Class 12 Physics Atoms Notes | Detailed Explanation, MCQs, Short & Long Questions

YAtoms are the fundamental building blocks of matter. Everything around us—solids, liquids, gases—is made up of atoms. The study of atomic structure helps us understand how atoms behave, how light interacts with matter, and how stability of matter arises. In this chapter, we explore different models of the atom (Thomson, Rutherford, Bohr), atomic spectra, quantisation, and how the Bohr model explains the hydrogen atom spectra. These concepts form the foundation for quantum mechanics and modern physics.

Class 12 Physics Atoms Notes | Detailed Explanation, MCQs, Short & Long Questions

1. Early Ideas and Atomic Models

1.1 What is an Atom

• The term atom comes from Greek atomos meaning indivisible.

• Modern understanding: atoms are not indivisible—they contain electrons, a nucleus (protons + neutrons), with most of mass in nucleus and volume largely empty space.

• Typical atomic radius ~ 10−1010^{-10}10−10 m.

1.2 Thomson’s “Plum Pudding” Model

• Proposed after discovery of electrons by J.J. Thomson.

• Model: atom is a sphere of positive charge, with electrons embedded in it like plums in a pudding.

• This model tried to explain neutrality of atoms and presence of electrons.

• Limitations: fails to explain experimental data like alpha‐particle scattering; cannot account for discrete spectral lines of atoms.

2. Rutherford’s Model of Atom

2.1 Rutherford’s Alpha Scattering Experiment

• Setup: Thin gold foil, source of alpha particles, detector screen.

• Observations:

  1. Most α‐particles pass through with little or no deflection.

  2. Some deflected at large angles.

  3. Very few come straight back.

• Conclusions:

  • Atom is mostly empty space (because most α pass through).

  • Most mass and all positive charge concentrated in a small central nucleus.

  • Electrons revolve around nucleus.

2.2 Key Terms

• Impact parameter (b): The perpendicular distance between the initial path of the α‐particle and the centre of nucleus if the α had not been deflected.

• Scattering angle (θ): Angle at which α‐particle deviates from its original straight line.

2.3 Distance of Closest Approach

• When an α‐particle approaches nucleus, kinetic energy is converted into electrostatic potential energy.

• At closest approach, all kinetic energy (relative) is momentarily converted. Formula:

  K=14πε0Z1Z2e2r0K = \frac{1}{4\pi\varepsilon_0} \frac{Z_1 Z_2 e^2}{r_0}K=4πε0​1​r0​Z1​Z2​e2​

  where r0r_0r0​ is distance of closest approach, $Z$ charges etc. This gives estimate of nuclear radius scale.

2.4 Limitations of Rutherford’s Model

• Unable to explain why atom is stable; according to classical electromagnetism, accelerating electrons (in orbit) should emit radiation and spiral into nucleus.

• Cannot explain atomic emission spectra—why atoms emit light only at specific discrete frequencies.

3. Bohr’s Model of Hydrogen Atom

Given Rutherford model’s deficiencies, Niels Bohr proposed a model incorporating early quantum ideas.

3.1 Bohr’s Postulates

1. Stationary Orbits: Electrons revolve in certain allowed orbits without radiating energy. These orbits are stable or “stationary”.

2. Quantisation of Angular Momentum: The angular momentum of an electron in these orbits is an integral multiple of h2π\dfrac{h}{2\pi}2πh​.

   L=mvr=nh2π,n=1,2,3,…L = m v r = \frac{nh}{2\pi}, \qquad n = 1,2,3,\dotsL=mvr=2πnh​,n=1,2,3,…

3. Radiation by Transition: Radiation is emitted or absorbed when an electron jumps between two permitted orbits. The frequency $\nu$ of the emitted or absorbed photon is given by

   hν=Ei−Efh\nu = E_i – E_fhν=Ei​−Ef​

   where EiE_iEi​ is the energy of initial orbit, EfE_fEf​ of final.

3.2 Mathematical Derivation for Hydrogen‐like Atom

• For a hydrogen atom (or any one‐electron ion with nuclear charge $+Ze$), consider electron of mass $m$, charge $-e$ orbiting at radius $r$ with speed $v$.

