Atoms — Class 12 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Physics chapter "Atoms" — 6 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Physics chapter "Atoms" — 6 important questions with detailed answers for CBSE board exam preparation.
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
NCERT Solutions for Class 12 Physics — Atoms. Step-by-step answers to all 6 textbook questions from this chapter, written for CBSE board preparation and free to use.
Atoms — All 6 Questions Solved
Q1. Explain Bohr's model of hydrogen atom. Derive expression for energy levels and radius of orbit.
Bohr's Model (1913):
Atom with nucleus at center and electrons in circular orbits, combining classical mechanics with quantum postulates.
Postulates:
1. Electron moves in circular orbit without radiating energy
2. Only orbits with angular momentum L = nℏ are allowed (n = 1, 2, 3, ...)
where ℏ = h/(2π) = 1.055 × 10⁻³⁴ J·s
3. Radiation emitted when electron transitions between orbits
Derivation of Orbital Radius:
Coulomb force provides centripetal force:
ke²/r² = m_e v²/r
ke²/r = m_e v² ... (1)
Angular momentum quantization:
m_e vr = nℏ
v = nℏ/(m_e r) ... (2)
Substituting (2) into (1):
ke²/r = m_e [nℏ/(m_e r)]²
ke² = n²ℏ²/r
r_n = n²ℏ²/(m_e ke²) = n² × a₀
Bohr radius: a₀ = ℏ²/(m_e ke²) = 0.529 Å = 0.0529 nm
Orbital radii:
r₁ = a₀ = 0.529 Å
r₂ = 4a₀ = 2.12 Å
r₃ = 9a₀ = 4.77 Å
Derivation of Energy Levels:
Total energy = Kinetic + Potential
E = ½m_e v² - ke²/r
From (1): m_e v² = ke²/r
E = ½(ke²/r) - ke²/r = -ke²/(2r)
E_n = -ke²/(2r_n) = -ke²/(2n²a₀)
E_n = -13.6 eV/n²
Energy levels:
E₁ = -13.6 eV (ground state)
E₂ = -3.4 eV
E₃ = -1.51 eV
E_∞ = 0 (ionization)
Ionization Energy = 13.6 eV
Q2. An electron in hydrogen atom transitions from n = 3 to n = 1. Find the energy and wavelength of photon emitted.
Given:
Initial state: n_i = 3
Final state: n_f = 1
Energy levels: E_n = -13.6 eV/n²
Energy of Initial Level:
E₃ = -13.6/3² = -13.6/9 = -1.51 eV
Energy of Final Level:
E₁ = -13.6/1² = -13.6 eV
Energy of Emitted Photon:
ΔE = E_f - E_i = E₁ - E₃
ΔE = -13.6 - (-1.51) = -13.6 + 1.51 = -12.09 eV
The energy released (emitted) = 12.09 eV
Alternatively:
ΔE = 13.6[1/n_f² - 1/n_i²] = 13.6[1/1 - 1/9]
ΔE = 13.6[1 - 1/9] = 13.6 × 8/9 = 12.09 eV
Frequency of Photon:
hν = ΔE = 12.09 eV = 12.09 × 1.6 × 10⁻¹⁹ J = 1.93 × 10⁻¹⁸ J
ν = (1.93 × 10⁻¹⁸)/(6.63 × 10⁻³⁴) = 2.91 × 10¹⁵ Hz
Wavelength:
λ = c/ν = (3 × 10⁸)/(2.91 × 10¹⁵) = 1.03 × 10⁻⁷ m = 103 nm
Or using: λ = hc/ΔE = (1240 eV·nm)/(12.09 eV) = 102.6 nm
This is in ultraviolet region (Lyman series, n → 1)
Answers:
- Energy of photon: 12.09 eV
- Wavelength: 102.6 nm (UV region)
- Frequency: 2.91 × 10¹⁵ Hz
Q3. What are X-rays? Explain characteristic and continuous X-ray spectra.
X-rays: Electromagnetic radiation produced when high-energy electrons strike a metal target, with wavelengths 0.1-100 nm.
