The answer in one line
For the (nucleus charge +e) with the electron jumping from orbit n + k = down to n = : -
The photon frequency is ω = . It lies between ω_n = and ω_(n+k) = . The ratios are ω/ω_n = and ω/ω_(n+k) = .
For neighbouring orbits (k = 1) the proof is exact for every n. The gap 1 − ω/ω_n equals (3n + 2)/(2(n + 1)²), which is about 3/(2n) and goes to 0 as n grows. So ω → ω_n.
Problem Statement
Show that the frequency ω of the photon sent out when an electron jumps between neighbouring circular orbits of a hydrogen-like ion obeys ω_n > ω > ω_(n+1). Here ω_n and ω_(n+1) are the frequencies of revolution of the electron on those two orbits. Also show that as n → ∞ the photon frequency ω → ω_n.
A hydrogen-like ion is an atom with one electron only, but a nucleus of charge +Ze. Z = 1 is hydrogen, Z = 2 is He⁺, Z = 3 is Li²⁺, and so on. Your values now: (Z = ), of mass kg, jump from n + k = to n = , and the nucleus is a .
We assume Bohr's two rules. Rule 1 says only some circular orbits are allowed: those where the angular momentum m v r equals n ħ, with n = 1, 2, 3 and so on. Rule 2 says the electron sends out one photon when it jumps, and the photon carries the energy difference between the two orbits. We use rule 1 in step 1 and rule 2 in step 2. This is the Bohr model, a quantum idea added to Newton's laws.
We assume the nucleus does not move (it is infinitely heavy), as the book does. The centre-of-mass tab shows what a movable nucleus changes.
We assume the electron is slow compared with light, so Newton's laws are fine. Its speed is v/c = Zα/n = now, where α = 1/137 is the fine-structure constant. The page warns above 0.1 c.
We assume the photon takes the whole energy difference. The nucleus recoils a little, but that takes only a fraction of the photon energy for your centre, so we ignore it.
For the proof we use neighbouring orbits, k = 1, as the problem says. The slider k lets you try bigger jumps, and then the inequality can fail.
The big idea in plain words
An electron going round a nucleus has a "turning rate", called the frequency of revolution. Orbit n has one rate, and orbit n + 1 has a slower one. Think of a runner on an inner lane and another on an outer lane.
When the electron jumps down one orbit, it sends out light. The light also has a frequency. The claim is that this light frequency sits between the two turning rates. It is faster than the outer orbit and slower than the inner orbit.
This is a clue found by Bohr. For big orbits the light frequency should match the turning rate, because that is what old physics says: an orbiting charge radiates at the frequency it turns. Here the two rates get closer and closer as n grows, so the squeezed light frequency must approach them both. This link between old and new physics is called the correspondence principle.
The plan has four steps. (1) Find the turning rate ω_n from Newton's law and Bohr's first rule. (2) Find the photon frequency from the energy difference. (3) Compare them by simple algebra. (4) Let n grow large and see what the ratio does.
Boxed derivation 1: the turning rate of orbit n
We want the radius r_n, speed v_n and turning rate ω_n of orbit n. We use Newton's law for a circle and Bohr's first rule. The symbols: m is the electron mass (9.109 × 10⁻³¹ kg), e = 1.602 × 10⁻¹⁹ C is the electron charge, ε₀ = 8.854 × 10⁻¹² F/m is the electric constant, ħ = 1.0546 × 10⁻³⁴ J s is Planck's constant divided by 2π, and k = e²/(4π ε₀) = 2.307 × 10⁻²⁸ J m is the strength of the electric pull.
Step 1. The pull of the nucleus, Z k/r², supplies the force that bends the path into a circle: m v²/r = Z k/r². So m v² = Z k/r.
Step 2. Bohr's first rule is m v r = n ħ, so v = n ħ/(m r).