• Centripetal force = Coulomb force:

  mv2r=14πε0Ze2r2\frac{m v^2}{r} = \frac{1}{4 \pi \varepsilon_0} \frac{Ze^2}{r^2}rmv2​=4πε0​1​r2Ze2​

• Quantisation condition:

  mvr=nh2πm v r = \frac{n h}{2\pi}mvr=2πnh​

• From these two equations, one derives:

  rn=4πε0 h2me2n2Z=a0n2Zr_n = \frac{4 \pi \varepsilon_0 \, h^2}{m e^2} \frac{n^2}{Z} = a_0 \frac{n^2}{Z}rn​=me24πε0​h2​Zn2​=a0​Zn2​

  where a0a_0a0​ is Bohr radius for $n=1$, $Z=1$:

  a0=4πε0 h2me2≈5.29×10−11 ma_0 = \frac{4\pi\varepsilon_0\,h^2}{m e^2} \approx 5.29 \times 10^{-11} \text{ m}a0​=me24πε0​h2​≈5.29×10−11 m

• Energy levels: Total energy EnE_nEn​ of an electron in the $n$th orbit:

  En=−me4Z28ε02h21n2=−13.6 eV  Z2n2E_n = – \frac{m e^4 Z^2}{8 \varepsilon_0^2 h^2} \frac{1}{n^2} = – \frac{13.6\,\mathrm{eV}\; Z^2}{n^2}En​=−8ε02​h2me4Z2​n21​=−n213.6eVZ2​

• So ground state (n=1, Z=1) for hydrogen is $-13.6$ eV.

3.3 Spectral Lines: Rydberg Formula

• When electron transitions from a higher level nin_ini​ to lower nfn_fnf​, photon of frequency

  hν=Ei−Ef=13.6 eV(Z2nf2−Z2ni2)h\nu = E_i – E_f = 13.6\,\mathrm{eV} \left(\frac{Z^2}{n_f^2} – \frac{Z^2}{n_i^2}\right)hν=Ei​−Ef​=13.6eV(nf2​Z2​−ni2​Z2​)

• In terms of wavelength $\lambda$ (since $\nu =c/\lambda$), one can write Rydberg formula:

  1λ=RZ2(1nf2−1ni2)\frac{1}{\lambda} = R Z^2 \left(\frac{1}{n_f^2} – \frac{1}{n_i^2}\right)λ1​=RZ2(nf2​1​−ni2​1​)

  where $R$ is Rydberg constant ≈1.097×107 m−1\approx 1.097 \times 10^7 \, \mathrm{m^{-1}}≈1.097×107m−1 for hydrogen.

3.4 Important Features & Terms

• Ground State: $n=1$, lowest energy, most stable.

• Excited State: any $n>1$.

• Ionisation Energy: Energy required to remove electron completely (from ground state to $n\to \infty$). For hydrogen, ~13.6 eV.

3.5 Limitations of Bohr’s Model

• Only works well for hydrogen or hydrogen‐like one‐electron systems (e.g. He⁺, Li²⁺, etc.).

• Cannot explain the fine structure of spectral lines (splitting due to spin, relativistic effects).

• Cannot account for Zeeman effect in detail, Stark effect, or multiple electron atoms’ behavior.

• Doesn’t give orbital shapes (only circular orbits considered), while quantum mechanics shows orbitals are probability distributions.

4. Atomic Spectra

4.1 What is Atomic Spectrum

• Emission Spectrum: When atoms absorb energy, electrons jump to higher energy levels; when they fall back, they emit photons of certain wavelengths → bright lines on dark background.

• Absorption Spectrum: When continuous light passes through a gas, atoms absorb specific wavelengths corresponding to transitions; result is dark lines on continuous spectrum.

4.2 Types of Spectral Series in Hydrogen Atom

• Lyman series: transitions down to nf=1n_f = 1nf​=1 (in UV region).

• Balmer series: transitions down to nf=2n_f = 2nf​=2 (visible light).

• Paschen: to nf=3n_f = 3nf​=3 (infrared), etc.