Production:
1. Electrons accelerated through high potential difference (10 kV - 100 kV)
2. Directed at metal target (Cu, Mo, W, etc.)
3. Electrons decelerate and lose energy
4. Energy converted to X-ray photons
Continuous X-ray Spectrum (Bremsstrahlung):
- Produced by deceleration of electrons
- Continuous distribution of wavelengths
- Minimum wavelength λ_min when entire electron energy converts to one photon:
eV = hc/λ_min
λ_min = hc/(eV) = 1240/(V in kV) nm
- For 50 kV: λ_min = 1240/50 = 24.8 pm
- Intensity increases with atomic number of target
- Increases with electron current
Characteristic X-ray Spectrum:
- Produced when electron knocks out inner shell electron
- Outer electron transitions down, emitting X-ray
- Discrete wavelengths characteristic of element
- Independent of accelerating voltage (above threshold)
- Named after transitions: K-line, L-line, M-line, etc.
For K-line: electron transitions from L to K shell
Energy = E_L - E_K = difference in ionization energies
Wavelength determined by Moseley's law:
ν = R_∞ c(Z - σ)²[1/n₁² - 1/n₂²]
Where σ = screening constant
Relationship:
- Characteristic lines superimposed on continuous spectrum
- Only appears above threshold voltage
- Minimum voltage V_min = characteristic X-ray energy/e
Applications:
1. Medical imaging (radiography)
2. Crystallography
3. Material analysis
4. Industrial inspection
Q4. What is the Franck-Hertz experiment? Explain how it validates Bohr's energy level concept.
Franck-Hertz Experiment (1914):
Demonstrated quantized energy levels in atoms by measuring electron-atom collisions.
Setup:
- Mercury vapor in glass tube
- Electrons emitted from cathode with adjustable voltage
- Electrons accelerated toward anode through mercury vapor
- Current measured vs accelerating voltage
Procedure:
1. Electrons accelerated from cathode by voltage V
2. Electron kinetic energy: K.E. = eV
3. Electrons collide with Hg atoms
4. At low voltages: elastic collisions (little energy transfer)
5. At specific voltages: inelastic collisions (energy loss)
6. Current drops when eV matches energy difference between levels
Observations:
1. Current increases smoothly initially
2. Sharp drops occur at regular voltage intervals
3. First drop at V₁ = 4.9 V
4. Subsequent drops at 2 × 4.9 V, 3 × 4.9 V, etc.
5. Drops become more pronounced with excitation
6. Ultraviolet light produced at certain voltages (fluorescence)
Explanation:
At V = 4.9 V:
Electron gains energy eV = 4.9 eV
This matches energy difference ΔE = E₂ - E₁ = 4.9 eV in Hg atom
Inelastic collision occurs, electron transfers energy
Hg atom excited to higher level
Current drops (fewer electrons reach anode)
Validation of Bohr Model:
1. Discrete energy levels confirmed (not continuous)
2. Energy difference = 4.9 eV for Hg
3. Excitation produces photon when atom de-excites
4. Wavelength of emitted light: λ = hc/ΔE = 1240/4.9 ≈ 253 nm (UV)
5. Matches observed mercury resonance line
Significance:
- Experimental proof of quantum energy levels
- Direct evidence for discrete orbits
- Validated Bohr's model
- Foundation for quantum mechanics
Q5. Explain spectral series in hydrogen atom. Write the formula and find the wavelengths of Hα and Hβ lines.
Spectral Series: Groups of spectral lines produced by transitions to the same lower level.
General Formula (Rydberg):
1/λ = R_H[1/n_f² - 1/n_i²]
Where:
R_H = Rydberg constant = 1.097 × 10⁷ m⁻¹ = 0.01097 nm⁻¹
n_f = final level (lower)
n_i = initial level (upper), n_i > n_f
Spectral Series:
1. Lyman series: n_f = 1 (ultraviolet region)
1/λ = R_H[1 - 1/n²], n = 2, 3, 4, ...
Lα: n = 2 → 1, λ = 121.6 nm
2. Balmer series: n_f = 2 (visible region)
1/λ = R_H[1/4 - 1/n²], n = 3, 4, 5, ...
Hα, Hβ, Hγ, Hδ, ...
3. Paschen series: n_f = 3 (infrared region)
1/λ = R_H[1/9 - 1/n²], n = 4, 5, 6, ...
