Atoms and Nuclei
Atoms and Nuclei for JEE Main & Advanced
Atomic Structure
Atomic Models (Thomson, Rutherford, Bohr)Topic 1
Thomson's Plum Pudding Model (1898): Atom = uniform positive sphere with electrons embedded like raisins in pudding. Failed to explain Rutherford's scattering and atomic spectra.
Rutherford's Gold Foil Experiment (1911): Alpha particles scattered by thin gold foil; most pass through, few deflected by large angles (some bounce back). Conclusions:
- Atom mostly empty space
- Tiny dense positive nucleus at center (radius $\sim 10^{-15}$ m vs atomic radius $10^{-10}$ m)
- Electrons orbit nucleus (planetary model)
Distance of Closest Approach: $r_0 = \dfrac{1}{4\pi\epsilon_0}\cdot\dfrac{2Ze^2}{K}$ for $\alpha$-particle of KE $K$ hitting nucleus head-on.
Impact Parameter: $b = \dfrac{Ze^2\cot(\theta/2)}{4\pi\epsilon_0 K}$ for scattering angle $\theta$.
Failures of Rutherford Model:
- Classical EM theory: accelerating electron should radiate, spiral into nucleus → atom unstable
- Cannot explain discrete atomic spectra
Bohr's Postulates (1913):
- Electron revolves only in certain stationary orbits where it does not radiate energy
- Angular momentum quantized: $L = mvr = nh/(2\pi) = n\hbar$, $n = 1, 2, 3, \ldots$
- Energy emitted/absorbed when electron jumps between orbits: $h\nu = E_i - E_f$
For Hydrogen-like Atom (charge $Ze$, one electron):
| Quantity | Formula |
|---|---|
| Radius of nth orbit | $r_n = \dfrac{n^2 h^2 \epsilon_0}{\pi m Z e^2} = 0.529\dfrac{n^2}{Z}$ Å |
| Velocity in nth orbit | $v_n = \dfrac{Ze^2}{2\epsilon_0 nh} = 2.18 \times 10^6 \cdot Z/n$ m/s |
| Energy of nth orbit | $E_n = -\dfrac{13.6\,Z^2}{n^2}$ eV |
| Total energy | $E_n = -K_n$ (PE = $-2K$) |
For hydrogen ($Z = 1$): $r_1 = 0.529$ Å (Bohr radius), $E_1 = -13.6$ eV, $v_1 = c/137$.
Find radius of 3rd Bohr orbit in hydrogen.
Show solution
$$r_3 = 0.529 \times 3^2 = 0.529 \times 9 = 4.76 \text{ Å}$$
Final Answer: $r_3 = 4.76$ Å.
Find ionization energy of He$^+$ ($Z = 2$).
Show solution
Ionization energy = energy to remove electron from $n = 1$ to $\infty$ = $|E_1|$: $$|E_1| = 13.6\,Z^2 = 13.6 \times 4 = 54.4 \text{ eV}$$
Final Answer: $54.4$ eV.
Rutherford's experiment proved that the atom:
According to Bohr, angular momentum is:
The energy of $n = 2$ orbit of H atom:
The radius of 1st Bohr orbit of H:
Bohr model is valid for:
Hydrogen Spectrum and Energy LevelsTopic 2
When electron jumps from higher orbit $n_i$ to lower $n_f$, photon emitted with: $$h\nu = E_{n_i} - E_{n_f}, \quad \frac{1}{\lambda} = R_H Z^2\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$$
$R_H = 1.097 \times 10^7$ m⁻¹ is Rydberg constant.
Spectral Series of Hydrogen:
| Series | $n_f$ | Region | First line $\lambda$ (approx) |
|---|---|---|---|
| Lyman | $1$ | UV | $121.6$ nm |
| Balmer | $2$ | Visible | $656.3$ nm ($H_\alpha$) |
| Paschen | $3$ | IR | $1875$ nm |
| Brackett | $4$ | IR | $4051$ nm |
| Pfund | $5$ | Far IR | $7458$ nm |
Series limit = wavelength when $n_i \to \infty$: $\lambda_{\text{limit}} = n_f^2/(R_H Z^2)$.
