Unit 08: Atoms & Nuclei • Chapter 12
Rutherford Scattering & Bohr's Quantized Atom
Geiger-Marsden scattering kinematics, distance of closest approach ($r_0$), impact parameter ($b$), classical collapse paradox, Bohr's postulates, and derivations of orbital radius ($r_n$), velocity ($v_n$), and total energy ($E_n$).
Socho ek bohot bade khali football stadium ke beechon-beech ek chota sa kancha (marble) rakha ho. Rutherford ne jab heavy $\alpha$-particles ko gold foil par fire kiya, toh $99.86\%$ kanche bina kisi rukawat ke seedhe nikal gaye, par $\approx 8000$ mein se ek $\alpha$-particle $90^\circ$ se zyada angle par ulti disha mein palat gaya! Rutherford ne kaha: "Yeh utna hi hairatangez tha jaise aap kisi 15-inch ke artillery shell ko tissue paper par fire karein aur woh wapas aakar aapko hi uda de!" Isse saabit hua ki atom ka $99.999\%$ hissa poori tarah khali hai aur sara positive charge stadium ke center mein ek tiny dense Nucleus mein band hai. Lekin classical physics ne sawal khada kiya: orbiting electron accelerate hokar radiation emit karega aur $10^{-8}\text{ s}$ mein nucleus mein girkar atom ko destroy kar dega! Tab Niels Bohr ne quantum magic introduce kiya: electron kisi bhi arbitrary raste par nahi, balki sirf Stationary Non-Radiating Orbits mein hi ghoom sakta hai jahan uska angular momentum $h/2\pi$ ka integral multiple ho!
1. Rutherford Scattering Kinematics
Closest approach distance and impact parameter ($b$):
Head-on collision: $b = 0 \implies \cot(\theta/2) = 0 \implies \theta = 180^\circ$ (Direct rebound). Large $b \implies \theta \approx 0^\circ$ (Undeviated).
2. Bohr's Postulates & Hydrogen Levels
Quantized parameters for hydrogen-like atom of atomic number $Z$:
Energy partition: Kinetic Energy $K = -E_n$, Potential Energy $U = 2E_n$, and Total Energy $E_n = K + U = -K$.
Deduction of the Distance of Closest Approach ($r_0$) in Head-On Collision
Fig 12.1A: Geiger-Marsden scattering geometry showing distance of closest approach ($r_0$), impact parameter ($b$), and scattering angle ($\theta$).
1. Consider an $\alpha$-particle of mass $m$, charge $q_1 = +2e$, initially fired from infinity with kinetic energy $K = \frac{1}{2}mv^2$ directly towards a target nucleus of charge $q_2 = +Ze$ in a head-on collision ($b = 0$).
2. Due to repulsive Coulomb electrostatic force, the $\alpha$-particle decelerates continuously until at a minimum distance $r_0$ (distance of closest approach), its instantaneous velocity drops to zero ($v = 0$).
3. By the Law of Conservation of Energy, the entire initial kinetic energy $K$ is transformed into electrostatic potential energy $U$ at $r_0$:
Nuclear Size Upper Bound: For a $7.7\text{ MeV}$ $\alpha$-particle fired on gold ($Z = 79$): $r_0 \approx 30\text{ fm} = 3 \times 10^{-14}\text{ m}$. Since the $\alpha$-particle turns back before touching the actual nuclear surface, the nuclear radius must be strictly less than $r_0$ ($R_{\text{nucleus}} < 10^{-14}\text{ m}$).
Derivations of Bohr's Orbital Radius ($r_n$), Orbital Velocity ($v_n$), and Total Energy ($E_n$)
1. Consider an electron of mass $m$ and charge $-e$ revolving in a circular orbit of radius $r_n$ with speed $v_n$ around a nucleus of charge $+Ze$.
2. The electrostatic Coulomb force provides the necessary centripetal force:
3. By Bohr's Second Postulate (Angular Momentum Quantization):
4. Substituting Eq. 2 into Eq. 1:
5. Substituting $r_n$ back into Eq. 2 gives Orbital Speed:
6. Total Energy Derivation:
• Kinetic Energy: $K_n = \frac{1}{2}m v_n^2 = \frac{1}{8\pi\varepsilon_0}\frac{Ze^2}{r_n}$.