4.3 Explanation via Bohr Model

• The lines correspond to difference in energy levels given by Bohr’s energy formula.

• Frequency $\nu$ of emitted photon = (Ei−Ef)/h(E_i – E_f)/h(Ei​−Ef​)/h.

• Wavelength $\lambda =c/\nu$.

5. De Broglie Hypothesis and Quantisation

• Louis de Broglie proposed that particles like electrons also have wave‐like property. Wavelength $\lambda =h/p$, where $p=mv$.

• In Bohr’s model, the second postulate (quantised angular momentum) can be viewed as the condition that the electron’s orbit circumference must contain an integer number of de Broglie wavelengths:

  2πr=nλ=nhmv2\pi r = n \lambda = n \frac{h}{m v}2πr=nλ=nmvh​

• This explains why only certain orbits are permitted: those for which this condition holds.

6. Detailed Definitions and Key Formulae

Below is a summary of important terms and formulae which students should memorize / understand well.

Key formulae (hydrogen‐like atoms, charge $Ze$):

1. Bohr radius:

   rn=4πε0 h2me2n2Z=a0n2Zr_n = \frac{4\pi\varepsilon_0\,h^2}{m e^2} \frac{n^2}{Z} = a_0 \frac{n^2}{Z}rn​=me24πε0​h2​Zn2​=a0​Zn2​

2. Energy of nth level:

   En=−13.6 eV  Z2n2E_n = -\frac{13.6\,\mathrm{eV}\; Z^2}{n^2}En​=−n213.6eVZ2​

3. Frequency of transition:

   ν=Ei−Efh\nu = \frac{E_i – E_f}{h}ν=hEi​−Ef​​

4. Rydberg formula:

   1λ=RZ2(1nf2−1ni2)\frac{1}{\lambda} = R Z^2 \left(\frac{1}{n_f^2} – \frac{1}{n_i^2}\right)λ1​=RZ2(nf2​1​−ni2​1​)

5. Relation of de Broglie wavelength:

   λ=hmv\lambda = \frac{h}{m v}λ=mvh​

6. Distance of closest approach (for α‐particle scattering):

   r0=14πε0Z1Z2e2Kr_0 = \frac{1}{4 \pi \varepsilon_0} \frac{Z_1 Z_2 e^2}{K}r0​=4πε0​1​KZ1​Z2​e2​

   Here $K$ = initial kinetic energy of α‐particle.

7. Rutherford Scattering Formula (Qualitative)

• Rutherford derived expression for number of α‐particles scattered at angle between θ and θ + dθ, showing that scattering probability falls off with increasing angle.

• Key result: dependence on 1sin⁡4(θ/2)\frac{1}{\sin^4(\theta/2)}sin4(θ/2)1​.

• This experimental match confirmed nucleus‐based model (small high charge concentrated at centre).

8. Comparative Summary

Here’s a comparison among the models and features:

9. More Advanced Topics (Beyond Bohr)

Though Bohr’s model is historically very important and works well for hydrogen‐like atoms, there are more advanced ideas:

• Fine structure: splitting of spectral lines due to relativistic corrections, spin‐orbit coupling.

• Quantum mechanical model: electron as wavefunction; orbitals of different shapes (s, p, d etc.).

• Uncertainty principle: cannot define exact position and momentum simultaneously.

10. Applications and Problems

Students should be comfortable solving problems such as:

• Finding radius of nth orbit for hydrogen or singly ionised element.

• Compute energy of electron in a given orbit.

• Find wavelength/frequency of emitted light for transition between orbits.

• Compute ionisation energy.

• Use Rydberg formula for different spectral series.