4. Brackett series: n_f = 4 (infrared region)
5. Pfund series: n_f = 5 (infrared region)
Hα Line (n = 3 → 2):
1/λ_Hα = R_H[1/4 - 1/9] = R_H[9 - 4]/(36) = R_H × 5/36
1/λ_Hα = 1.097 × 10⁷ × 5/36 = 1.522 × 10⁶ m⁻¹
λ_Hα = 1/(1.522 × 10⁶) = 6.563 × 10⁻⁷ m = 656.3 nm (red light)
Hβ Line (n = 4 → 2):
1/λ_Hβ = R_H[1/4 - 1/16] = R_H[4 - 1]/16 = R_H × 3/16
1/λ_Hβ = 1.097 × 10⁷ × 3/16 = 2.056 × 10⁶ m⁻¹
λ_Hβ = 1/(2.056 × 10⁶) = 4.861 × 10⁻⁷ m = 486.1 nm (cyan-blue light)
Answers:
- Hα wavelength: 656.3 nm (red, visible)
- Hβ wavelength: 486.1 nm (cyan, visible)
- Both are part of Balmer series (visible spectrum)
Q6. What is the difference between emission and absorption spectra? Explain continuous, line, and band spectra.
Emission Spectrum: Spectrum of light emitted directly from a hot source.
Absorption Spectrum: Spectrum of light after passing through absorbing material; shows dark lines on bright background.
Types of Spectra:
1. Continuous Spectrum:
- Unbroken distribution of all wavelengths
- Produced by incandescent solid, liquid, or dense gas
- Thermal radiation (blackbody radiation)
- Temperature determines color and intensity
- Examples: sunlight, tungsten bulb filament
2. Line Spectrum (Atomic Spectrum):
Emission:
- Discrete bright lines on dark background
- Produced by excited isolated atoms or ions
- Gas discharge tubes (neon, mercury)
- Each line represents specific transition
Absorption:
- Dark lines at same wavelengths on continuous background
- When white light passes through cool gas
- Hydrogen absorption in sun's spectrum (Fraunhofer lines)
- Frequencies absorbed = frequencies emitted
3. Band Spectrum (Molecular Spectrum):
- Groups of closely-spaced lines appearing as bands
- Produced by molecules (O₂, N₂, CO₂, etc.)
- Combination of electronic, vibrational, rotational transitions
- Many levels → many lines → band appearance
- Examples: diatomic molecule spectra
Characteristics:
Line Spectrum:
- Sharp, well-defined lines
- Characteristic of element (unique fingerprint)
- Used for identification
- Regular pattern (hydrogen: Balmer series)
Band Spectrum:
- Blurred edges, band-like structure
- Characteristic of molecule
- More complex than atomic
- Overlapping lines
Atomic vs Molecular:
Atoms: simple energy levels → discrete lines
Molecules: electronic + vibrational + rotational → many transitions → bands
Fraunhofer Lines:
- Dark lines in sun's spectrum
- Cool outer atmosphere absorbs wavelengths
- Same wavelengths as emission lines from that element
- Used to determine solar composition
- Examples: Hα line, sodium D lines
Atoms — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. What observation from Rutherford's alpha-particle scattering experiment led to the conclusion that the atom has a tiny, dense, positively charged nucleus?
- A. Most alpha particles passed straight through the foil.
- B. Some alpha particles were deflected by small angles.
- C. A very small fraction of alpha particles were scattered back at angles greater than 90°.
- D. Alpha particles were found to be negatively charged.
Answer: C. A very small fraction of alpha particles were scattered back at angles greater than 90°. — The scattering of a very small fraction of alpha particles at large angles indicated that the positive charge and most of the mass of the atom are concentrated in a tiny central region called the nucleus.
MCQ 2. Which of the following is NOT a postulate of Bohr's model of the atom?
- A. Electrons revolve around the nucleus in certain stable orbits without radiating energy.
- B. The angular momentum of an electron in a stable orbit is an integral multiple of h/2π.
- C. When an electron jumps from a higher energy orbit to a lower energy orbit, it emits radiation of a specific frequency.
- D. Electrons can revolve in any orbit, but only those with specific energies are stable.
Answer: D. Electrons can revolve in any orbit, but only those with specific energies are stable. — Bohr's model states that electrons can only revolve in *certain specific* stable orbits (stationary states), not any orbit. These orbits correspond to discrete energy levels.
