Q.A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Bohr Model Energy Levels
The Intuition: Why Can't an Electron Just Sit Anywhere?
Imagine you're rolling a marble on a staircase. The marble can rest on any step — step 1, step 2, step 3 — but it can never float halfway between two steps. The staircase forces the marble into specific, fixed positions.
That's the core idea of the Bohr model. Before Bohr, physicists thought electrons orbited the nucleus like planets around the sun — they could be at any distance, any energy. But experiments showed something strange: atoms only emit or absorb light at very specific colours (wavelengths), not a continuous rainbow. That meant electrons could only have certain, fixed energies — like the steps of a staircase.
Bohr's genius was to say: an electron in an atom cannot have any arbitrary energy. It can only occupy certain "allowed" energy levels. When it jumps from one level to another, it either absorbs or emits a photon of light whose energy exactly matches the difference between those levels.
The Precise Statement
In the Bohr model of the hydrogen atom (and hydrogen-like ions with one electron), the electron moves in circular orbits around the nucleus. But only those orbits are allowed where the electron's angular momentum is an integer multiple of 2πh (where h is Planck's constant).
This quantisation condition leads to a simple formula for the energy of the electron in the n-th orbit:
En=−n213.6 eV
Here:
- n is the principal quantum number — a positive integer (n=1,2,3,…)
- En is the energy of the electron in that level (in electronvolts)
- The negative sign means the electron is bound to the nucleus — you need to add energy to free it
The lowest energy state (n=1) is called the ground state. Its energy is −13.6 eV. The higher states (n=2,3,4,…) are excited states — they have less negative (higher) energies.
As n increases, the energy levels get closer together. At n=∞, the energy becomes 0 eV — the electron is completely free from the atom (ionisation).
What This Explains
When an electron jumps from a higher level (ni) to a lower level (nf), it emits a photon of energy:
ΔE=Enf−Eni=13.6(nf21−ni21) eV
This single formula predicts all the spectral lines of hydrogen — the Lyman series (jumps to n=1), Balmer series (to n=2), Paschen series (to n=3), and so on. Each series corresponds to a different "final step" on the staircase. …
Why this formula?
Why the Bohr Model Gives Those Energy Levels
The Bohr model is a beautiful piece of physics because it takes a simple, almost desperate idea — "electrons only exist in certain orbits" — and derives the entire hydrogen spectrum from it. The key is that Bohr didn't just assume the energy levels; he forced them to be consistent with classical physics in one specific way, then broke with it in another.
The Two Non-Negotiable Pieces
First, the electron moves in a circle around the proton. That's pure classical mechanics: the Coulomb attraction provides the centripetal force.
4πε01r2e2=rmv2
This gives you a relation between speed v and radius r:
v2=4πε0mre2
Second, the total energy of the electron is the sum of its kinetic and potential energies. Potential energy for a Coulomb force is −4πε01re2 (negative because the force is attractive, and we set zero at infinity).
E=21mv2−4πε01re2
Substitute v2 from above:
E=21(4πε0re2)−4πε0re2=−214πε0re2
So far, nothing is quantised. Any radius r gives a valid classical orbit, and the energy just follows from that radius. The problem is that a classical electron in a curved path radiates energy and spirals into the nucleus — atoms should collapse. Bohr needed a rule to pick out stable orbits.
The Quantisation Condition
Bohr's revolutionary step was to postulate that the angular momentum of the electron is quantised in units of ℏ=h/2π:
mvr=nℏ,n=1,2,3,…
Why this particular rule? Bohr later said it was the simplest way to get the right answer. But there's a deeper physical motivation: if you think of the electron as a wave (de Broglie's idea, which came a decade later), the condition that a standing wave fits exactly around the circumference 2πr=nλ gives mvr=nℏ directly. So the quantisation condition is really a wave condition imposed on a particle picture.
The angular momentum quantisation is the only non-classical assumption in the Bohr model. Everything else follows from classical mechanics and electromagnetism.
Deriving the Allowed Radii and Energies
From mvr=nℏ, we get v=nℏ/(mr). Substitute this into the centripetal force equation:
4πε01r2e2=rm(mrnℏ)2=mr3n2ℏ2
Solve for r:
rn=me24πε0ℏ2n2
The constant in front is the Bohr radius a0≈0.529A˚. So the radii are rn=a0n2.