Number of Spectral Lines when electron transitions from $n$th level to ground: $$N = \frac{n(n-1)}{2}$$
Ionization Energy and Potential:
- Ionization energy: $|E_1| = 13.6 Z^2$ eV
- Ionization potential: $V_i = 13.6 Z^2$ V
- Excitation energy from $n_1$ to $n_2$: $E_{n_2} - E_{n_1}$
- Excitation potential: corresponding $V$
Limitations of Bohr Model:
- Cannot explain fine structure (splitting due to spin-orbit coupling)
- Cannot explain Zeeman/Stark effects (magnetic/electric field splitting)
- Cannot account for relative intensities of spectral lines
- Cannot extend to multi-electron atoms
- Inconsistent with quantum mechanics' uncertainty principle
These led to the modern quantum mechanical model (Schrödinger, Heisenberg).
Find wavelength of $H_\alpha$ line in hydrogen (Balmer series: $n_i = 3 \to n_f = 2$).
Show solution
$$\frac{1}{\lambda} = R_H\left(\frac{1}{4} - \frac{1}{9}\right) = R_H \cdot \frac{5}{36}$$ $$\lambda = \frac{36}{5R_H} = \frac{36}{5 \times 1.097 \times 10^7} \approx 6.563 \times 10^{-7} \text{ m} = 656.3 \text{ nm}$$
Final Answer: $\lambda \approx 656.3$ nm.
Find total number of spectral lines emitted when electron drops from $n = 4$ to $n = 1$.
Show solution
$$N = \frac{n(n-1)}{2} = \frac{4 \times 3}{2} = 6 \text{ lines}$$
Final Answer: $6$ lines.
Balmer series of hydrogen lies in:
The Lyman series corresponds to transitions to:
Number of spectral lines when an electron drops from $n = 5$ to ground:
Ionization energy of hydrogen atom:
The wavelength of first Lyman line ($n: 2 \to 1$) for hydrogen:
Nuclear Physics
Nuclear Composition, Binding EnergyTopic 1
Nucleus consists of protons ($p$, charge $+e$) and neutrons ($n$, neutral); collectively called nucleons.
Notation: $^A_Z X$, where $A$ = mass number (total nucleons), $Z$ = atomic number (protons), $N = A - Z$ = neutrons.
Isotopes: Same $Z$, different $A$ (e.g., $^1$H, $^2$H, $^3$H). Isobars: Same $A$, different $Z$ (e.g., $^{14}$C, $^{14}$N). Isotones: Same $N$, different $Z$ (e.g., $^{13}$C, $^{14}$N).
Nuclear Size: $R = R_0 A^{1/3}$, where $R_0 \approx 1.2 \times 10^{-15}$ m = $1.2$ fm. Density: $\rho \approx 2.3 \times 10^{17}$ kg/m³ (constant for all nuclei).
Mass-Energy Equivalence: $E = mc^2$. Atomic mass unit: $1$ u = $1.66 \times 10^{-27}$ kg = $931.5$ MeV/$c^2$.
Mass Defect: $\Delta m = Zm_p + (A-Z)m_n - M_{\text{nucleus}}$ (always positive).
Binding Energy (BE): Energy needed to dissociate nucleus into individual nucleons. $$\text{BE} = \Delta m \cdot c^2 = \Delta m \cdot 931.5 \text{ MeV (if } \Delta m \text{ in u)}$$
Binding Energy per Nucleon (BE/A): Measure of nuclear stability.
BE/A curve features:
- Rises rapidly for light nuclei
- Peaks at $\sim 8.8$ MeV/nucleon near $^{56}$Fe
- Slowly decreases for heavy nuclei
- Explains why fission of heavy / fusion of light → energy release
Nuclear Forces:
- Strong, short-range (~$1$ fm), attractive (charge independent)
- Much stronger than EM force (factor $\sim 100$)
- Saturated (each nucleon interacts with few neighbors only)
Find binding energy of $^4_2$He nucleus. Atomic mass: $4.0026$ u. ($m_p = 1.0073$, $m_n = 1.0087$ u)
Show solution
$\Delta m = 2(1.0073) + 2(1.0087) - 4.0026 = 2.0146 + 2.0174 - 4.0026 = 0.0294$ u. $$\text{BE} = 0.0294 \times 931.5 \approx 27.4 \text{ MeV}$$ BE/nucleon $= 27.4/4 = 6.85$ MeV.