11. Sample Derivation: Hydrogen Atom Radius and Energy

Let’s do a step‐by‐step derivation for hydrogen (Z = 1):

1. Coulomb force provides centripetal force:

   mv2r=14πε0e2r2\frac{m v^2}{r} = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2}rmv2​=4πε0​1​r2e2​

   → mv2=14πε0e2rm v^2 = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r}mv2=4πε0​1​re2​ … (i)

2. Quantisation of angular momentum:

   m v r = \frac{n h}{2\pi} \implies v = \frac{n h}{2\pi m r} \] … (ii)

3. Substitute (ii) into (i):

   m(nh2πmr)2=14πε0e2rm \left(\frac{n h}{2\pi m r}\right)^2 = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r}m(2πmrnh​)2=4πε0​1​re2​

   Simplify:

   n2h24π2mr2=e24πε0r\frac{n^2 h^2}{4\pi^2 m r^2} = \frac{e^2}{4\pi\varepsilon_0 r}4π2mr2n2h2​=4πε0​re2​

   Solve for $r$:

   r=4πε0 h2me2n2r = \frac{4\pi\varepsilon_0\, h^2}{m e^2} n^2r=me24πε0​h2​n2

   That is rn=a0n2r_n = a_0 n^2rn​=a0​n2 with a0=4πε0h2me2≈5.29×10−11 a_0 = \frac{4\pi\varepsilon_0 h^2}{m e^2} \approx 5.29 \times 10^{-11}a0​=me24πε0​h2​≈5.29×10−11 m.

4. Kinetic energy T=12mv2T = \frac{1}{2} m v^2T=21​mv2, potential energy U=−14πε0e2rU = – \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r}U=−4πε0​1​re2​.

   Total energy E=T+U=−1214πε0e2rE = T + U = – \frac{1}{2} \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r}E=T+U=−21​4πε0​1​re2​.

   Substitute $r$ from above → yields

   En=−me48ε02h21n2E_n = – \frac{m e^4}{8 \varepsilon_0^2 h^2} \frac{1}{n^2}En​=−8ε02​h2me4​n21​

   Which works out to −13.6 eV/n2-13.6 \, \mathrm{eV} / n^2−13.6eV/n2.

12. Important Constants and Values

• Planck’s constant h≈6.626×10−34  J⋅sh \approx 6.626 \times 10^{-34}\; \mathrm{J\cdot s}h≈6.626×10−34J⋅s.

• Elementary charge e≈1.602×10−19  Ce \approx 1.602 \times 10^{-19} \; \mathrm{C}e≈1.602×10−19C.

• Permittivity of free space ε0≈8.854×10−12  F/m\varepsilon_0 \approx 8.854 \times 10^{-12} \; \mathrm{F/m}ε0​≈8.854×10−12F/m.

• Bohr radius a0≈5.29×10−11  ma_0 \approx 5.29 \times 10^{-11} \; \mathrm{m}a0​≈5.29×10−11m.

• Rydberg constant for hydrogen R≈1.097×107  m−1R \approx 1.097 \times 10^7 \; \mathrm{m^{-1}}R≈1.097×107m−1.

13. Practice Questions (with Answers in Brief)

1. Find the radius of the 3rd orbit (n=3) of hydrogen atom.
   Using rn=a0n2r_n = a_0 n^2rn​=a0​n2, r3=5.29×10−11×9≈4.76×10−10r_3 = 5.29 \times 10^{-11} \times 9 ≈ 4.76 \times 10^{-10}r3​=5.29×10−11×9≈4.76×10−10 m.

2. What is the energy emitted when electron falls from n=3 to n=2 in hydrogen?

   E3=−13.6/9=−1.511 eV,  E2=−13.6/4=−3.4 eVE_3 = -13.6 / 9 = -1.511\ \mathrm{eV},\; E_2 = -13.6 / 4 = -3.4\ \mathrm{eV} E3​=−13.6/9=−1.511 eV,E2​=−13.6/4=−3.4 eV

   Difference = E2−E3=1.889 eVE_2 – E_3 = 1.889\ \mathrm{eV}E2​−E3​=1.889 eV. Photon of this energy emitted; find wavelength = λ=hcΔE≈\lambda = \frac{hc}{ΔE} ≈λ=ΔEhc​≈ (plug values) 656 nm (Balmer‐alpha line).

3. If the nucleus has charge +2e (e.g. He⁺), how do energy levels change?
   Energy scales with Z2Z^2Z2. So for Z = 2, En=−13.6 ×4/n2=−54.4/n2E_n = -13.6\, \times 4 / n^2 = -54.4 / n^2En​=−13.6×4/n2=−54.4/n2 eV.

