of electric field lines passing normally through that surface. Mathematically:

$$\Phi = \vec{E} \cdot \vec{A} = E A \cos \theta$$

Where $\vec{E}$ is the electric field vector, $\vec{A}$ is the area vector, and $\theta$ is the angle between $\vec{E}$ and the surface normal. SI unit: $\text{N}\cdot\text{m}^2/\text{C}$ or $\text{V}\cdot\text{m}$.

2. Coulomb's Law and Vector Formulation

Question: State Coulomb's Law. Write its vector form.

Answer: Coulomb's Law: The electrostatic force of attraction or repulsion between two stationary point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them:

$$F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}$$

Vector Form: The force exerted by charge $q_1$ on charge $q_2$ separated by position vector $\vec{r}_{12}$ is:

$$\vec{F}_{21} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2} \hat{r}_{12} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{|\vec{r}_2 - \vec{r}_1|^3} (\vec{r}_2 - \vec{r}_1)$$

3. Equipotential Surfaces and Properties

Question: What is an equipotential surface? List its properties.

Answer: An equipotential surface is an imaginary or real surface locus of all points having the same electric potential.

Key Properties:

1. No work is done in moving a test charge between any two points on an equipotential surface ($W = q_0 \Delta V = 0$).
2. The electric field is always perpendicular to the equipotential surface at every point ($\vec{E} \perp d\vec{r}$).
3. Two equipotential surfaces can never intersect each other (if they did, there would be two different potentials at the intersection point, which is impossible).
4. Equipotential surfaces are closer together in regions of strong electric fields and farther apart in weak fields ($E = -dV/dr$).

4. Drift Velocity and Electric Current Relation

Question: Define drift velocity. How is it related to current?

Answer: Drift velocity ($v_d$): The average velocity with which free electrons in a conductor drift under the influence of an applied external electric field. It is given by $v_d = \frac{e E \tau}{m}$, where $\tau$ is relaxation time.

Relation with Current ($I$):

$$I = n A e v_d$$

Where $n$ is electron number density, $A$ is cross-sectional area, $e$ is electronic charge ($1.6 \times 10^{-19}\text{ C}$), and $v_d$ is drift velocity.

5. Kirchhoff's Laws for Electrical Circuits

Question: State Kirchhoff's laws for electrical circuits.

Answer:

1. Kirchhoff's First Law / Junction Rule (KCL): The algebraic sum of currents meeting at any electrical junction is zero ($\sum I = 0$). Total current entering a node equals total current leaving it. Based on the Law of Conservation of Charge.
2. Kirchhoff's Second Law / Loop Rule (KVL): In any closed electrical loop, the algebraic sum of changes in potential (including EMFs and potential drops across resistors) is zero ($\sum \Delta V = 0$ or $\sum \mathcal{E} = \sum IR$). Based on the Law of Conservation of Energy.

6. Total Internal Reflection (TIR)

Question: What is total internal reflection? State the conditions.

Answer: Total Internal Reflection (TIR) is the phenomenon in which a light ray traveling from an optically denser medium to a rarer medium is completely reflected back into the denser medium at the interface without any refraction.

Conditions for TIR:

1. The light ray must travel from an optically denser medium to an optically rarer medium.
2. The angle of incidence ($i$) in the denser medium must be strictly greater than the critical angle ($i_c$) for the given pair of media ($\sin i_c = 1/\mu$).

7. Electric Field of a Charged Spherical Conductor

Question: What is the electric field at the surface of a charged spherical conductor of radius $R$ and charge $Q$?

Answer: At the surface of a charged spherical conductor of radius $R$ carrying total charge $Q$, the electric field is directed radially outward (for positive $Q$) or inward (for negative $Q$) and is given by:

$$E = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R^2} = \frac{\sigma}{\varepsilon_0}$$

Where $\sigma = \frac{Q}{4\pi R^2}$ is the surface charge density. The field outside behaves as if the entire charge were concentrated at the center of the sphere.

8. Electric Dipole Moment Calculation

Question: Two point charges $+2\,\mu\text{C}$ and $-2\,\mu\text{C}$ are placed $5\text{ cm}$ apart. What is the electric dipole moment?

Solution:

Given: Charge magnitude $q = 2\,\mu\text{C} = 2 \times 10^{-6}\text{ C}$, separation distance $2a = 5\text{ cm} = 0.05\text{ m}$.

Formula: $p = q \times 2a$

Calculation:

$$p = (2 \times 10^{-6}\text{ C}) \times (0.05\text{ m}) = 1.0 \times 10^{-7}\text{ C}\cdot\text{m}$$

Answer: The electric dipole moment is $1.0 \times 10^{-7}\text{ C}\cdot\text{m}$, directed along the axis from the negative charge ($-2\,\mu\text{C}$) to the positive charge ($+2\,\mu\text{C}$).

Common Mistakes and Exam Preparation Tips

Examiners frequently penalize students for minor oversights in Modern Physics. Keep these critical points in mind:

• Sign Errors in Energy Calculations: Remember that bound electron total energy is always negative ($E_n = -13.6/n^2\text{ eV}$). Kinetic energy is strictly positive ($K = +13.6/n^2\text{ eV}$), and potential energy is negative and double the total energy ($U = -27.2/n^2\text{ eV}$).
• Excitation Energy vs. Ionization Energy: Excitation energy to state $n$ is $E_n - E_1$ (e.g., $10.2\text{ eV}$ for $n=2$). Ionization energy from state $n$ is $-E_n$ (e.g., $+13.6\text{ eV}$ from ground state, $+3.4\text{ eV}$ from $n=2$).
• Wavelength Extremes:
  • Longest wavelength ($\lambda_{\text{max}}$) corresponds to minimum energy transition ($\Delta n = 1$, e.g., $n_2 = n_1 + 1$).
  • Shortest wavelength ($\lambda_{\text{min}}$ / Series Limit) corresponds to maximum energy transition ($n_2 = \infty$).
• Rydberg Constant Units: The Rydberg constant $R \approx 1.097 \times 10^7\text{ m}^{-1}$ has units of reciprocal meters ($\text{m}^{-1}$), not meters.
• De Broglie Relationship: Remember that an integer number of de Broglie wavelengths fit into the electron's circumference: $2\pi r_n = n\lambda$.

CBSE Board Exam Relevance and Marks Weightage

In the CBSE Class 12 Physics syllabus, Chapter 12 (Atoms) along with Chapter 13 (Nuclei) forms the Unit VIII: Atoms and Nuclei, which carries a combined weightage of 7 to 8 marks out of 70 in the theory examination.

Question Format	Typical Marks	Common Focus Areas
Multiple Choice (MCQ) / Assertion-Reason	1 Mark	Ratios of radii/velocities ($r \propto n^2$, $v \propto 1/n$), spectral series regions, potential vs kinetic energy relations.
Short Answer (SA-I)	2 Marks	Impact parameter definition, limitations of Rutherford model, de Broglie explanation of Bohr's postulate.
Short Answer (SA-II)	3 Marks	Derivation of Bohr radius or energy levels, numericals calculating shortest/longest wavelengths in Balmer/Lyman series.
Case-Based / Long Answer	4 / 5 Marks	Comprehensive questions on Geiger-Marsden experiment and Bohr's spectral series transitions.

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