Final Answer: BE $\approx 27.4$ MeV; BE/A $\approx 6.85$ MeV.
Find radius of $^{27}$Al nucleus.
Show solution
$$R = R_0 A^{1/3} = 1.2 \times (27)^{1/3} = 1.2 \times 3 = 3.6 \text{ fm}$$
Final Answer: $R = 3.6$ fm.
The strong nuclear force is:
$1$ u of mass is equivalent to:
Isotopes have same:
Nuclear density is:
Peak of BE/A curve occurs near:
Radioactivity, Fission and FusionTopic 2
Radioactivity: Spontaneous emission of particles/radiation from unstable nuclei. Discovered by Becquerel (1896); systematically studied by Marie and Pierre Curie.
Types of Radiation:
| Type | Identity | Charge | Penetration | Effect on $A, Z$ |
|---|---|---|---|---|
| Alpha ($\alpha$) | $^4_2$He nucleus | $+2e$ | Lowest (stopped by paper) | $A \to A-4, Z \to Z-2$ |
| Beta ($\beta^-$) | electron | $-e$ | Medium (stopped by aluminium) | $Z \to Z+1$, $A$ same |
| Beta ($\beta^+$) | positron | $+e$ | Medium | $Z \to Z-1$, $A$ same |
| Gamma ($\gamma$) | photon (high-energy EM) | $0$ | High (needs lead) | No change |
In $\beta^-$ decay: $n \to p + e^- + \bar\nu$; in $\beta^+$: $p \to n + e^+ + \nu$.
Radioactive Decay Law: $$\frac{dN}{dt} = -\lambda N, \quad N(t) = N_0 e^{-\lambda t}$$ $\lambda$ = decay constant. Activity: $A = \lambda N$ (decays per second; unit: becquerel, Bq, or curie: $1$ Ci $= 3.7 \times 10^{10}$ Bq).
Half-life: $T_{1/2} = \ln 2/\lambda = 0.693/\lambda$.
Mean life: $\tau = 1/\lambda = T_{1/2}/\ln 2 = T_{1/2}/0.693$.
After $n$ half-lives: $N = N_0/2^n$.
Nuclear Fission: Heavy nucleus splits into lighter ones, releasing energy and neutrons.
Example: $^{235}\text{U} + n \to ^{141}\text{Ba} + ^{92}\text{Kr} + 3n + \sim 200$ MeV.
Chain reaction: neutrons released cause further fission. Critical mass = minimum mass for sustained chain reaction. Used in nuclear reactors (controlled) and atomic bombs (uncontrolled).
Reactor components: fuel ($^{235}$U, Pu), moderator (slows neutrons: water, graphite), coolant (water, sodium), control rods (Cd, B; absorb neutrons), shielding (concrete).
Nuclear Fusion: Light nuclei combine to form heavier (e.g., H → He in stars). Requires very high temperature ($\sim 10^7$ K) to overcome Coulomb repulsion.
Example: $^2\text{H} + ^3\text{H} \to ^4\text{He} + n + 17.6$ MeV.
Powers the Sun and stars. Hydrogen bomb uses fusion (uncontrolled). Controlled fusion is still under research (tokamaks, ITER project).
A radioactive sample has $T_{1/2} = 8$ days. Fraction remaining after $24$ days:
Show solution
$n = t/T_{1/2} = 24/8 = 3$ half-lives. $$N/N_0 = 1/2^n = 1/8$$
Final Answer: $1/8$ of original.
A radioactive nuclide has decay constant $\lambda = 0.05$/year. Find half-life and mean life.
Show solution
$T_{1/2} = 0.693/0.05 = 13.86$ years. $\tau = 1/\lambda = 1/0.05 = 20$ years.
Final Answer: $T_{1/2} \approx 13.86$ years; $\tau = 20$ years.
In $\beta^-$ decay:
Half-life and mean life relation:
Highest penetration:
Critical mass refers to:
The fusion reaction primarily powers:
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