Now plug rn back into the energy expression E=−214πε0re2:
En=−214πε0e2⋅a0n21=−214πε0a0e2⋅n21
Substitute a0=me24πε0ℏ2:
En=−214πε0e2⋅4πε0ℏ2me2⋅n21=−8ε02h2me4⋅n21
En=−n213.6 eV …
Concept: Bohr Model Energy Levels — a 12.5 eV electron beam can only excite hydrogen atoms (ground state E1=−13.6 eV) to levels whose energy gap it can fully supply.
Reasoning:
- ΔE1→2=10.2 eV and ΔE1→3≈12.09 eV are both ≤12.5 eV — reachable. ΔE1→4=12.75 eV is not — n=4 is out of reach.
- So atoms get excited to n=2 or n=3 only.
- Downward transitions: 3→1, 2→1 (both Lyman series, UV), and 3→2 (Balmer series, the Hα red line) — three lines total. …
A 12.5 eV electron beam can supply enough energy to excite ground-state hydrogen atoms up to n=3 (which needs 12.09 eV) but not to n=4 (which needs 12.75 eV). The excited atoms then de-excite via three possible downward jumps — 3→2, 3→1, and 2→1 — emitting three distinct wavelengths: two in the Lyman series (102.6 nm and 121.5 nm) and one in the Balmer series (656.3 nm).
Why the Bohr model is the right tool
At room temperature, essentially all hydrogen atoms sit in the ground state (n=1). When the 12.5 eV electron beam collides with these atoms, a beam electron can transfer some of its kinetic energy to the bound atomic electron — but only in the exact discrete amounts that match the gap between two Bohr energy levels. If the beam energy is short of the next gap, no excitation happens (the collision is elastic). So the first job is to find which excited levels are actually reachable.
The hydrogen energy levels are:
En=−n213.6 eV
Step 1 — Which levels can 12.5 eV reach?
Ground state: E1=−13.6 eV.
ΔE1→2=E2−E1=(−413.6)−(−13.6)=10.2 eV
ΔE1→3=E3−E1=(−913.6)−(−13.6)≈12.09 eV
ΔE1→4=E4−E1=(−1613.6)−(−13.6)=12.75 eV
The 12.5 eV beam can supply 10.2 eV (reaching n=2) and 12.09 eV (reaching n=3), but not 12.75 eV (reaching n=4). So the highest level any atom can be excited to is n=3; the beam electron that caused a 1→3 excitation keeps the leftover 12.5−12.09=0.41 eV as its own kinetic energy — the atom only ever absorbs a whole discrete quantum, never a fraction.
Step 2 — List every allowed downward transition
Some atoms end up excited to n=2, some to n=3. Each can then fall to any lower level:
- From n=3: 3→2 and 3→1
- From n=2: 2→1
That gives three distinct spectral lines in total (an atom excited to n=3 may cascade 3→2→1, emitting two photons, or jump directly 3→1, emitting one — across many atoms, all three lines appear).
Step 3 — Compute each wavelength
Using the Rydberg relation λ1=R(nf21−ni21) with R=1.097×107 m−1:
3→1 (Lyman):
λ1=R(1−91)=98R⇒λ=8R9≈1.026×10−7 m=102.6 nm
2→1 (Lyman): …
Method: Energy-Level Transition Analysis Using the Bohr Model
The core idea is that an incoming electron can transfer only discrete amounts of energy to a hydrogen atom — exactly the gaps between Bohr energy levels. Once the atom is excited, it de-excites in steps, emitting photons whose wavelengths correspond to the Lyman, Balmer, or Paschen series.
Step 1: Write the Bohr energy levels for hydrogen
For hydrogen (Z=1), the energy of the n-th orbit is:
En=−n213.6 eV
So the first few levels are:
| n | En (eV) |
|---|---|
| 1 | −13.6 |
| 2 | −3.4 |
| 3 | −1.51 |
| 4 | −0.85 |
| 5 | −0.54 |
Step 2: Find the maximum excitation possible
The incoming electron has 12.5 eV of kinetic energy. A ground-state hydrogen atom (E1=−13.6 eV) can absorb at most this much energy. The highest energy level reachable is the one where:
En−E1≤12.5 eV
Check n=3: E3−E1=(−1.51)−(−13.6)=12.09 eV — this is less than 12.5 eV, so n=3 is reachable.
