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Structure of Atom - Bohr's Model of Atom

Grade 11ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Bohr's model is based on the quantization of energy. He proposed that electrons revolve around the nucleus in specific circular paths called 'orbits' or 'stationary states' without radiating energy. These orbits are associated with a fixed amount of energy and are designated by the principal quantum number n=1,2,3,n = 1, 2, 3, \dots (or K,L,M,NK, L, M, N shells).

The angular momentum of an electron in a given orbit is quantized. An electron can revolve only in those orbits for which its angular momentum is an integral multiple of h2π\frac{h}{2\pi}. This is expressed as mvr=nh2πmvr = \frac{nh}{2\pi}, where mm is the mass, vv is the velocity, and rr is the radius of the orbit.

Energy is emitted or absorbed only when an electron jumps from one stationary state to another. Energy is absorbed (+ΔE+\Delta E) when jumping to a higher level (nlowernhighern_{lower} \to n_{higher}) and emitted (ΔE-\Delta E) when falling to a lower level (nhighernlowern_{higher} \to n_{lower}).

The energy difference between two levels is given by ΔE=E2E1=hν\Delta E = E_2 - E_1 = h\nu, where hh is Planck's constant (6.626×1034 J s6.626 \times 10^{-34} \text{ J s}) and ν\nu is the frequency of the radiation.

Bohr's theory successfully explains the line spectrum of Hydrogen and Hydrogen-like species (ions with only one electron like He+He^+, Li2+Li^{2+}, and Be3+Be^{3+}).

📐Formulae

mvr=nh2πmvr = \frac{nh}{2\pi}

rn=a0n2Z=0.529n2Z A˚r_n = a_0 \frac{n^2}{Z} = 0.529 \frac{n^2}{Z} \text{ \AA}

vn=v0Zn=2.18×106Zn m/sv_n = v_0 \frac{Z}{n} = 2.18 \times 10^6 \frac{Z}{n} \text{ m/s}

En=13.6Z2n2 eV/atomE_n = -13.6 \frac{Z^2}{n^2} \text{ eV/atom}

En=2.18×1018Z2n2 J/atomE_n = -2.18 \times 10^{-18} \frac{Z^2}{n^2} \text{ J/atom}

1λ=RHZ2(1n121n22) where RH1.09677×107 m1\frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \text{ where } R_H \approx 1.09677 \times 10^7 \text{ m}^{-1}

💡Examples

Problem 1:

Calculate the radius of the third orbit (n=3n=3) of a He+He^+ ion.

Solution:

For He+He^+, the atomic number Z=2Z = 2. The formula for the radius is rn=0.529n2Z A˚r_n = 0.529 \frac{n^2}{Z} \text{ \AA}. Substituting the values: r3=0.529×322=0.529×92=0.529×4.5=2.3805 A˚r_3 = 0.529 \times \frac{3^2}{2} = 0.529 \times \frac{9}{2} = 0.529 \times 4.5 = 2.3805 \text{ \AA}

Explanation:

The radius of a Bohr orbit is directly proportional to the square of the principal quantum number n2n^2 and inversely proportional to the atomic number ZZ.

Problem 2:

Calculate the energy of an electron in the second excited state of a Hydrogen atom.

Solution:

The second excited state corresponds to n=3n=3 (since n=1n=1 is the ground state and n=2n=2 is the first excited state). For Hydrogen, Z=1Z=1. Using the formula En=13.6Z2n2 eVE_n = -13.6 \frac{Z^2}{n^2} \text{ eV}: E3=13.6×1232=13.6×191.51 eVE_3 = -13.6 \times \frac{1^2}{3^2} = -13.6 \times \frac{1}{9} \approx -1.51 \text{ eV}

Explanation:

Energy levels in a Hydrogen atom are negative, indicating that the electron is bound to the nucleus. As nn increases, the energy becomes less negative (higher energy).

Bohr's Model of Atom - Revision Notes & Key Formulas | ICSE Class 11 Chemistry