Some Properties of Atoms — Question 9

Section 40-1 · Physics pages 6588–6589

Question 9 A spectroscopic energy network distinguishes direct and cascade photons An atom has nondegenerate levels at 0, 2.4 and 5.9 eV. Assume all downward transitions are allowed and neglect recoil. A preparation puts each of N atoms in the top level. A fraction b decays directly to the ground; the rest follow the two-step cascade. n = 0 n = 1 n = 2 E Schematic; labels, not drawing lengths, specify the data. (a) Find all three photon energies. (b) Find the mean number of photons and total radiated energy per initially excited atom.
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Question 9 - Solution A spectroscopic energy network distinguishes direct and cascade photons n = 0 n = 1 n = 2 E (a) Method and calculation E20 = 5.9 −0 = 5.9 eV . E21 = 3.5 eV, E10 = 2.4 eV . (b) Interpretation and further calculation ⟨nγ⟩= b + 2(1 −b) = 2 −b . ⟨Etotal γ ⟩= b(5.9) + (1 −b)(3.5 + 2.4) = 5.9 eV . Physical check Branching changes the photon count and spectrum, but not the total energy released between the same initial and final levels.

Question 9

A spectroscopic energy network distinguishes direct and cascade photons

An atom has nondegenerate levels at 00, 2.42.4 and 5.9eV5.9\,\mathrm{eV}. Assume all downward transitions are allowed and neglect recoil. A preparation puts each of N atoms in the top level. A fraction b decays directly to the ground; the rest follow the two-step cascade.

Diagram shown in the original SVG worksheet.

  1. Find all three photon energies.

  2. Find the mean number of photons and total radiated energy per initially excited atom.

Question 9 - Solution

A spectroscopic energy network distinguishes direct and cascade photons

Diagram shown in the original SVG worksheet.

(a) Method and calculation

E20=5.9−0=5.9eV.E_{20}=5.9-0=\boxed{5.9\,\mathrm{eV}}.E21=3.5eV,E10=2.4eV.\boxed{E_{21}=3.5\,\mathrm{eV},\quad E_{10}=2.4\,\mathrm{eV}}.

(b) Interpretation and further calculation

⟨nγ⟩=b+2(1−b)=2−b.\langle n_\gamma\rangle=b+2(1-b)=\boxed{2-b}.⟨Eγtotal⟩=b(5.9)+(1−b)(3.5+2.4)=5.9eV.\langle E_\gamma^{\mathrm{total}}\rangle=b(5.9)+(1-b)(3.5+2.4)=\boxed{5.9\,\mathrm{eV}}.

Physical check

Branching changes the photon count and spectrum, but not the total energy released between the same initial and final levels.