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Atoms NCERT Solutions & Important Questions
If you are preparing for your CBSE Board Exams 2026, JEE, or NEET, understanding the structure of an atom is absolutely crucial. In this post, you will get the Updated NCERT Solutions, simplified explanations, and board-style Important Questions. Tension mat lijiye, we have broken down complex topics like Bohr's model and hydrogen spectra into easy, bite-sized concepts. Let's start learning!
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Chapter NameAtoms (Chapter 12)
SubjectPhysics
ClassClass 12
BoardCBSE (2026 Syllabus)
Important TopicsRutherford's Model, Bohr's Postulates, Hydrogen Line Spectrum
Difficulty LevelModerate
Exam Weightage~4-5 Marks (Part of Dual Nature & Atoms/Nuclei Unit)
Learning Objectives
After completing this chapter, students will be able to:
- Understand Rutherford's alpha-particle scattering experiment and its conclusions.
- Explain the limitations of Rutherford's atomic model.
- State and apply Bohr's postulates for the hydrogen atom.
- Calculate the radius, velocity, and total energy of an electron in different orbits.
- Derive the frequencies and wavelengths of various spectral series (Lyman, Balmer, Paschen, etc.) of the hydrogen atom.
Key Concepts & Formulas
- Alpha-Particle Scattering: Discovered that atoms have a tiny, dense, positively charged core called the nucleus.
- Distance of Closest Approach ($r_0$): The minimum distance an alpha particle reaches before being deflected back by the nucleus.
$$r_0 = \frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{K}$$
(where $K$ is kinetic energy). - Bohr's Quantization Condition: The angular momentum of an electron in a stationary orbit is an integral multiple of $h/2\pi$.
$$mvr = \frac{nh}{2\pi}$$ - Radius of $n^{th}$ Orbit ($r_n$):
$$r_n = 0.53 \times \frac{n^2}{Z} \text{ (\AA)}$$ - Energy of an Electron ($E_n$): The total energy of an electron in the $n^{th}$ orbit of a hydrogen atom.
$$E_n = -\frac{13.6}{n^2} \text{ eV}$$ - Rydberg Formula: Used to calculate the wavelength of emitted radiation during electron transitions.
$$\frac{1}{\lambda} = R \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$$
(where $R = 1.097 \times 10^7 \text{ m}^{-1}$ is the Rydberg constant).
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Full NCERT Solutions
Question 12.1: Choose the correct alternative from the clues given at the end of each statement: (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 solid atom based on ......... is doomed to collapse. (Thomson's model/Rutherford's model)
Answer:
(a) no different from (Both models assume the atomic size is roughly $10^{-10}\text{ m}$).
(b) Thomson's model, Rutherford's model (In Thomson's model, electrons are embedded and stable; in Rutherford's, they revolve and experience centripetal force).
(c) Rutherford's model (A revolving electron is accelerating, hence it must radiate energy and spiral into the nucleus according to classical electromagnetism).
(a) no different from (Both models assume the atomic size is roughly $10^{-10}\text{ m}$).
(b) Thomson's model, Rutherford's model (In Thomson's model, electrons are embedded and stable; in Rutherford's, they revolve and experience centripetal force).
(c) Rutherford's model (A revolving electron is accelerating, hence it must radiate energy and spiral into the nucleus according to classical electromagnetism).
Question 12.2: 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 a solid at temperatures below 14 K.) What results do you expect?
Answer:
Step 1: Analyze the Nucleus. If we use solid hydrogen, the nucleus of a hydrogen atom consists of a single proton.
Step 2: Compare Masses. The mass of an alpha particle is about 4 times the mass of a proton.
Step 3: Collision Mechanics. During a collision, a heavier body (alpha particle) will not rebound when hitting a lighter body (proton).
Conclusion: Therefore, we would not observe any large-angle scattering or rebounding (scattering $> 90^\circ$). The alpha particles would just pass straight through or deviate very slightly.
