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Atoms - Bohr’s Model of Hydrogen Atom

Grade 12CBSEPhysics

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

🔑Concepts

Bohr's First Postulate: Electrons revolve in certain stable, non-radiating circular orbits called 'stationary states'. The centripetal force is provided by the electrostatic attraction: mv2r=14πϵ0Ze2r2\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r^2}

Bohr's Second Postulate (Quantization Condition): An electron can revolve only in those orbits for which its orbital angular momentum (LL) is an integral multiple of h2π\frac{h}{2\pi}. Thus, L=mvr=n=nh2πL = mvr = n\hbar = \frac{nh}{2\pi}, where n=1,2,3,...n = 1, 2, 3, ... is the principal quantum number.

Bohr's Third Postulate (Frequency Condition): An electron might make a transition from a higher energy state (EiE_i) to a lower energy state (EfE_f) by emitting a photon of energy hν=EiEfh\nu = E_i - E_f.

The radius of the nthn^{th} orbit is proportional to n2n^2: rnn2r_n \propto n^2. For a Hydrogen atom (Z=1Z=1), the first Bohr radius is a00.529 A˚a_0 \approx 0.529 \text{ \AA}.

The velocity of an electron in the nthn^{th} orbit is inversely proportional to nn: vn1nv_n \propto \frac{1}{n}.

Energy of the electron: The total energy EnE_n is negative, indicating that the electron is bound to the nucleus. Relationship between energies: K.E.=EnK.E. = -E_n and P.E.=2EnP.E. = 2E_n.

Hydrogen Spectrum: When an electron jumps from n2n_2 to n1n_1, the wavelength λ\lambda of emitted radiation is given by the Rydberg formula: 1λ=R(1n121n22)\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right), where RR is the Rydberg constant (1.097×107 m11.097 \times 10^7 \text{ m}^{-1}).

📐Formulae

mvr=nh2πmvr = \frac{nh}{2\pi}

rn=ϵ0n2h2πme2Z=0.529n2Z A˚r_n = \frac{\epsilon_0 n^2 h^2}{\pi m e^2 Z} = 0.529 \frac{n^2}{Z} \text{ \AA}

vn=Ze22ϵ0nh=(c137)Znv_n = \frac{Ze^2}{2\epsilon_0 nh} = \left( \frac{c}{137} \right) \frac{Z}{n}

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

1λ=RZ2(1n121n22)\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)

💡Examples

Problem 1:

Calculate the energy required to excite an electron from the ground state (n=1n=1) to the second excited state (n=3n=3) in a Hydrogen atom.

Solution:

The energy of the nthn^{th} level is given by En=13.6n2 eVE_n = -\frac{13.6}{n^2} \text{ eV}. For ground state (n=1n=1): E1=13.6 eVE_1 = -13.6 \text{ eV}. For the second excited state (n=3n=3): E3=13.632=13.691.51 eVE_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 \text{ eV}. Energy required ΔE=E3E1=1.51(13.6)=12.09 eV\Delta E = E_3 - E_1 = -1.51 - (-13.6) = 12.09 \text{ eV}.

Explanation:

To move an electron to a higher orbit, energy equal to the difference between the final and initial states must be provided.

Problem 2:

Find the ratio of the radius of the 2nd2^{nd} orbit to the 3rd3^{rd} orbit of a Hydrogen atom.

Solution:

According to Bohr's model, the radius rnr_n is given by rnn2r_n \propto n^2. Therefore, the ratio is r2r3=2232=49\frac{r_2}{r_3} = \frac{2^2}{3^2} = \frac{4}{9}.

Explanation:

The orbital radius increases as the square of the principal quantum number nn.

Bohr’s Model of Hydrogen Atom - Revision Notes & Key Formulas | CBSE Class 12 Physics