Check n=4: E4−E1=(−0.85)−(−13.6)=12.75 eV — this exceeds 12.5 eV, so n=4 is not reachable.
A common mistake is to think the atom can absorb the full 12.5 eV and jump to n=4. But 12.75 eV is needed for n=4 — the extra 0.25 eV cannot be absorbed, so the atom can only reach n=3.
Thus the atom can be excited to n=2 or n=3 only.
Step 3: List all possible downward transitions
From n=3:
- 3→2 (Balmer series)
- 3→1 (Lyman series) …
Students often lose marks on this exact problem because they rush past two key ideas: the energy levels of hydrogen and the meaning of "room temperature". Let me walk through the mistakes one by one.
Mistake 1: Forgetting that hydrogen at room temperature is in the ground state
At room temperature, almost all hydrogen atoms are in n=1. Students sometimes assume atoms are already excited, or they try to use the Maxwell-Boltzmann distribution unnecessarily. The exam expects you to know: room temperature means kBT≈0.025 eV, far too small to excite hydrogen from n=1 to n=2 (which needs 10.2 eV). So every atom starts at n=1.
How to avoid: Always check the initial state. If the problem says "gaseous hydrogen at room temperature", the answer is n=1 for all atoms. Write it down explicitly before doing anything else.
Mistake 2: Using the beam energy as the exact excitation energy
The electron beam has 12.5 eV of kinetic energy. Students often think this means the atom absorbs exactly 12.5 eV and jumps to some level. But an electron can transfer any amount up to its full kinetic energy — it doesn't have to give all of it. The atom can absorb 10.2 eV (to reach n=2) or 12.09 eV (to reach n=3), and the leftover energy stays with the scattered electron.
How to avoid: List the excitation energies from n=1:
- n=1→2: 13.6−3.4=10.2 eV
- n=1→3: 13.6−1.51=12.09 eV
- n=1→4: 13.6−0.85=12.75 eV
Now compare with 12.5 eV. The atom can reach n=2 and n=3 (both below 12.5 eV), but not n=4 (needs 12.75 eV>12.5). So the maximum excited state is n=3.
A common trap: students see 12.5 eV and think "close to 12.75" and include n=4. But 12.5<12.75, so n=4 is impossible. Precision matters.
Mistake 3: Only listing one transition per atom
After excitation, the atom de-excites. A single atom excited to n=3 can return to ground via multiple paths: 3→2, 2→1, or directly 3→1. Each path emits a photon of a specific wavelength. Students sometimes list only the direct 3→1 transition, forgetting the cascade.
How to avoid: For each excited state reached, list all possible downward transitions. For n=3:
- 3→2 (Balmer series, visible)
- 2→1 (Lyman series, UV)
- 3→1 (Lyman series, UV)
For n=2 (also reached by some atoms):
- 2→1 (Lyman series)
So the emitted wavelengths come from three distinct transitions: 3→2, 2→1, and 3→1. That's three different wavelengths.
Mistake 4: Confusing series names with specific lines
Students sometimes say "Lyman series will be emitted" without specifying which lines. The question asks "What series of wavelengths?" — the answer is: two lines from the Lyman series (2→1 and 3→1) and one line from the Balmer series (3→2). Don't just name the series; mention how many lines in each. …
Showing the 12 most recent of 51 on this concept.
- AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQQ.The wave number of a spectral line in Brackett series of hydrogen atom is 4009R. The electron has transmitted from the orbit having quantum number (A) 5 (B) 6 (C) 4 (D) 7
›Reveal solutionSolution
This tests the Rydberg formula for the Brackett series (transitions ending at n=4) to find the initial orbit's quantum number from the given wave number. Answer: n=5.
Concept and Intuition
Each spectral series of hydrogen corresponds to electron transitions ending on a fixed lower energy level: Lyman (nf=1), Balmer (nf=2), Paschen (nf=3), Brackett (nf=4), Pfund (nf=5). The wave number of the emitted photon is given by the Rydberg formula, νˉ=R(nf21−ni21), where ni>nf is the initial (higher) orbit the electron falls from.