Step 1: Analyze the Nucleus. If we use solid hydrogen, the nucleus of a hydrogen atom consists of a single proton.
Step 2: Compare Masses. The mass of an alpha particle is about 4 times the mass of a proton.
Step 3: Collision Mechanics. During a collision, a heavier body (alpha particle) will not rebound when hitting a lighter body (proton).
Conclusion: Therefore, we would not observe any large-angle scattering or rebounding (scattering $> 90^\circ$). The alpha particles would just pass straight through or deviate very slightly.
Question 12.3: What is the shortest wavelength present in the Paschen series of spectral lines?
Answer:
Step 1: Identify orbits for shortest wavelength. For the Paschen series, the final orbit is $n_f = 3$. The shortest wavelength corresponds to the maximum energy transition, which happens when the electron jumps from infinity ($n_i = \infty$).
Step 2: Apply Rydberg's formula.
$$\frac{1}{\lambda} = R \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$$
$$\frac{1}{\lambda} = 1.097 \times 10^7 \left(\frac{1}{3^2} - \frac{1}{\infty^2}\right)$$
$$\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{1}{9}$$
Step 3: Solve for Wavelength.
$$\lambda = \frac{9}{1.097 \times 10^7} = 8.204 \times 10^{-7}\text{ m}$$
Final Answer: The shortest wavelength in the Paschen series is $820.4\text{ nm}$.
Step 1: Identify orbits for shortest wavelength. For the Paschen series, the final orbit is $n_f = 3$. The shortest wavelength corresponds to the maximum energy transition, which happens when the electron jumps from infinity ($n_i = \infty$).
Step 2: Apply Rydberg's formula.
$$\frac{1}{\lambda} = R \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$$
$$\frac{1}{\lambda} = 1.097 \times 10^7 \left(\frac{1}{3^2} - \frac{1}{\infty^2}\right)$$
$$\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{1}{9}$$
Step 3: Solve for Wavelength.
$$\lambda = \frac{9}{1.097 \times 10^7} = 8.204 \times 10^{-7}\text{ m}$$
Final Answer: The shortest wavelength in the Paschen series is $820.4\text{ nm}$.
Question 12.4: A difference of 2.3 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?
Answer:
Step 1: Extract and convert energy values.
Energy difference, $\Delta E = 2.3\text{ eV}$
Convert eV to Joules: $\Delta E = 2.3 \times 1.6 \times 10^{-19}\text{ J} = 3.68 \times 10^{-19}\text{ J}$
Planck's constant, $h = 6.63 \times 10^{-34}\text{ Js}$
Step 2: Apply frequency formula.
Formula: $\Delta E = h\nu$
$$\nu = \frac{\Delta E}{h} = \frac{3.68 \times 10^{-19}}{6.63 \times 10^{-34}}$$
$$\nu = 5.55 \times 10^{14}\text{ Hz}$$
Final Answer: The frequency of the emitted radiation is $5.55 \times 10^{14}\text{ Hz}$.
Step 1: Extract and convert energy values.
Energy difference, $\Delta E = 2.3\text{ eV}$
Convert eV to Joules: $\Delta E = 2.3 \times 1.6 \times 10^{-19}\text{ J} = 3.68 \times 10^{-19}\text{ J}$
Planck's constant, $h = 6.63 \times 10^{-34}\text{ Js}$
Step 2: Apply frequency formula.
Formula: $\Delta E = h\nu$
$$\nu = \frac{\Delta E}{h} = \frac{3.68 \times 10^{-19}}{6.63 \times 10^{-34}}$$
$$\nu = 5.55 \times 10^{14}\text{ Hz}$$
Final Answer: The frequency of the emitted radiation is $5.55 \times 10^{14}\text{ Hz}$.
Question 12.5: The ground state energy of a hydrogen atom is -13.6 eV. What are the kinetic and potential energies of the electron in this state?
