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Structure of Atom - Bohr's Model of Atom

Grade 11CBSEChemistry

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

🔑Concepts

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Electrons revolve around the nucleus in specific circular paths called orbits or stationary states. These orbits are associated with a fixed amount of energy, labeled as n=1,2,3...n = 1, 2, 3... or K,L,M,N...K, L, M, N....

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Energy of an electron in an orbit does not change with time. However, an electron can move from a lower energy state to a higher energy state by absorbing energy, or from a higher to a lower state by emitting energy as electromagnetic radiation.

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The angular momentum of an electron is quantized. An electron can move only in those orbits for which its angular momentum (mvrmvr) is an integral multiple of h2π\frac{h}{2\pi}.

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The model explains the stability of the atom and the line spectrum of Hydrogen and Hydrogen-like species (e.g., He+He^+, Li2+Li^{2+}, Be3+Be^{3+}).

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Limitations: It fails to explain the spectra of multi-electron atoms, the splitting of spectral lines in magnetic fields (Zeeman effect) or electric fields (Stark effect), and contradicts the Heisenberg Uncertainty Principle.

📐Formulae

mvr=nh2π (Quantization of Angular Momentum)mvr = \frac{nh}{2\pi} \text{ (Quantization of Angular Momentum)}

rn=a0n2Z=0.529×n2Z A˚ (Radius of nth orbit)r_n = a_0 \frac{n^2}{Z} = 0.529 \times \frac{n^2}{Z} \text{ \AA} \text{ (Radius of } n^{th} \text{ orbit)}

En=−2.18×10−18(Z2n2) J/atomE_n = -2.18 \times 10^{-18} \left( \frac{Z^2}{n^2} \right) \text{ J/atom}

En=−13.6(Z2n2) eV/atomE_n = -13.6 \left( \frac{Z^2}{n^2} \right) \text{ eV/atom}

vn=2.18×106×Zn m/s (Velocity of electron in nth orbit)v_n = 2.18 \times 10^6 \times \frac{Z}{n} \text{ m/s (Velocity of electron in } n^{th} \text{ orbit)}

ΔE=E2−E1=hν=hcλ\Delta E = E_2 - E_1 = h\nu = \frac{hc}{\lambda}

νˉ=1λ=RHZ2(1n12−1n22) where RH=109,677 cm−1\bar{\nu} = \frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \text{ where } R_H = 109,677 \text{ cm}^{-1}

💡Examples

Problem 1:

Calculate the radius of the second orbit (n=2n=2) of Li2+Li^{2+} ion.

Solution:

For Li2+Li^{2+}, the atomic number Z=3Z = 3. Given n=2n = 2. Using the formula rn=0.529×n2Z A˚r_n = 0.529 \times \frac{n^2}{Z} \text{ \AA}: r2=0.529×223=0.529×43≈0.705 A˚r_2 = 0.529 \times \frac{2^2}{3} = 0.529 \times \frac{4}{3} \approx 0.705 \text{ \AA}

Explanation:

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

Problem 2:

What is the energy associated with the first orbit of He+He^+ ion in Joules?

Solution:

For He+He^+, Z=2Z = 2 and n=1n = 1. Using the formula En=−2.18×10−18(Z2n2) JE_n = -2.18 \times 10^{-18} \left( \frac{Z^2}{n^2} \right) \text{ J}: E1=−2.18×10−18×(2212)=−2.18×10−18×4=−8.72×10−18 JE_1 = -2.18 \times 10^{-18} \times \left( \frac{2^2}{1^2} \right) = -2.18 \times 10^{-18} \times 4 = -8.72 \times 10^{-18} \text{ J}

Explanation:

The negative sign indicates that the electron is bound to the nucleus. As ZZ increases, the energy becomes more negative, implying the electron is more tightly held.

Problem 3:

Calculate the frequency of radiation emitted when an electron falls from n=4n=4 to n=2n=2 in a Hydrogen atom (Z=1Z=1).

Solution:

ΔE=2.18×10−18(1n12−1n22) J\Delta E = 2.18 \times 10^{-18} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \text{ J}. Here n1=2,n2=4n_1=2, n_2=4. ΔE=2.18×10−18(14−116)=2.18×10−18(316)=4.0875×10−19 J\Delta E = 2.18 \times 10^{-18} \left( \frac{1}{4} - \frac{1}{16} \right) = 2.18 \times 10^{-18} \left( \frac{3}{16} \right) = 4.0875 \times 10^{-19} \text{ J}. Frequency ν=ΔEh=4.0875×10−19 J6.626×10−34 J s≈6.17×1014 Hz\nu = \frac{\Delta E}{h} = \frac{4.0875 \times 10^{-19} \text{ J}}{6.626 \times 10^{-34} \text{ J s}} \approx 6.17 \times 10^{14} \text{ Hz}.

Explanation:

When an electron transitions from a higher energy level to a lower one, energy is released as a photon. The frequency is determined by the energy difference ΔE=hν\Delta E = h\nu.

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