MCQ 3. The total energy of an electron in the nth stationary orbit of the hydrogen atom is proportional to:
- A. n
- B. n²
- C. 1/n
- D. 1/n²
Answer: D. 1/n² — For a hydrogen atom, the total energy of an electron in the nth orbit is given by E_n = -13.6/n² eV. Thus, E_n is proportional to 1/n².
MCQ 4. Which spectral series of the hydrogen atom lies in the visible region of the electromagnetic spectrum?
- A. Lyman series
- B. Balmer series
- C. Paschen series
- D. Brackett series
Answer: B. Balmer series — The Balmer series corresponds to electron transitions to the n=2 orbit, with its lines (Hα, Hβ, etc.) falling within the visible light spectrum.
MCQ 5. The series limit wavelength of the Lyman series for a hydrogen atom is λ₀. What is the series limit wavelength of the Balmer series for the hydrogen atom?
- A. 2λ₀
- B. 4λ₀
- C. λ₀/2
- D. λ₀/4
Answer: B. 4λ₀ — For Lyman series, 1/λ₀ = R(1/1² - 1/∞²) = R. For Balmer series, 1/λ'_0 = R(1/2² - 1/∞²) = R/4. Therefore, λ'_0 = 4/R = 4λ₀.
MCQ 6. Rutherford's alpha-particle scattering experiment showed that most of the alpha particles passed through the gold foil undeflected. This observation leads to the conclusion that:
- A. The nucleus is positively charged.
- B. The nucleus is very dense.
- C. Most of the space in an atom is empty.
- D. Electrons revolve around the nucleus.
Answer: C. Most of the space in an atom is empty. — The fact that most alpha particles passed through undeflected indicates that the atom is mostly empty space, allowing the particles to pass without interaction.
MCQ 7. If the radius of the first Bohr orbit of a hydrogen atom is r, what is the radius of the third Bohr orbit for the hydrogen atom?
- A. r/3
- B. 3r
- C. 9r
- D. r/√3
Answer: C. 9r — The radius of the nth Bohr orbit is given by r_n = r₁ * n², where r₁ is the radius of the first orbit. For n=3, r₃ = r₁ * 3² = 9r₁.
MCQ 8. According to de Broglie's hypothesis, the condition for an electron to be in a stable orbit in an atom is that the circumference of the orbit must be:
- A. Equal to the electron's wavelength.
- B. Half of the electron's wavelength.
- C. An integral multiple of the electron's wavelength.
- D. A fractional multiple of the electron's wavelength.
Answer: C. An integral multiple of the electron's wavelength. — De Broglie's hypothesis, when applied to Bohr's model, states that stable orbits are those where the circumference (2πr) is an integral multiple (n) of the electron's de Broglie wavelength (λ), i.e., 2πr = nλ.
MCQ 9. An electron in a hydrogen atom makes a transition from n=4 to n=2. The energy of the emitted photon corresponds to:
- A. Lyman series
- B. Balmer series
- C. Paschen series
- D. Brackett series
Answer: B. Balmer series — Transitions ending in the n=2 orbit belong to the Balmer series.
MCQ 10. J.J. Thomson's 'plum pudding' model of the atom failed to explain which experimental observation?
- A. The discrete line spectra of elements.
- B. The existence of electrons.
- C. The overall neutrality of atoms.
- D. The large-angle scattering of alpha particles by a thin metal foil.
Answer: D. The large-angle scattering of alpha particles by a thin metal foil. — Thomson's model predicted that alpha particles would pass through with minimal deflection. Rutherford's alpha scattering experiment, showing large-angle deflections, contradicted this and led to the nuclear model.
MCQ 11. The ionization potential of the hydrogen atom is 13.6 V. What is the ionization potential of a He+ ion?
- A. 13.6 V
- B. 27.2 V
- C. 54.4 V
- D. 6.8 V
Answer: C. 54.4 V — The ionization energy for a hydrogen-like atom is given by E = 13.6 * Z²/n² eV. For He+, Z=2, n=1. Ionization energy = 13.6 * (2)²/1² = 54.4 eV. The ionization potential is numerically equal to this value in Volts.