Answer:
Given: Total Energy ($E$) = $-13.6\text{ eV}$.
In a hydrogen atom, the relationships between Total Energy ($E$), Kinetic Energy ($K$), and Potential Energy ($U$) are:
$K = -E$
$U = 2E$
Step 1: Kinetic Energy.
$$K = -(-13.6\text{ eV}) = +13.6\text{ eV}$$
Step 2: Potential Energy.
$$U = 2 \times (-13.6\text{ eV}) = -27.2\text{ eV}$$
Given: Total Energy ($E$) = $-13.6\text{ eV}$.
In a hydrogen atom, the relationships between Total Energy ($E$), Kinetic Energy ($K$), and Potential Energy ($U$) are:
$K = -E$
$U = 2E$
Step 1: Kinetic Energy.
$$K = -(-13.6\text{ eV}) = +13.6\text{ eV}$$
Step 2: Potential Energy.
$$U = 2 \times (-13.6\text{ eV}) = -27.2\text{ eV}$$
Question 12.6: 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.
Answer:
Step 1: Find energy difference.
Initial state, $n_i = 1$
Final state, $n_f = 4$
Energy in $n^{th}$ state: $E_n = -\frac{13.6}{n^2}\text{ eV}$
Energy of ground state, $E_1 = -13.6\text{ eV}$
Energy of $n=4$ state, $E_4 = -\frac{13.6}{4^2} = -0.85\text{ eV}$
Energy absorbed, $\Delta E = E_4 - E_1 = -0.85 - (-13.6) = +12.75\text{ eV}$
Step 2: Convert to Joules.
$\Delta E = 12.75 \times 1.6 \times 10^{-19} = 2.04 \times 10^{-18}\text{ J}$
Step 3: Calculate Frequency ($\nu$).
$$\nu = \frac{\Delta E}{h} = \frac{2.04 \times 10^{-18}}{6.63 \times 10^{-34}} = 3.08 \times 10^{15}\text{ Hz}$$
Step 4: Calculate Wavelength ($\lambda$).
$$\lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{3.08 \times 10^{15}} = 9.74 \times 10^{-8}\text{ m} = 97.4\text{ nm}$$
Step 1: Find energy difference.
Initial state, $n_i = 1$
Final state, $n_f = 4$
Energy in $n^{th}$ state: $E_n = -\frac{13.6}{n^2}\text{ eV}$
Energy of ground state, $E_1 = -13.6\text{ eV}$
Energy of $n=4$ state, $E_4 = -\frac{13.6}{4^2} = -0.85\text{ eV}$
Energy absorbed, $\Delta E = E_4 - E_1 = -0.85 - (-13.6) = +12.75\text{ eV}$
Step 2: Convert to Joules.
$\Delta E = 12.75 \times 1.6 \times 10^{-19} = 2.04 \times 10^{-18}\text{ J}$
Step 3: Calculate Frequency ($\nu$).
$$\nu = \frac{\Delta E}{h} = \frac{2.04 \times 10^{-18}}{6.63 \times 10^{-34}} = 3.08 \times 10^{15}\text{ Hz}$$
Step 4: Calculate Wavelength ($\lambda$).
$$\lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{3.08 \times 10^{15}} = 9.74 \times 10^{-8}\text{ m} = 97.4\text{ nm}$$
Question 12.7: (a) Using the 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.
