The Bohr Model of the Hydrogen Atom — Question 4

Section 39-7 · Physics pages 6538–6539

Question 4 A spectral series has a longest line and a shorter limiting wavelength Consider every hydrogen emission transition n →2 with integer n ≥3. Use the nonrelativistic Bohr model with an infinitely heavy nucleus unless stated otherwise: En = −13.6Z2/n2 eV, rn = a0n2/Z, a0 = .0529 nm, and hc = 1240 eV nm. Here Z is the nuclear charge number. n = 1 n = 2 n = 3 n = 4 E E = 0 Schematic; labels, not drawing lengths, specify the data. (a) Find the longest wavelength in the series. (b) Find the short-wavelength limit as n →∞and identify which endpoint is approached rather than attained by a finite n.
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Question 4 - Solution A spectral series has a longest line and a shorter limiting wavelength n = 1 n = 2 n = 3 n = 4 E E = 0 (a) Method and calculation Eγ = 13.6 (1 4 −1 n2 ) eV; Eγ smallest at n = 3. λmax = 1240 13.6(1/4 −1/9) = 656.471 nm . (b) Interpretation and further calculation n →∞: Eγ →3.40 eV ⇒λlimit = 364.706 nm . 364.706 nm < λ ≤656.471 nm for finite bound-state indices. Physical check The series limit corresponds to the continuum threshold as the initial binding tends to zero. Finite upper levels produce wavelengths strictly above the limit.

Question 4

A spectral series has a longest line and a shorter limiting wavelength

Consider every hydrogen emission transition n→2n\to2 with integer n≥3n\ge3. Use the nonrelativistic Bohr model with an infinitely heavy nucleus unless stated otherwise: En=−13.6Z2/n2eVE_n=-13.6Z^2/n^2\,\mathrm{eV}, rn=a0n2/Zr_n=a_0n^2/Z, a0=.0529nma_0=.0529\,\mathrm{nm}, and hc=1240eVnmhc=1240\,\mathrm{eV\,nm}. Here Z is the nuclear charge number.

Diagram shown in the original SVG worksheet.

  1. Find the longest wavelength in the series.

  2. Find the short-wavelength limit as n→∞n\to\infty and identify which endpoint is approached rather than attained by a finite n.

Question 4 - Solution

A spectral series has a longest line and a shorter limiting wavelength

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Eγ=13.6(14−1n2)eV;Eγ smallest at n=3.E_\gamma=13.6\left(\frac14-\frac1{n^2}\right)\mathrm{eV};\quad E_\gamma\text{ smallest at }n=3.λmax=124013.6(1/4−1/9)=656.471nm.\boxed{\lambda_{\max}=\frac{1240}{13.6(1/4-1/9)}=656.471\,\mathrm{nm}}.

(b) Interpretation and further calculation

n→∞:Eγ→3.40eV⇒λlimit=364.706nm.n\to\infty:\quad E_\gamma\to3.40\,\mathrm{eV}\Rightarrow\boxed{\lambda_{\mathrm{limit}}=364.706\,\mathrm{nm}}.364.706nm<λ≤656.471nmfor finite bound-state indices.\boxed{364.706\,\mathrm{nm}<\lambda\le656.471\,\mathrm{nm}}\quad\text{for finite bound-state indices}.

Physical check

The series limit corresponds to the continuum threshold as the initial binding tends to zero. Finite upper levels produce wavelengths strictly above the limit.