The Bohr Model Cheat Sheet: Solving Atomic Structure Problems Fast
Master essential Bohr model formulas for radius, energy, velocity and quick calculation methods for electron transitions.
Why Bohr Model is Crucial for JEE
Bohr's atomic model appears in 2-3 questions per JEE paper. Mastering these formulas can help you solve problems in under 60 seconds:
- Direct formula application problems
- Electron transition energy calculations
- Hydrogen-like ions (He⁺, Li²⁺ etc.)
- Wavelength and frequency of emitted photons
🚀 Bohr Model Cheat Sheet
Radius of nth Orbit
$r_n = \frac{n^2h^2}{4\pi^2mkZe^2}$
Simplified: $r_n = \frac{n^2}{Z}a_0$
Energy of nth Orbit
$E_n = -\frac{2\pi^2mk^2Z^2e^4}{n^2h^2}$
Simplified: $E_n = -\frac{13.6Z^2}{n^2}$ eV
Electron Velocity
$v_n = \frac{2\pi kZe^2}{nh}$
Simplified: $v_n = \frac{Z}{n}v_1$
Transition Energy
$\Delta E = 13.6Z^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$
In eV (n₂ > n₁)
Radius of nth Orbit
Where $a_0 = 0.529 × 10^{-10}$ m (Bohr radius)
🎯 Quick Application Method:
For Hydrogen (Z=1): $r_n = n^2 × 0.529$ Å
For Hydrogen-like ions: $r_n = \frac{n^2}{Z} × 0.529$ Å
Ratio problems: $\frac{r_{n1}}{r_{n2}} = \frac{n_1^2}{n_2^2}$
⚡ JEE Example:
Problem: Find radius ratio for 2nd orbit of He⁺ to 3rd orbit of Li²⁺
Quick Solution:
$\frac{r_2(\text{He}^+)}{r_3(\text{Li}^{2+})} = \frac{\frac{2^2}{2}}{\frac{3^2}{3}} = \frac{2}{3}$
Solved in 15 seconds!
Energy of nth Orbit
Ground state energy for Hydrogen: -13.6 eV
🎯 Quick Application Method:
For Hydrogen (Z=1): $E_n = -\frac{13.6}{n^2}$ eV
Ionization energy: $E_{\infty} - E_n = \frac{13.6Z^2}{n^2}$ eV
Energy ratios: $\frac{E_{n1}}{E_{n2}} = \frac{n_2^2}{n_1^2}$
⚡ JEE Example:
Problem: Ionization energy of He⁺ in ground state
Quick Solution:
$E_1 = -\frac{13.6 × 2^2}{1^2} = -54.4$ eV
Ionization energy = $0 - (-54.4) = 54.4$ eV
Solved in 10 seconds!
Electron Transition Energy Calculation
Common Transitions (Hydrogen):
n=2→1
10.2 eV
n=3→1
12.1 eV
n=3→2
1.89 eV
n=4→2
2.55 eV
⚡ JEE Example:
Problem: Energy released when electron jumps from n=4 to n=2 in Hydrogen
Quick Solution:
$\Delta E = 13.6\left(\frac{1}{2^2} - \frac{1}{4^2}\right) = 13.6\left(\frac{1}{4} - \frac{1}{16}\right)$
$= 13.6 × \frac{3}{16} = 2.55$ eV
Memorize common transitions for instant answers!
🚀 Problem-Solving Strategies
Memory Techniques:
- $r_n ∝ n^2/Z$ - Radius increases with n²
- $E_n ∝ -Z^2/n^2$ - Energy becomes less negative
- $v_n ∝ Z/n$ - Velocity decreases with n
- Ground state H: -13.6 eV (memorize!)
JEE Exam Tips:
- Use eV units for faster calculation
- Memorize common transition energies
- For wavelength: $\lambda = \frac{12400}{\Delta E(eV)}$ Å
- Practice hydrogen-like ions (He⁺, Li²⁺)
Advanced Applications Available
Includes spectral series, wavelength calculations, and JEE Advanced level problems
📝 Quick Self-Test
Try these JEE-level problems to test your speed:
1. Radius of 3rd orbit in He⁺ ion (in terms of a₀)
2. Energy required to excite electron from n=1 to n=4 in Li²⁺
3. Ratio of velocities in 1st and 3rd orbits of Hydrogen
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