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Q.Prove that radius of the nthn^{th} Bohr orbit of an atom is directly proportional to the square of the principal quantum number.

Nagaland NbseNagaland Board of School Education 2016Subjective· 3mImportance★★★★★
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Combining the Coulomb-force-provides-centripetal-force condition with Bohr's angular-momentum quantisation gives rn∝n2r_n \propto n^2.

Bohr's postulates used.

  1. The electron revolves in a circular orbit under the Coulomb attraction of the nucleus, which provides the centripetal force.
  2. Angular momentum of the electron is quantised: mvr=nh2πmvr = \dfrac{nh}{2\pi}, n=1,2,3,…n=1,2,3,\dots

Step 1 -- Coulomb force = centripetal force.

For an electron (mass mm, charge −e-e) orbiting a nucleus of charge +Ze+Ze at radius rr with speed vv:

mv2r=14πε0Ze2r2...(1)\dfrac{mv^2}{r} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze^2}{r^2} \qquad \text{...(1)}

Step 2 -- quantisation of angular momentum.

mvr=nh2π  ⟹  v=nh2πmr...(2)mvr = \dfrac{nh}{2\pi} \implies v = \dfrac{nh}{2\pi m r} \qquad\text{...(2)}

Step 3 -- substitute (2) into (1).

m(nh2πmr)21r=Ze24πε0r2m\left(\dfrac{nh}{2\pi m r}\right)^2 \dfrac1r = \dfrac{Ze^2}{4\pi\varepsilon_0 r^2}

n2h24π2mr3=Ze24πε0r2\dfrac{n^2h^2}{4\pi^2 m r^3} = \dfrac{Ze^2}{4\pi\varepsilon_0 r^2}

Multiplying both sides by r3r^3 and simplifying:

n2h24π2m=Ze2r4πε0\dfrac{n^2h^2}{4\pi^2 m} = \dfrac{Ze^2 r}{4\pi\varepsilon_0} …

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