Step-by-Step Solution
- For Brackett series, nf=4, so νˉ=R(161−ni21).
- Given νˉ=4009R, so 161−ni21=4009.
- Convert 161 to a denominator of 400: 161=40025. …
- AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQQ.The atomic number (Z) of a hydrogen like atom whose shortest wavelength of Brackett Series is same as the shortest wavelength of Balmer series of hydrogen atom is (A) Z=1 (B) Z=2 (C) Z=3 (D) Z=4
›Reveal solutionSolution
Matching the series-limit wavelengths of hydrogen's Balmer series and a hydrogen-like atom's Brackett series gives Z=2 (i.e. singly-ionized helium).
Concept and Intuition
The Rydberg formula for a hydrogen-like ion of atomic number Z is λ1=RZ2(n121−n221). The shortest wavelength of any series (its "series limit") corresponds to the electron falling from n2=∞ down to the series' lower level n1, since that transition carries the largest possible energy jump within the series:
λmin1=RZ2(n121).
For hydrogen's Balmer series, n1=2 and Z=1. For the unknown hydrogen-like atom's Brackett series, n1=4 with its own Z. Setting the two series-limit wavelengths equal lets us solve directly for Z.
Step-by-Step Solution
- Hydrogen Balmer series limit (n1=2, Z=1): λBalmer1=R(1)2⋅221=4R.
- Hydrogen-like Brackett series limit (n1=4, atomic number Z): λBrackett1=RZ2⋅421=16RZ2. …
- AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQQ.The minimum frequency of light which can ionize a hydrogen atom is approximately (A) 3.3×1015 Hz (B) 5×1015 Hz (C) 91.1 Hz (D) 30.5 Hz
›Reveal solutionSolution
Ionizing a ground-state hydrogen atom requires supplying at least its binding energy, 13.6 eV. Converting this energy to a photon frequency via E=hf gives about 3.3×1015 Hz.
Concept and Intuition
The electron in a hydrogen atom's ground state is bound with energy −13.6 eV. To ionize it (remove it to infinity with zero kinetic energy, the minimum condition), a photon must supply exactly this much energy. The minimum frequency is the one where the photon energy just equals the ionization energy.
Step-by-Step Solution
- Ionization energy of hydrogen: E=13.6 eV =13.6×1.6×10−19 J=2.176×10−18 J.
- Using E=hf: f=hE=6.63×10−342.176×10−18. …
- AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQQ.In hydrogen spectrum, the ratio of the shortest wavelength of Balmer series and longest wavelength of pfund series is (A) 36:125 (B) 11:225 (C) 11:125 (D) 36:11
›Reveal solutionSolution
Comparing the shortest Balmer wavelength (series limit) with the longest Pfund wavelength (n=6→5) gives a ratio of 11:225.
Concept and Intuition
For hydrogen, λ1=R(n121−n221). A series's shortest wavelength corresponds to the transition from n2=∞ (series limit, the maximum possible energy gap for that series). A series's longest wavelength corresponds to the transition from the very next higher level (the minimum energy gap).
Step-by-Step Solution
- Balmer series (n1=2) shortest λ: transition from n2=∞. λBalmer1=R(41−0)=4R⇒λBalmer=R4. …
- AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQQ.The energy required to excite an electron from 1st to 3rd Bohr orbit in Li2+ ion is (A) 212.1 eV (B) 136.3 eV (C) 108.8 eV (D) 122.4 eV
›Reveal solutionSolution
Bohr model energy levels scale as Z2/n2; for a hydrogen-like ion, the excitation energy between two orbits is just the difference of these level energies.
Concept and Intuition
Li²⁺ is hydrogen-like (one electron, nuclear charge Z=3), so its orbit energies follow the same Bohr formula as hydrogen but scaled by Z2: En=−13.6n2Z2 eV. Because the charge is 3 times that of hydrogen's, all the level spacings are 9× larger — a scaling factor that's often forgotten.
Step-by-Step Solution
- En=−13.6×n232=−n2122.4 eV.
- E1=−122.4 eV.
- E3=−122.4/9=−13.6 eV. …
- AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQQ.In hydrogen atom, if the kinetic energy of an electron in an orbit having angular momentum π2h is E, then the potential energy of the electron in the first orbit of hydrogen atom is (h – Planck's constant) (A) −32E (B) −8E (C) −4E (D) −16E
›Reveal solutionSolution
This uses Bohr's quantization of angular momentum to identify the orbit number, then the 1/n2 scaling of kinetic energy and the virial-theorem relation between KE and PE.
