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Q.Using postulates of Bohr's theory of H-atom, show that the radii of the orbits in hydrogen atom varies as n2n^2, where n is the principal quantum number of the atom.

Nagaland NbseNagaland Board of School Education 2024Subjective· 3mImportance★★★★★
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Setting the Coulomb attraction equal to the required centripetal force, then eliminating velocity using Bohr's angular-momentum quantization rule, directly yields r∝n2r\propto n^2.

Bohr's postulates used:

  1. Electrons revolve in circular orbits around the nucleus, with the electrostatic (Coulomb) force of attraction providing the necessary centripetal force.
  2. The angular momentum of the electron is quantized: mvr=nh2πmvr = \dfrac{nh}{2\pi}, where n=1,2,3,…n=1,2,3,\dots is the principal quantum number.

Step 1 — Coulomb force = centripetal force (for a nucleus of charge +Ze+Ze, electron charge −e-e, orbit radius rr, speed vv):

14πϵ0Ze2r2=mv2r\frac{1}{4\pi\epsilon_0}\frac{Ze^2}{r^2} = \frac{mv^2}{r}

v2=Ze24πϵ0mr...(1)v^2 = \frac{Ze^2}{4\pi\epsilon_0 m r} \quad\text{...(1)}

Step 2 — Quantization of angular momentum:

mvr=nh2π⇒v=nh2πmr...(2)mvr = \frac{nh}{2\pi} \quad\Rightarrow\quad v = \frac{nh}{2\pi m r} \quad\text{...(2)}

Step 3 — Eliminate vv: square equation (2) and substitute into (1):

(nh2πmr)2=Ze24πϵ0mr\left(\frac{nh}{2\pi m r}\right)^2 = \frac{Ze^2}{4\pi\epsilon_0 m r}

n2h24π2m2r2=Ze24πϵ0mr\frac{n^2h^2}{4\pi^2m^2r^2} = \frac{Ze^2}{4\pi\epsilon_0 m r}

Cross-multiplying and simplifying (multiply both sides by 4πmr2ϵ04\pi m r^2 \epsilon_0 and rearrange):

n2h2ϵ0=πmrZe2n^2h^2\epsilon_0 = \pi m r Z e^2 …

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