• Potential Energy: $U_n = -\frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r_n}$.
• Total Energy: $E_n = K_n + U_n = -\frac{1}{8\pi\varepsilon_0}\frac{Ze^2}{r_n}$. Substituting $r_n$:
Negative Sign Significance: Total energy is strictly negative, which proves that the electron is bound to the nucleus. An infinite energy ($E = 0$) corresponds to an unbound, completely free electron at rest ($n \to \infty$).
Trap 1: Energy Inter-Relations ($K, U, E$) Signs
*Board Trap:* If total energy of an electron in an orbit is $-3.4\text{ eV}$, find its kinetic and potential energy.
• Kinetic Energy is strictly POSITIVE: $\mathbf{K = -E = +3.4\text{ eV}}$.
• Potential Energy is NEGATIVE and DOUBLE the total energy: $\mathbf{U = 2E = -6.8\text{ eV}}$.
Verify: $E = K + U = (+3.4) + (-6.8) = -3.4\text{ eV}$.
Trap 2: Classical Collapse Paradox of Rutherford Model
*Why did Rutherford's model fail?*
1. Instability Paradox: Orbiting electron has centripetal acceleration ($a = v^2/r$). According to classical Maxwellian electrodynamics, an accelerating charge radiates EM energy continuously. Its orbit should shrink into a spiral, collapsing into the nucleus within $\approx 10^{-8}\text{ s}$!
2. Continuous Spectrum Paradox: Spiral radiation should emit a continuous spectrum, contradicting the discrete line spectrum observed in experiments.
Trap 3: Orbital Frequency vs Revolution Time Scaling
Time period of orbital revolution: $$T_n = \frac{2\pi r_n}{v_n} \propto \frac{n^2}{1/n} \propto \mathbf{n^3}$$ Orbital frequency: $f_n = \frac{1}{T_n} \propto \mathbf{\frac{1}{n^3}}$. As principal quantum number $n$ increases, the time period scales with $n^3$ while velocity drops as $1/n$.
Trap 4: Applicability of Bohr's Theory
Bohr's model applies strictly to SINGLE-ELECTRON (Hydrogen-like) atoms:
• Applicable: $\text{H } (Z=1)$, $\text{He}^+ (Z=2)$, $\text{Li}^{2+} (Z=3)$, $\text{Be}^{3+} (Z=4)$.
• NOT applicable to neutral Helium ($\text{He}$), neutral Lithium, or multi-electron atoms because it completely ignores electron-electron mutual repulsion!
Problem: A $5.0\text{ MeV}$ $\alpha$-particle is projected towards a target nucleus of Gold ($Z = 79$) in a head-on collision. Calculate the distance of closest approach. How will this distance change if the kinetic energy of the $\alpha$-particle is doubled?
Solution: Kinetic energy $K = 5.0\text{ MeV} = 5.0 \times 10^6 \times 1.6 \times 10^{-19}\text{ J} = 8.0 \times 10^{-13}\text{ J}$, $Z = 79$.
Since $r_0 \propto \frac{1}{K}$:
Problem: The ground state energy of a hydrogen atom is $-13.6\text{ eV}$. (a) What are the kinetic and potential energies of the electron in the second excited state? (b) If the electron jumps from the third orbit to the first orbit, calculate the frequency and wavelength of the emitted photon.
Solution:
1. "Second excited state" corresponds to principal quantum number $n = 3$ (since $n=1$ is ground state, $n=2$ is 1st excited state).
2. Transition from $n = 3$ to $n = 1$ ($E_1 = -13.6\text{ eV}$):
Frequency of emitted photon ($\nu = \Delta E/h$):
Closest Approach
$r_0 = \frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{K}$
Bohr Radius
$r_n \propto \frac{n^2}{Z}$ ($r_1 = 0.53\text{ \AA}$)
Total Energy
$E_n = -13.6\frac{Z^2}{n^2}\text{ eV}$
Energy Triangle
$K = -E, U = 2E$