The Bohr Model of the Hydrogen Atom — Question 8

Section 39-7 · Physics pages 6546–6547

Question 8 The high-index photon frequency approaches the classical orbital frequency For hydrogen write En = −B/n2. In the Bohr model the orbital frequency is forb,n = 2B/(hn3). Consider a photon emitted in the transition n + 1 →n. +Ze v r Schematic; labels, not drawing lengths, specify the data. (a) Derive the ratio of photon frequency to orbital frequency at the lower orbit. (b) Find the large-n limit and evaluate the ratio at n = 20.
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Question 8 - Solution The high-index photon frequency approaches the classical orbital frequency +Ze v r (a) Method and calculation fγ = B h ( 1 n2 − 1 (n + 1)2 ) = B h 2n + 1 n2(n + 1)2 . fγ forb,n = n(2n + 1) 2(n + 1)2 . (b) Interpretation and further calculation n →∞: fγ/forb,n →1 . n = 20 : fγ/forb,20 = 820/882 = .929705 . Physical check Agreement is asymptotic. At low quantum number, the emitted frequency is an energy-difference frequency and need not equal the orbital revolution frequency.

Question 8

The high-index photon frequency approaches the classical orbital frequency

For hydrogen write En=−B/n2E_n=-B/n^2. In the Bohr model the orbital frequency is forb,n=2B/(hn3)f_{{\mathrm{orb}},n}=2B/(hn^3). Consider a photon emitted in the transition n+1→nn+1\to n.

Diagram shown in the original SVG worksheet.

  1. Derive the ratio of photon frequency to orbital frequency at the lower orbit.

  2. Find the large-n limit and evaluate the ratio at n=20n=20.

Question 8 - Solution

The high-index photon frequency approaches the classical orbital frequency

Diagram shown in the original SVG worksheet.

(a) Method and calculation

fγ=Bh(1n2−1(n+1)2)=Bh2n+1n2(n+1)2.f_\gamma=\frac Bh\left(\frac1{n^2}-\frac1{(n+1)^2}\right)=\frac Bh\frac{2n+1}{n^2(n+1)^2}.fγforb,n=n(2n+1)2(n+1)2.\boxed{\frac{f_\gamma}{f_{{\mathrm{orb}},n}}=\frac{n(2n+1)}{2(n+1)^2}}.

(b) Interpretation and further calculation

n→∞:fγ/forb,n→1.n\to\infty:\quad\boxed{f_\gamma/f_{{\mathrm{orb}},n}\to1}.n=20:fγ/forb,20=820/882=.929705.n=20:\quad\boxed{f_\gamma/f_{{\mathrm{orb}},20}=820/882=.929705}.

Physical check

Agreement is asymptotic. At low quantum number, the emitted frequency is an energy-difference frequency and need not equal the orbital revolution frequency.