Step 3. Put v into step 1: m (n ħ/(m r))² = Z k/r. This gives n² ħ²/(m r²) = Z k/r, so r_n = n² ħ²/(m Z k) = n² a₀/Z, where a₀ = ħ²/(m k) = 0.529 Å = 52.9 pm is the Bohr radius. Now r_n = and r_(n+k) = .
Step 4. The speed is v_n = n ħ/(m r_n) = Z k/(n ħ). Numbers: v_n/c = Z α/n = .
Step 5. The turning rate is ω_n = v_n/r_n = [Z k/(n ħ)] × [m Z k/(n² ħ²)] = Z² m k²/(ħ³ n³). Define ω₁ = m k²/ħ³ = (the turning rate of the first orbit of hydrogen). Then ω_n = Z² ω₁/n³. For your ion, Z² ω₁ = , so ω_n = and ω_(n+k) = . One turn takes T = 2π/ω_n = s.
Boxed derivation 2: the energy of orbit n and the photon frequency
We want the energy of each orbit, because Bohr's second rule needs the energy difference.
Step 1. The motion energy is ½ m v² = Z k/(2r) (from m v² = Z k/r). The electric energy is −Z k/r. The total is E = −Z k/(2r).
Step 2. Put r_n = n² ħ²/(m Z k): E_n = −Z² m k²/(2 ħ² n²) = −13.6 eV × Z²/n². Now E_n = − and E_(n+k) = −. (The value 13.6 eV is the Rydberg energy, m k²/(2ħ²).)
Step 3. Bohr's second rule: when the electron falls from orbit n + k to orbit n, the photon takes ħω = E_(n+k) − E_n. So ω = (Z² m k²/(2 ħ³)) (1/n² − 1/(n+k)²) = (Z² ω₁/2)(1/n² − 1/(n+k)²).
Step 4. For k = 1, 1/n² − 1/(n+1)² = ((n+1)² − n²)/(n²(n+1)²) = (2n + 1)/(n²(n+1)²). So ω = (Z² ω₁/2)(2n + 1)/(n²(n+1)²).
Step 5. Numbers: ω = . The photon energy is ħω = , its frequency is f = ω/2π = Hz, and its wavelength is λ = 2πc/ω = . Check with hc = 1239.84 eV nm: λ = 1239.84/( eV) = nm.
Boxed derivation 3: the proof that ω_n > ω > ω_(n+1)
We want to show that ω sits between the two turning rates, for every n = 1, 2, 3 and so on. We show it two ways.
Route A: the book's way. Divide everything by Z² ω₁ and write the claim as 1/n³ > ½ (1/n² − 1/(n+1)²) > 1/(n+1)³. Multiply every part by n²(n+1)², which is positive, so the signs stay the same. The left part becomes (n+1)²/n. The middle becomes ½((n+1)² − n²) = (2n+1)/2. The right part becomes n²/(n+1). So we must show (n+1)²/n > (2n+1)/2 > n²/(n+1).
Now (n+1)²/n = n + 2 + 1/n. And (2n+1)/2 = n + ½. And n²/(n+1) = n − 1 + 1/(n+1), because n² = (n+1)(n−1) + 1. So the claim is n + 2 + 1/n > n + ½ > n − 1 + 1/(n+1). Subtract n from each part: 2 + 1/n > ½ > −1 + 1/(n+1). The left one is true because 2 + 1/n > 2. The right one is true because 1/(n+1) ≤ ½, so −1 + 1/(n+1) ≤ −½ < ½. Both hold for every n ≥ 1. The claim is proved.
Route B: the ratios. From derivations 1 and 2, ω/ω_n = (n³/2)(2n+1)/(n²(n+1)²) = n(2n+1)/(2(n+1)²). Subtract from 1: 1 − ω/ω_n = [2(n+1)² − n(2n+1)]/(2(n+1)²) = (3n + 2)/(2(n+1)²), which is positive for every n. So ω < ω_n.