4. Compute distance of closest approach for α‐particles of given KE to nucleus of gold (Z ~79) (using formula r0r_0r0​).

14. Summary and Conclusion

• The atomic models evolved from “plum pudding” to Rutherford’s nuclear model to Bohr’s quantised model.

• Bohr’s model introduced quantisation and successfully explained spectral lines of hydrogen‐like atoms.

• Atomic spectra (emission & absorption) provide evidence for discrete energy levels.

• De Broglie idea adds wave nature of electron, giving deeper justification to Bohr’s angular momentum quantisation.

  1. It fails for multi-electron atoms.

  2. Cannot explain fine structure splitting and Zeeman effect.

  3. Does not consider the wave nature of electrons.

     1. It could not explain the stability of the atom as accelerating electrons should radiate energy and fall into the nucleus.

It failed to explain the line spectra of atoms.Cannot explain intensity variations in spectral lines.

Despite its success, Bohr’s model has limitations—multi‐electron atoms, fine structure, quantum mechanical behaviour etc.

Objective Questions (MCQs) – Class 12 Physics: Atoms

Q1. Who proposed the plum pudding model of the atom?

Options:
(a) Rutherford (b) J.J. Thomson (c) Bohr (d) Dalton
Answer: Thomson

Q2. In Rutherford’s scattering experiment, most α-particles passed through the foil because:

Options:
(a) Nucleus is large (b) Atom is mostly empty space (c) Electrons are heavy (d) Alpha particles are small
Answer: Atom is mostly empty space

Q3. The distance of closest approach depends on:

Options:
(a) Mass of α-particle (b) Kinetic energy of α-particle (c) Velocity of α-particle (d) None
Answer: Kinetic energy of α-particle

Q4. The unit of Rydberg constant $R$ is:

Options:
(a) m (b) m⁻¹ (c) Hz (d) J
Answer: m⁻¹

Q5. The radius of nth orbit of hydrogen atom is proportional to:

Options:
(a) n2n^2n2 (b) n (c) 1/n21/n^21/n2 (d) $1/n$
Answer: n2n^2n2

Q6. The energy of electron in the nth orbit of hydrogen atom is given by:

Options:
(a) −13.6/n2-13.6/n^2−13.6/n2 eV (b) 13.6/n213.6/n^213.6/n2 eV (c) −n2/13.6-n^2/13.6−n2/13.6 eV (d) None
Answer: −13.6/n2-13.6/n^2−13.6/n2 eV

Q7. The energy required to remove an electron from the ground state of hydrogen atom is called:

Options:
(a) Excitation energy (b) Ionisation energy (c) Binding energy (d) Work function
Answer: Ionisation energy

Q8. The first orbit of hydrogen atom has energy:

Options:
(a) $-3.4$ eV (b) $-1.5$ eV (c) $-13.6$ eV (d) $-0.85$ eV
Answer: $-13.6$ eV

Q9. The spectral series in the visible region of hydrogen spectrum is:

Options:
(a) Lyman (b) Balmer (c) Paschen (d) Brackett
Answer: Balmer

Q10. For hydrogen atom, the Bohr radius is:

Options:
(a) 5.29×10−115.29 \times 10^{-11}5.29×10−11 m (b) 1×10−101 \times 10^{-10}1×10−10 m (c) 1×10−91 \times 10^{-9}1×10−9 m (d) 2×10−112 \times 10^{-11}2×10−11 m
Answer: 5.29×10−115.29 \times 10^{-11}5.29×10−11 m

Q11. The speed of electron in the first Bohr orbit is:

Options:
(a) 2.2×1062.2 \times 10^62.2×106 m/s (b) 3×1083 \times 10^83×108 m/s (c) 1×1051 \times 10^51×105 m/s (d) 1×1071 \times 10^71×107 m/s
Answer: 2.2×1062.2 \times 10^62.2×106 m/s

Q12. According to Bohr, angular momentum of electron is quantised as:

Options:
(a) $mvr=nh$ (b) $mvr=n\hslash$ (c) $mvr=n/2\pi$ (d) mvr=n2hmvr = n^2 hmvr=n2h
Answer: $mvr=n\hslash$

Q13. The value of Rydberg constant for hydrogen is:

Options:
(a) 1.097×1071.097 \times 10^71.097×107 m⁻¹ (b) 3×1083 \times 10^83×108 m⁻¹ (c) 6.626×10−346.626 \times 10^{-34}6.626×10−34 m⁻¹ (d) None
Answer: 1.097×1071.097 \times 10^71.097×107 m⁻¹

Q14. The energy of nth orbit varies as:

Options:
(a) $n$ (b) $1/n$ (c) 1/n21/n^21/n2 (d) n2n^2n2
Answer: 1/n21/n^21/n2

Q15. In hydrogen atom, wavelength of spectral line is given by:

Options:
(a) Bohr formula (b) Rutherford formula (c) Rydberg formula (d) None
Answer: Rydberg formula

Q16. The series lying in the ultraviolet region is:

Options:
(a) Balmer (b) Paschen (c) Lyman (d) Brackett
Answer: Lyman

Q17. The energy of electron in the second orbit of hydrogen atom is:

Options:
(a) $-13.6$ eV (b) $-3.4$ eV (c) $-1.51$ eV (d) $-0.85$ eV
Answer: $-3.4$ eV

Q18. The Paschen series lies in:

Options:
(a) UV region (b) IR region (c) Visible region (d) X-ray region
Answer: IR region

Q19. De Broglie wavelength of electron in nth orbit is given by:

Options:
(a) $\lambda =h/p$ (b) $\lambda =2\pi r/n$ (c) $\lambda =h/mv$ (d) Both (a) & (c)
Answer: Both (a) & (c)

Q20. The hydrogen atom consists of:

Options:
(a) One proton only (b) One electron only (c) One proton & one electron (d) One neutron & one electron
Answer: One proton & one electron

Short Answer Questions – Class 12 Physics: Atoms

Q1. Define atomic number and mass number.

Answer:
Atomic number (Z) is the number of protons present in the nucleus of an atom and determines its positive charge and identity. Mass number (A) is the total number of protons and neutrons (nucleons) in the nucleus.

Q2. What is the distance of closest approach in Rutherford scattering?

Answer:
The distance of closest approach is the minimum distance between the center of the nucleus and the α-particle before it comes to rest momentarily due to repulsion before retracing its path. It is calculated using energy conservation between kinetic energy and electrostatic potential energy.

Q3. What is the significance of Bohr’s quantisation condition?

Answer:
Bohr’s quantisation condition mvr=nh2πmvr = \frac{nh}{2\pi}mvr=2πnh​ implies that only those orbits are allowed for electrons where the angular momentum is an integral multiple of h2π\frac{h}{2\pi}2πh​. This explains why electrons do not spiral into the nucleus and why atoms are stable.

Q4. Write the expression for the radius of the nth orbit of a hydrogen atom.

Answer:
The radius of the nth orbit is given by

rn=a0n2r_n = a_0 n^2rn​=a0​n2

where a0=5.29×10−11a_0 = 5.29 \times 10^{-11}a0​=5.29×10−11 m is the Bohr radius for n=1.

Q5. What is the ionisation energy of hydrogen?

Answer:
The ionisation energy is the energy required to remove the electron from the ground state (n=1) of hydrogen atom to infinity (n → ∞). Its value is 13.6 eV.

Q6. Name the spectral series of hydrogen atom lying in the visible region.

Answer:
The Balmer series of hydrogen atom lies in the visible region of the electromagnetic spectrum.

Q7. What is meant by emission spectrum?

Answer:
The emission spectrum is obtained when the excited electron in an atom returns to a lower energy level, emitting photons of discrete wavelengths. These wavelengths appear as bright lines on a dark background.

Q8. State two limitations of Rutherford’s atomic model.

Answer:

• It could not explain the stability of the atom as accelerating electrons should radiate energy and fall into the nucleus.