Answer:
(a) Speed of electron ($v_n$):
Formula: $v_n = \frac{c}{137 \times n} \approx \frac{2.18 \times 10^6}{n}\text{ m/s}$
* For $n=1$: $v_1 = 2.18 \times 10^6\text{ m/s}$
* For $n=2$: $v_2 = \frac{2.18 \times 10^6}{2} = 1.09 \times 10^6\text{ m/s}$
* For $n=3$: $v_3 = \frac{2.18 \times 10^6}{3} = 7.27 \times 10^5\text{ m/s}$
(b) Orbital Period ($T_n$):
Formula: $T_n = \frac{2\pi r_n}{v_n}$ (where $r_n = 0.53 \times 10^{-10} \times n^2\text{ m}$)
* For $n=1$: $T_1 = 1.52 \times 10^{-16}\text{ s}$
* For $n=2$: $T_2 = 1.22 \times 10^{-15}\text{ s}$
* For $n=3$: $T_3 = 4.12 \times 10^{-15}\text{ s}$
(a) Speed of electron ($v_n$):
Formula: $v_n = \frac{c}{137 \times n} \approx \frac{2.18 \times 10^6}{n}\text{ m/s}$
* For $n=1$: $v_1 = 2.18 \times 10^6\text{ m/s}$
* For $n=2$: $v_2 = \frac{2.18 \times 10^6}{2} = 1.09 \times 10^6\text{ m/s}$
* For $n=3$: $v_3 = \frac{2.18 \times 10^6}{3} = 7.27 \times 10^5\text{ m/s}$
(b) Orbital Period ($T_n$):
Formula: $T_n = \frac{2\pi r_n}{v_n}$ (where $r_n = 0.53 \times 10^{-10} \times n^2\text{ m}$)
* For $n=1$: $T_1 = 1.52 \times 10^{-16}\text{ s}$
* For $n=2$: $T_2 = 1.22 \times 10^{-15}\text{ s}$
* For $n=3$: $T_3 = 4.12 \times 10^{-15}\text{ s}$
Question 12.8: 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?
Answer:
Step 1: State the proportionality.
Given: $r_1 = 5.3 \times 10^{-11}\text{ m}$.
According to Bohr's theory, the radius is proportional to the square of the principal quantum number ($r \propto n^2$).
Formula: $r_n = n^2 \times r_1$
Step 2: Calculate for $n=2$.
$r_2 = (2)^2 \times 5.3 \times 10^{-11} = 4 \times 5.3 \times 10^{-11} = 2.12 \times 10^{-10}\text{ m}$
Step 3: Calculate for $n=3$.
$r_3 = (3)^2 \times 5.3 \times 10^{-11} = 9 \times 5.3 \times 10^{-11} = 4.77 \times 10^{-10}\text{ m}$
Step 1: State the proportionality.
Given: $r_1 = 5.3 \times 10^{-11}\text{ m}$.
According to Bohr's theory, the radius is proportional to the square of the principal quantum number ($r \propto n^2$).
Formula: $r_n = n^2 \times r_1$
Step 2: Calculate for $n=2$.
$r_2 = (2)^2 \times 5.3 \times 10^{-11} = 4 \times 5.3 \times 10^{-11} = 2.12 \times 10^{-10}\text{ m}$
Step 3: Calculate for $n=3$.
$r_3 = (3)^2 \times 5.3 \times 10^{-11} = 9 \times 5.3 \times 10^{-11} = 4.77 \times 10^{-10}\text{ m}$
EXTRA IMPORTANT QUESTIONS (BOARD STYLE 2026)
Here are 15 highly predicted questions for the 2026 Board Exams!
Multiple Choice Questions (MCQs)
1. The ratio of the energies of the hydrogen atom in its first to second excited state is:
(a) $1/4$
(b) $4/9$
(c) $9/4$
(d) $4$
Answer: (c) $9/4$. (First excited state = $n=2$, second = $n=3$. Energy $\propto 1/n^2$)
2. Which series of the hydrogen spectrum lies in the visible region?
(a) Lyman
(b) Balmer
(c) Paschen
(d) Brackett
Answer: (b) Balmer
3. According to Bohr's model, the angular momentum of an electron is proportional to:
(a) $n$
(b) $1/n$
(c) $n^2$
(d) $1/n^2$
Answer: (a) $n$ (Since $L = nh/2\pi$)
Short Answer Questions (2 Marks)
4. Why is the classical Rutherford model considered unstable?
Answer: A revolving electron accelerates continuously. According to classical electromagnetism, an accelerating charge emits radiation, loses energy, and would eventually spiral into the nucleus.