In the same way ω/ω_(n+1) = ((n+1)³/2)(2n+1)/(n²(n+1)²) = (n+1)(2n+1)/(2n²). Then ω/ω_(n+1) − 1 = [(n+1)(2n+1) − 2n²]/(2n²) = (3n + 1)/(2n²), which is positive. So ω > ω_(n+1).
Numbers now (k = ): ω/ω_n = and ω/ω_(n+k) = . The gap 1 − ω/ω_n = , and 3/(2n) = . The verdict for your values: .
The page computes the gap from the exact formula (3n + 2)/(2(n+1)²) and not from a subtraction of two almost equal numbers. A subtraction would lose all its digits for large n.
Boxed derivation 4: the limit n → ∞
We want to show ω → ω_n for large n. We use the gap from derivation 3.
Step 1. 1 − ω/ω_n = (3n + 2)/(2(n+1)²). Divide top and bottom by n²: = (3/n + 2/n²)/(2(1 + 1/n)²).
Step 2. As n → ∞, 1/n → 0 and 1/n² → 0, so the top → 0 and the bottom → 2. The gap → 0. So ω/ω_n → 1, which means ω → ω_n.
Step 3. The book shows the same thing another way: ω_n/ω_(n+1) = ((n+1)/n)³ = (1 + 1/n)³ ≈ 1 + 3/n → 1. The two turning rates squeeze together, and ω is trapped between them, so it must approach both.
Step 4. A bonus. For k = 1, ω is even closer to the turning rate of the middle orbit n + ½: ω_(n+½) = Z² ω₁/(n + ½)³ = for your values. The ratio ω/ω_(n+½) = (n + ½)⁴/(n²(n+1)²) = (1 + 1/(4n(n+1)))², and this is now, which is very close to 1.
Step 5. Why bigger jumps can fail. For a jump of k, ω/ω_n = n k (2n + k)/(2(n + k)²). For large n this goes to k. So a jump of 2 gives ω ≈ 2 ω_n, which is ABOVE ω_n. Then ω_n > ω is false. Only k = 1 keeps ω below ω_n for all n. Try the preset "Fails: jump of 2 from n = 10".
The solution viewed in each frame
Lab frame: the nucleus is at rest and the electron circles it
This is the frame used above. The nucleus is fixed, so all the motion belongs to the electron. We found ω_n = , ω = and ω_(n+k) = . The ratios are and , and the verdict is .
Centre-of-mass frame: the origin moves with the centre of mass
We want the frequencies when the nucleus can move too. Let the electron mass be m and the nucleus mass be M. Both bodies circle their common centre of mass.
Step 1: total momentum is zero. With the origin at the centre of mass, m v₁ + M v₂ = 0. So the two momenta are equal and opposite. The frame has no outside force, so no pseudo-force is needed.
Step 2: the separation. Let r be the distance between the two bodies. The electron is at M r/(m + M) = r from the centre of mass, and the nucleus is at the rest of r on the other side.
Step 3: Newton's law for the separation. The relative motion obeys μ r ω² = Z k/r², where μ = m M/(m + M) is the reduced mass. Bohr's first rule applies to the angular momentum of the relative motion, μ v r = n ħ. Following the same algebra as derivation 1, all the formulas stay the same with m replaced by μ.
Step 4: the result. Every frequency is multiplied by μ/m = M/(M + m) = . The ratios ω/ω_n and ω/ω_(n+k) do not contain m, so they do not change. For your centre () the CM photon frequency is the lab one times . For the book values (n = 5, Z = 1) this gives .
So the inequality is the same in this frame. A heavy nucleus makes the frames almost the same. For equal masses (positronium) all frequencies are halved. Try the preset "Flips: positronium (CM)".
Electron rest frame
We ride along with the electron at one instant. The frame is accelerating (the electron moves on a circle), so a pseudo-force −m a acts on the electron. It is equal and opposite to the pull of the nucleus, so the electron stays at rest. The nucleus now circles the electron, with the same turning rate ω_n. So the two turning rates are the same as in the lab frame.