MCQ 12. The wavelength of the first line of the Balmer series in the hydrogen spectrum is λ. What is the wavelength of the second line of the Balmer series?
- A. 20λ/27
- B. 27λ/20
- C. 2λ
- D. λ/2
Answer: A. 20λ/27 — For the first line of Balmer series (n=3 to n=2), 1/λ₁ = R(1/2² - 1/3²) = 5R/36. For the second line (n=4 to n=2), 1/λ₂ = R(1/2² - 1/4²) = 3R/16. Therefore, λ₂/λ₁ = (16/3R) / (36/5R) = (16/3) * (5/36) = 20/27. So, λ₂ = 20λ/27.
MCQ 13. The velocity of an electron in the first Bohr orbit of the hydrogen atom is v. What is the velocity of the electron in the second Bohr orbit of the hydrogen atom?
- A. v/2
- B. 2v
- C. v/4
- D. 4v
Answer: A. v/2 — The velocity of an electron in the nth Bohr orbit is proportional to 1/n (v_n proportional to Z/n). So, v₂ = v₁/2.
MCQ 14. The Lyman series of the hydrogen spectrum lies in the:
- A. Visible region
- B. Infrared region
- C. Ultraviolet region
- D. X-ray region
Answer: C. Ultraviolet region — The Lyman series corresponds to transitions to the ground state (n=1) and produces photons with the highest energies, placing them in the ultraviolet region.
MCQ 15. For a hydrogen atom, if the potential energy of an electron in a specific orbit is V, what is its kinetic energy in that orbit?
- A. -V
- B. -V/2
- C. V/2
- D. V
Answer: B. -V/2 — For an electron in a stable orbit of a hydrogen atom, the total energy E = -KE = V/2, where KE is kinetic energy and V is potential energy. Thus, KE = -V/2 (since V is negative).
MCQ 16. Which of the following phenomena cannot be explained by Bohr's model of the atom?
- A. The stability of atoms.
- B. The line spectrum of hydrogen.
- C. The splitting of spectral lines in a magnetic field (Zeeman effect).
- D. The calculation of energy of electron in hydrogen atom.
Answer: C. The splitting of spectral lines in a magnetic field (Zeeman effect). — Bohr's model successfully explained stability and hydrogen spectrum but failed to account for the fine structure of spectral lines, the Zeeman effect (splitting in magnetic field), or the spectra of multi-electron atoms.
MCQ 17. According to Bohr's model, the angular momentum of an electron in a stationary orbit is quantized. It is an integral multiple of:
- A. h
- B. h/π
- C. h/2π
- D. 2πh
Answer: C. h/2π — Bohr's second postulate states that the angular momentum (L) of an electron in a stable orbit is L = n(h/2π), where n is an integer.
MCQ 18. If the wavelength of the first line of the Lyman series is λ₁, and the wavelength of the first line of the Paschen series is λ₂, what is the ratio λ₁/λ₂?
- A. 7/108
- B. 108/7
- C. 5/108
- D. 108/5
Answer: A. 7/108 — Lyman series first line (n=2 to n=1): 1/λ₁ = R(1/1² - 1/2²) = 3R/4. Paschen series first line (n=4 to n=3): 1/λ₂ = R(1/3² - 1/4²) = 7R/144. Ratio λ₁/λ₂ = (4/3R) / (144/7R) = (4/3) * (7/144) = 7/108.
MCQ 19. The energy of an electron in the ground state of the hydrogen atom is -13.6 eV. What is the energy required to excite the electron from the ground state to the first excited state?
- A. 13.6 eV
- B. 3.4 eV
- C. 10.2 eV
- D. 1.51 eV
Answer: C. 10.2 eV — Ground state energy (n=1) E₁ = -13.6 eV. First excited state (n=2) energy E₂ = -13.6/2² = -3.4 eV. Energy required for excitation = E₂ - E₁ = -3.4 - (-13.6) = 10.2 eV.
MCQ 20. An alpha particle consists of:
- A. One proton and one neutron.
- B. Two protons and two neutrons.
- C. Two protons and one electron.
- D. One proton and two electrons.
Answer: B. Two protons and two neutrons. — An alpha particle is a helium nucleus, which contains two protons and two neutrons.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.