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Atomic Structure & Bohr-Sommerfeld Model: Master Notes

Chapter Overview

Atomic physics models the quantized energy states of hydrogenic atoms and characteristic X-ray emissions. In IIT-JEE (Advanced), core problems center on Bohr parameter scaling (rnn2/Z,vnZ/n,EnZ2/n2), reduced mass corrections for exotic atoms (Muonic Hydrogen & Positronium), spectral line series transitions, and Moseley's Law for characteristic X-rays.


1. Bohr Postulates & Hydrogen-Like Atom Scaling

For any single-electron atom of nuclear charge +Ze (H,He+,Li2+,Be3+):

  1. Quantization of Angular Momentum:L=mvnrn=n=nh2π(n=1,2,3,)
  2. Electrostatic Centripetal Balance:mvn2rn=14πε0Ze2rn2

1.1 Master Scaling Laws

VariableGeneral FormulaScaling with (n,Z)Value for Hydrogen (Z=1,n=1)
Orbital Radius (rn)rn=ε0h2πme2n2Zrnn2Zr1=0.529 \AA
Orbital Velocity (vn)vn=e22ε0hZnvnZnv1=c1372.18×106 m/s
Kinetic Energy (Kn)Kn=12mvn2KnZ2n2K1=+13.6 eV
Potential Energy (Un)Un=14πε0Ze2rnUnZ2n2U1=27.2 eV=2K1
Total Energy (En)En=Kn+Un=Kn=Un2En=13.6Z2n2 eVE1=13.6 eV
Orbital Time Period (Tn)Tn=2πrnvnTnn3Z2T11.52×1016 s

2. Hydrogen Spectral Series & Rydberg Formula

1λ=RZ2(1n121n22)

(where R=me48ε02h3c1.097×107 m1 is the Rydberg Constant).

Multi-Mode DiagramAtomic Structure: Bohr Quantized Energy Ladder
Option 1: Publication-Grade Scientific Vector SVG

Hydrogen quantized energy levels E_n = -13.6 / n^2 eV with Lyman, Balmer, and Paschen spectral emission series.

n = 1 (-13.6 eV) n = 2 (-3.4 eV) n = 3 (-1.51 eV) Lyman
Bohr Energy: En=13.6Z2n2 eVE_n = -\frac{13.6 Z^2}{n^2} \text{ eV}
Rydberg Formula: 1λ=RZ2(1n121n22)\frac{1}{\lambda} = R Z^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)
  • Number of Spectral Lines emitted from state n to ground state:Nlines=n(n1)2

3. Reduced Mass Corrections (Finite Nuclear Mass M)

When nuclear mass M is not infinitely greater than orbiting particle mass m:

μ=mMm+M
  • Replace mμ in all Bohr energy and Rydberg formulas:En=(μme)En,R=(μme)R
  • Positronium (e+e+): μ=mememe+me=me2E1=13.62=6.8 eV.

4. X-Rays & Moseley's Law

4.1 Continuous X-Rays (Duane-Hunt Law)

λmin=hceV12400V(volts) \AA

4.2 Characteristic X-Rays & Moseley's Law

ν=a(Zb)
  • For Kα transition (n=2n=1): Screening constant b1, a=34Rc.
  • For Lα transition (n=3n=2): Screening constant b7.4.

5. Authentic Previous Years Questions (PYQs)

PYQ 1: JEE Advanced 2021 (Paper 2) — Muonic Hydrogen Rydberg Wavelength

Question:
A muonic hydrogen atom consists of a muon of mass mμ=207me orbiting a proton (mp=1836me). Find the ground state binding energy E1 of muonic hydrogen in eV.

Step-by-Step Solution:

  1. Reduced Mass of Muonic Hydrogen:μ=mμmpmμ+mp=(207me)(1836me)207me+1836me=3800522043me186.03me
  2. Ground State Energy:E1=(μme)Ehydrogen, 1=186.03×(13.6 eV)=2530 eV=2.53 keV

PYQ 2: JEE Advanced 2019 — Moseley's Law & Atomic Number Determination

Question:
The Kα X-ray wavelength of element A (ZA=43) is λA=0.70 \AA. Find the atomic number ZB of an element B whose Kα wavelength is λB=0.54 \AA. (Use screening constant b=1).

Step-by-Step Solution:

From Moseley's Law for Kα emission:

cλ=ν=a2(Z1)21λ(Z1)ZB1ZA1=λAλB=0.700.54=35271.296=1.138ZB1=(431)×1.138=42×1.13847.8ZB=49(Indium)

6. High-Yield Formula Sheet

EntityFormulaNotes
Bohr Radiusrn=0.529n2Z \AAHydrogenic ions
Total EnergyEn=13.6Z2n2 eVBound state E<0
X-Ray Cutoffλmin=12400V \AADuane-Hunt Law
Moseley's Lawν=a(Z1)For Kα emission
Reduced Massμ=m1m2m1+m2Finite nuclear mass