The photon is not the same. The speed of the electron is v/c = . A photon of frequency ω in the nucleus frame appears with frequency between ω(1 − v/c) and ω(1 + v/c) in the electron frame, depending on the direction. These are and rad/s.
We need the shifted photon to stay below ω_n. That needs (1 + v/c) ω < ω_n, which means v/c < ω_n/ω − 1 = . Now v/c = . For hydrogen-like ions with Zα below 1 the condition holds for every n, so the inequality survives. The lab values are the answer.
What changes and what stays the same
The picture changes from frame to frame: the nucleus stays fixed, or both bodies circle the centre, or the electron stands still. The frequencies change by the factor M/(M + m) in the centre-of-mass frame. What stays the same is the pair of ratios ω/ω_n and ω/ω_(n+k), and so the verdict. The Doppler shift in the electron frame is tiny compared with the gap.
The solution by each method
Energy method
- We want to compare ω with the two turning rates. We use the energy levels E_n = −13.6 eV × Z²/n² and conservation of energy: the photon takes the lost energy.
- The photon energy is ħω = E_(n+k) − E_n = .
- The turning-rate energies are ħω_n = and ħω_(n+k) = . The photon energy lies between them.
- In symbols, ħω_n = 2 × 13.6 eV × Z²/n³ and ħω = 13.6 eV × Z² (2n + 1)/(n²(n+1)²) for k = 1. Their ratio is n(2n+1)/(2(n+1)²) = , below 1.
Momentum method (angular momentum)
- We want ω_n from angular momentum. Bohr's first rule says L = m v r = n ħ.
- For circular motion, ω = L/(m r²). Put L = n ħ and r_n = n² ħ²/(m Z k): ω_n = n ħ m² Z² k²/(m n⁴ ħ⁴) = Z² m k²/(ħ³ n³). This is the same result as before.
- From orbit n + 1 to orbit n, the angular momentum drops by ħ. A photon can carry away ħ of angular momentum, so a one-orbit jump fits. For k jumps the drop is k ħ, which needs more than one photon (the Bohr model hides this; this is a hint from later physics).
- The photon also carries a momentum ħω/c. The electron momentum is m v. The photon momentum divided by the electron momentum is , so the photon barely changes the electron motion, and the nucleus recoil is negligible. Then the energy method gives the inequality, as above.
Graph method
- Graph G2 plots ω/ω_n (amber) and ω/ω_(n+k) (cyan) against n. The dashed line is 1.
- For k = 1 the amber curve rises from 0.375 at n = 1 and stays below 1. The cyan curve falls from 3.0 at n = 1 and stays above 1. Both approach 1 from opposite sides. This shows the inequality for every n and also the limit.
- Graph G3 plots the gap 1 − ω/ω_n on log axes. It follows the purple dotted line 3/(2n) for large n and has slope −1. So the gap falls as 1/n.
- At your n = the gap is . If you pick k = 2 or more, red stripes show where the inequality fails.
Quantum-rule method
- We apply the quantum relations directly. Photon energy is E = ħω = h f, with h = 6.626 × 10⁻³⁴ J s. Bohr's second rule gives h f = E_(n+k) − E_n.
- The de Broglie wavelength of the electron is λ_dB = h/(m v). Bohr's first rule is the same as 2π r = n λ_dB: exactly n electron waves fit round the orbit.
- The photon has wavelength λ = hc/ΔE = 1239.84 eV nm/ eV = nm.
- The photon is not at the turning rate of either orbit, but between. For large n it is close to the turning rate of the middle orbit n + ½. This is the link between the quantum photon and the classical turning rate.
Level-diagram method
- The energy ladder (top right of the simulation) shows orbits n and n + k and the arrow between them. The arrow length is ΔE = .
- The turning-rate energies ħω_n = and ħω_(n+k) = are the two ends of a bracket.
