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Atoms and Nuclei - Rutherford and Bohr’s Atomic Models

Grade 12ICSEPhysics

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

🔑Concepts

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Rutherford's α\alpha-particle Scattering Experiment: Observations showed that most α\alpha-particles passed undeviated, but a few were deflected at large angles, leading to the discovery of the nucleus. The distance of closest approach r0r_0 provides an estimate of nuclear size: 10−1510^{-15} m.

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Rutherford's Model Drawbacks: It could not explain the stability of the atom (accelerated electrons should radiate energy and spiral into the nucleus) or the discrete nature of atomic spectra.

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Bohr’s Postulate of Quantization: Electrons can only revolve in certain non-radiating 'stationary orbits' where the orbital angular momentum LL is an integral multiple of h2π\frac{h}{2\pi}.

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Bohr's Energy States: Electrons emit or absorb energy only when jumping from one orbit to another. The energy of the emitted photon is given by hν=Einitial−Efinalh\nu = E_{initial} - E_{final}.

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Hydrogen Spectral Series: Transitions to n=1n=1 (Lyman series, UV), n=2n=2 (Balmer series, Visible), n=3n=3 (Paschen series, IR), n=4n=4 (Brackett series, IR), and n=5n=5 (Pfund series, IR).

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Radius and Velocity scaling: The radius of the nthn^{th} orbit rn∝n2Zr_n \propto \frac{n^2}{Z} and the orbital velocity vn∝Znv_n \propto \frac{Z}{n}.

📐Formulae

r0=14πϵ02Ze2Kr_0 = \frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{K}

mvr=nh2πmvr = \frac{nh}{2\pi}

rn=ϵ0n2h2πmZe2=0.529n2Z A˚r_n = \frac{\epsilon_0 n^2 h^2}{\pi m Z e^2} = 0.529 \frac{n^2}{Z} \text{ \AA}

En=−me4Z28ϵ02n2h2=−13.6Z2n2 eVE_n = -\frac{m e^4 Z^2}{8 \epsilon_0^2 n^2 h^2} = -13.6 \frac{Z^2}{n^2} \text{ eV}

1λ=RZ2(1n12−1n22)\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)

R=me48ϵ02ch3≈1.097×107 m−1R = \frac{me^4}{8\epsilon_0^2 ch^3} \approx 1.097 \times 10^7 \text{ m}^{-1}

💡Examples

Problem 1:

Calculate the radius of the second excited state of the He+He^+ ion.

Solution:

For He+He^+, Z=2Z = 2. The second excited state corresponds to n=3n = 3. Using rn=a0n2Zr_n = a_0 \frac{n^2}{Z}, where a0=0.529 A˚a_0 = 0.529 \text{ \AA}, we get r3=0.529×322=0.529×4.5=2.3805 A˚r_3 = 0.529 \times \frac{3^2}{2} = 0.529 \times 4.5 = 2.3805 \text{ \AA}.

Explanation:

In Bohr's model, the radius depends on the square of the principal quantum number nn and is inversely proportional to the atomic number ZZ.

Problem 2:

Find the energy required to excite an electron from the ground state (n=1n=1) to the first excited state (n=2n=2) in a Hydrogen atom.

Solution:

Using En=−13.6n2 eVE_n = -\frac{13.6}{n^2} \text{ eV}, E1=−13.6 eVE_1 = -13.6 \text{ eV} and E2=−13.64=−3.4 eVE_2 = -\frac{13.6}{4} = -3.4 \text{ eV}. Energy required ΔE=E2−E1=−3.4−(−13.6)=10.2 eV\Delta E = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \text{ eV}.

Explanation:

Excitation energy is the difference in energy levels between the final and initial stationary states.

Problem 3:

Determine the shortest wavelength in the Balmer series of Hydrogen spectrum.

Solution:

For the Balmer series, n1=2n_1 = 2. The shortest wavelength (series limit) occurs when n2=∞n_2 = \infty. Using 1λ=R(122−1∞2)=R4\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = \frac{R}{4}. Thus, λ=4R=41.097×107≈364.6 nm\lambda = \frac{4}{R} = \frac{4}{1.097 \times 10^7} \approx 364.6 \text{ nm}.

Explanation:

The shortest wavelength corresponds to the maximum energy transition, which happens when the electron transitions from n=∞n = \infty to the base level of the series.

Rutherford and Bohr’s Atomic Models Revision - Class 12 Physics ICSE