5. State Bohr's quantization condition of angular momentum.
Answer: Bohr postulated that electrons can only revolve in specific orbits where their angular momentum ($mvr$) is an integral multiple of $\frac{h}{2\pi}$.
6. Define ionization energy. What is its value for a hydrogen atom?
Answer: The minimum energy required to completely remove an electron from its ground state to infinity. For hydrogen, it is $+13.6\text{ eV}$.
7. Draw the energy level diagram of a hydrogen atom showing the Lyman and Balmer series.
Answer: (Students must draw horizontal lines representing $n=1, 2, 3...$ and vertical downward arrows terminating at $n=1$ for Lyman and $n=2$ for Balmer).
Long Answer Questions (3-5 Marks)
8. Using Bohr's postulates, derive the expression for the radius of the $n^{th}$ orbit of a hydrogen atom.
Answer:
Step 1: Equate electrostatic force to centripetal force: $\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}$.
Step 2: Use Bohr's quantization condition $mvr = \frac{nh}{2\pi}$ to substitute $v$.
Step 3: Solving yields $r = \frac{n^2h^2\epsilon_0}{\pi m e^2}$.
Step 1: Equate electrostatic force to centripetal force: $\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}$.
Step 2: Use Bohr's quantization condition $mvr = \frac{nh}{2\pi}$ to substitute $v$.
Step 3: Solving yields $r = \frac{n^2h^2\epsilon_0}{\pi m e^2}$.
9. Explain the origin of spectral lines in a hydrogen atom. Derive the Rydberg formula.
Answer:
Step 1: Origin. When an electron jumps from a higher energy state ($n_i$) to a lower one ($n_f$), it emits a photon. $h\nu = E_i - E_f$.
Step 2: Derivation. Substituting $E_n = -\frac{me^4}{8\epsilon_0^2h^2n^2}$ leads directly to the formulation of the Rydberg formula $\frac{1}{\lambda} = R \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$.
Step 1: Origin. When an electron jumps from a higher energy state ($n_i$) to a lower one ($n_f$), it emits a photon. $h\nu = E_i - E_f$.
Step 2: Derivation. Substituting $E_n = -\frac{me^4}{8\epsilon_0^2h^2n^2}$ leads directly to the formulation of the Rydberg formula $\frac{1}{\lambda} = R \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$.
10. In a hydrogen atom, an electron makes a transition from $n=5$ to $n=1$. Calculate the maximum number of spectral lines emitted.
Answer:
Step 1: Use Formula. $\frac{n(n-1)}{2}$
Step 2: Substitute. $\frac{5(5-1)}{2} = \frac{20}{2} = 10$ spectral lines.
Step 1: Use Formula. $\frac{n(n-1)}{2}$
Step 2: Substitute. $\frac{5(5-1)}{2} = \frac{20}{2} = 10$ spectral lines.
Case-Based Questions (4 Marks)
Read the passage: In 1913, Niels Bohr introduced his atomic model incorporating Planck's quantum theory. He proposed that electrons do not radiate energy while in "stationary orbits."
11. What is the energy of an electron in a stationary orbit if its kinetic energy is $3.4\text{ eV}$?
Answer: Total Energy = -Kinetic Energy = $-3.4\text{ eV}$.
12. What does the negative sign in the total energy of an electron indicate?
Answer: It indicates that the electron is bound to the nucleus by attractive forces and requires external energy to be freed.
13. If an electron moves from $n=1$ to $n=3$ does it absorb or emit energy?
Answer: It absorbs energy because it is moving to a higher energy level.
Assertion-Reason Questions (1 Mark each)
Directions: (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true, R is false. (D) A is false, R is true.
14. Assertion (A): The Lyman series lies in the ultraviolet region.
Reason (R): The energy difference between states is maximum when an electron drops to $n=1$.