- The zoomed bar below the ladder shows the bracket stretched to fill the width, with ω at of the way from ω_(n+k) to ω_n. A value between 0 and 1 means inside.
Check method
- Units: ω_n = Z² m k²/(ħ³ n³). Here k is in J m, so m k² is kg J² m² and ħ³ is J³ s³. Together: kg m²/(J s³) = (kg m²/s²)/(J s) = 1/s. So ω_n has units 1/s. Good.
- Limits: n → ∞ gives a ratio of 1. For n = 1, ω/ω₁ = 3/8 = 0.375 (the gap is 5/8), so the photon is well inside the bracket. Z cancels in the ratios, and so does the mass. So the inequality holds for any hydrogen-like ion and for muonic atoms alike.
- Sanity: for k = 2 and large n the ratio goes to 2, so the inequality is broken. This shows that "neighbouring orbits" matters.
- Size: for hydrogen with n = 5 and k = 1, ω = 2.526 × 10¹⁴ rad/s, a photon of 0.166 eV, in the infrared. The real hydrogen line for n = 6 to 5 (the Pfund series, first line) lies at 7.46 µm. They agree.
The ratio method (Route B) is fastest. The book's chain with n + 2 + 1/n is the most reliable in an exam, because every step is plain algebra with no division by a small number.
What each slider changes
| Control | Changes | How |
|---|---|---|
| Lower orbit n | ω_n, ω, ω_(n+k), both ratios, the gap, radii, speed, photon energy and wavelength, ladder, G1 to G3, ruler | ω_n goes as 1/n³; radius as n²; speed as 1/n; the gap as about 3/(2n) |
| Nuclear charge Z | All three frequencies, photon energy, radii, speed, element label | Frequencies go as Z²; radii as 1/Z; speed as Z. The ratios and the verdict do NOT change: Z² cancels in the division |
| Jump size k | Photon frequency, both ratios, verdict, upper orbit | For k = 1 the verdict is always true; for large k the inequality ω_n > ω breaks |
| Gap (given-gap mode) | Solves n | n = (3 + √(9 − 8g))/(4g) − 1 for k = 1. It has no effect when the mode is off |
| Particle | Mass, all frequencies, radii | Frequencies go as the mass; radii as 1/mass. The ratios do not change |
| Centre | CM frame frequencies | Multiplies all frequencies by M/(M + m); the ratios do not change |
| Zoom, frame | Picture scale; positions and arrows | Zoom has no effect on the answer; the frame changes the frequencies only through recoil |
Parameter Table
| Quantity | Symbol | Now |
|---|---|---|
| Ion | Z | , Z = |
| Orbits | n, n + k | , |
| Radii | r_n, r_(n+k) | , |
| Turning rate, orbit n | ω_n | |
| Photon frequency | ω | |
| Turning rate, orbit n + k | ω_(n+k) | |
| Ratios | ω/ω_n, ω/ω_(n+k) | , |
| Gap | 1 − ω/ω_n | |
| Photon energy, wavelength | ħω, λ | , |
| Electron speed | v/c |
Common Exam Traps
- Thinking the photon frequency equals the turning rate. It is between the two turning rates. It only approaches them for large n.
- Mixing up which orbit is faster. Orbit n is closer to the nucleus, so ω_n is larger than ω_(n+1).
- Forgetting the factor ½ in E_n = −Z k/(2r). The total energy is half of the electric energy.
- Using ω = ω_n − ω_(n+1) for the photon. The photon frequency comes from the energy difference, not from the difference of turning rates.
- Subtracting two almost equal numbers for large n and getting 0 on a calculator. Use the exact gap (3n + 2)/(2(n+1)²) instead.
- Using the inequality for a jump of 2 or more. It can fail, because ω is close to k ω_n for large n.
- Saying the answer depends on Z. Z cancels in the ratios. Z changes only the size of the frequencies.