• It failed to explain the line spectra of atoms.

Q9. What is the Rydberg formula for the hydrogen spectrum?

Answer:
The Rydberg formula is

1λ=R(1nf2−1ni2)\frac{1}{\lambda} = R \left(\frac{1}{n_f^2} – \frac{1}{n_i^2}\right)λ1​=R(nf2​1​−ni2​1​)

where ni>nfn_i > n_fni​>nf​ and $R$ is the Rydberg constant.

Q10. What is Bohr radius?

Answer:
Bohr radius is the radius of the lowest energy orbit (n=1) of the hydrogen atom. Its value is 5.29×10−115.29 \times 10^{-11}5.29×10−11 m.

Long Answer Questions – Class 12 Physics: Atoms

Q1. Explain Rutherford’s α-particle scattering experiment and its conclusions.

Answer:
In Rutherford’s experiment, a thin gold foil was bombarded with α-particles. It was observed that most particles passed straight through, some were deflected at small angles, and a few deflected at large angles. This led to the conclusion that the atom has a small, dense, positively charged nucleus at the center, most of the atom is empty space, and electrons revolve around the nucleus.

Q2. Derive the expression for the radius and energy of the nth orbit of hydrogen atom using Bohr’s model.

Answer:
Using Coulomb’s law and the centripetal force equation, and applying Bohr’s quantisation mvr=nh2πmvr = \frac{nh}{2\pi}mvr=2πnh​, we derive the radius as

rn=4πϵ0h2me2⋅n2=a0n2r_n = \frac{4\pi\epsilon_0 h^2}{m e^2} \cdot n^2 = a_0 n^2rn​=me24πϵ0​h2​⋅n2=a0​n2

and the energy of nth orbit as

En=−13.6 eVn2.E_n = -\frac{13.6 \, \text{eV}}{n^2}.En​=−n213.6eV​.

Q3. Explain the spectral series of the hydrogen atom with formulae.

Answer:
Spectral series are obtained when an electron transitions from higher energy levels to a fixed lower level. For nf=1n_f = 1nf​=1, Lyman series (UV); nf=2n_f = 2nf​=2, Balmer (visible); nf=3n_f = 3nf​=3, Paschen (IR); nf=4n_f = 4nf​=4, Brackett; nf=5n_f = 5nf​=5, Pfund series. Formula:

1λ=R(1nf2−1ni2)\frac{1}{\lambda} = R \left(\frac{1}{n_f^2} – \frac{1}{n_i^2}\right)λ1​=R(nf2​1​−ni2​1​)

Q4. Discuss the limitations of Bohr’s atomic model.

Answer:

• It fails for multi-electron atoms.

• Cannot explain fine structure splitting and Zeeman effect.

• Does not consider the wave nature of electrons.

• Cannot explain intensity variations in spectral lines.

 Q5.What is de Broglie’s hypothesis and how does it justify Bohr’s quantisation condition?

Answer:
De Broglie proposed that particles like electrons have a wavelength λ=hp\lambda = \frac{h}{p}λ=ph​. For an electron in orbit, the circumference $2\pi r$ must be an integer multiple of $\lambda$: $2\pi r=n\lambda$. This leads directly to Bohr’s quantisation condition mvr=nh2πmvr = \frac{nh}{2\pi}mvr=2πnh​.

Conclusion

The study of Atoms in Class 12 Physics marks a significant step in understanding the microscopic world. Beginning with early atomic models like Thomson’s and Rutherford’s, we gradually arrive at Bohr’s quantised model, which successfully explains the stability of atoms and the discrete spectral lines observed in hydrogen and hydrogen-like species.

The chapter highlights:

• How Rutherford’s experiment revealed the nuclear structure of the atom.

• How Bohr’s quantisation of angular momentum resolved the stability issue and explained atomic spectra.

• The Rydberg formula and spectral series, which directly connect theory to experiment.

• The limitations of classical physics and the need for quantum mechanics to explain complex atomic phenomena.

Overall, this chapter bridges the gap between classical physics and modern quantum physics, laying the foundation for future studies in atomic, nuclear, and quantum physics

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