Answer: (A) Both A and R are true, and R is the correct explanation.
15. Assertion (A): The total energy of an electron in a hydrogen atom is positive.
Reason (R): Potential energy is double the kinetic energy with a negative sign.
Answer: (D) A is false, R is true. (Total energy is negative, not positive).
COMMON MISTAKES STUDENTS MAKE
- Sign Errors: Students frequently forget the negative sign when writing the Total Energy formula ($E_n = -13.6/n^2$).
- Mixing $n_i$ and $n_f$: In the Rydberg formula, always remember $n_i$ (initial) and $n_f$ (final). To get a positive wavelength, ensure the term in the brackets evaluates to a positive number.
- Confusing the Series:
* Lyman $\rightarrow$ ends at $n=1$ (UV region)
* Balmer $\rightarrow$ ends at $n=2$ (Visible region)
* Paschen $\rightarrow$ ends at $n=3$ (IR region) - Unit Conversions: Forgetting to convert electron volts (eV) into Joules (J) when calculating frequency using Planck's constant. Multiply eV by $1.6 \times 10^{-19}$ to get Joules.
EXAM PREPARATION TIPS
- Derivations are Key: Practice the derivations for the Radius of Orbit ($r_n$) and Total Energy ($E_n$) at least 3 times. They are frequently asked as 3-mark questions.
- Draw Clear Diagrams: In the board exam, always use a pencil and scale to draw the Hydrogen energy level diagram. Label the eV values for $n=1, 2, 3, 4$ clearly.
- Memorize Constants: Keep the values of $13.6\text{ eV}$, Rydberg constant ($R = 1.097 \times 10^7$), and Planck's constant ($h$) on your fingertips to save calculation time.
- PYQs: Solve the last 5 years' Previous Year Questions specifically targeting numericals on the Balmer and Lyman series.
FAQ SECTION
Is Chapter 12 Atoms important for CBSE Class 12 boards?
Yes, it is highly scoring and forms a combined unit with Dual Nature and Nuclei carrying approximately 12-14 marks total in the board exam.
Which is the most important derivation in the Atoms chapter?
The derivation for the radius of the $n^{th}$ Bohr orbit and the expression for the total energy of an electron are the most frequently asked derivations.
What is the difference between Rutherford's and Bohr's atomic models?
Rutherford could not explain the stability of atoms or discrete line spectra. Bohr fixed this by introducing the concept of quantized angular momentum and stationary orbits.
How do I calculate the shortest wavelength in any spectral series?
For the shortest wavelength, the energy gap must be maximum. Therefore, put the initial orbit $n_i = \infty$ and calculate using the Rydberg formula.
Where can I download the Class 12 Physics Chapter 12 NCERT PDF?
You can download the official and latest updated NCERT textbook PDF directly from the official NCERT website (ncert.nic.in).
Conclusion: Chapter 12, Atoms, is one of the most fascinating and scoring chapters in Class 12 Physics. By mastering Bohr's postulates and the energy level transitions, you can easily secure full marks in this section. Make sure to practice the numericals on the Rydberg formula and revise your notes regularly. Download our chapter notes, practice the PYQs, and face your 2026 board exams with ultimate confidence. Best of luck with your studies!
More Class 12 Physics Chapters
Ch 1: Electric Charges and Fields
Ch 2: Electrostatic Potential and Capacitance
Ch 3: Current Electricity
Ch 4: Moving Charges and Magnetism
Ch 5: Magnetism and Matter
Ch 6: Electromagnetic Induction
Ch 7: Alternating Current
Ch 8: Electromagnetic Waves
Ch 9: Ray Optics
Ch 10: Wave Optics
Ch 11: Dual Nature of Radiation and Matter
Ch 12: Atoms
Ch 13: Nuclei
Class 12 Mock QUIZs
Ch 14: Semiconductor Electronics