Concept and Intuition
In the Bohr model, Ln=2πnh identifies the orbit. Kinetic energy scales as KEn∝Z2/n2, and for the Coulomb potential the virial theorem gives PEn=−2KEn (total energy En=KEn+PEn=−KEn).
Step-by-Step Solution
- Given L=π2h; setting 2πnh=π2h gives n=4.
- So KE4=E (as given). Since KEn∝n21: KE4KE1=1242=16⇒KE1=16E.
- By the virial theorem for the Coulomb (1/r) potential: PEn=−2KEn. …
- AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQQ.In the pfund series of hydrogen spectrum, the wavelength of first spectral line is (Rydberg constant =1.097×107m−1) (A) 8547 nm (B) 6574 nm (C) 3729 nm (D) 7458 nm
›Reveal solutionSolution
The Pfund series (transitions to n=5) has its first line at n=6→n=5; the Rydberg formula gives λ≈7458 nm, in the infrared.
Concept and Intuition
The hydrogen spectral series are named by the lower energy level n1 the electron falls to: Lyman (n1=1), Balmer (n1=2), Paschen (n1=3), Brackett (n1=4), Pfund (n1=5). The "first line" (longest wavelength, smallest energy gap) of any series is the transition from the very next higher level, n2=n1+1.
Step-by-Step Solution
- Pfund series: n1=5. First line: n2=6→n1=5.
- Rydberg formula: λ1=R(n121−n221)=R(251−361).
- 251−361=90036−25=90011. …
- AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQQ.The difference between the frequencies of the first and second Lyman lines of hydrogen atom is (R - Rydberg constant and c - speed of light in vacuum) (A) 289Rc (B) 127Rc (C) 83Rc (D) 365Rc
›Reveal solutionSolution
The Lyman series frequencies follow ν=Rc(1−1/n2) for transitions to n=1; the difference between the first two lines is 365Rc.
Concept and Intuition
The Lyman series corresponds to transitions from higher levels n=2,3,4,… down to the ground state n=1. Using the Rydberg formula for wavenumber, νˉ=R(121−n21), and multiplying by c gives frequency ν=Rc(1−n21). The "first" Lyman line is the n=2→1 transition and the "second" is n=3→1.
Step-by-Step Solution
- First Lyman line (n=2→1): ν1=Rc(1−41)=43Rc.
- Second Lyman line (n=3→1): ν2=Rc(1−91)=98Rc.
- Difference: ν2−ν1=98Rc−43Rc. …
- AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQQ.The ratio of the wavelengths of the first Lyman line and the second Balmer line of hydrogen atom is (A) 3 : 4 (B) 1 : 4 (C) 2 : 3 (D) 1 : 3
›Reveal solutionSolution
This tests the Rydberg formula applied to specific named spectral lines. The ratio of wavelengths (Lyman-first : Balmer-second) is 1 : 4.
Concept and Intuition
Each spectral series of hydrogen corresponds to transitions ending on a fixed lower level (n=1 for Lyman, n=2 for Balmer), and the "first", "second", etc. lines within a series correspond to the electron starting from successively higher levels. The Rydberg formula converts each specific transition into a wavenumber 1/λ, and comparing two lines is just comparing these wavenumbers.
Step-by-Step Solution
- Rydberg formula: λ1=R(n121−n221).
- First Lyman line: transition n=2→n=1:
λL11=R(121−221)=R(1−41)=43R
- Second Balmer line: transition n=4→n=2: λB21=R(221−421)=R(41−161)=163R …
- AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQQ.The difference between the frequencies of second and first Paschen lines of hydrogen atom is (R - Rydberg constant and c - speed of light in vacuum) (A) 169Rc (B) 2516Rc (C) 4009Rc (D) 2003Rc
›Reveal solutionSolution
The Paschen series lines all terminate on n=3; working out the frequencies of the 4→3 and 5→3 transitions and subtracting gives 4009Rc.
Concept and Intuition
The hydrogen spectral series are named by the final (lower) energy level the electron falls to: Lyman (nf=1), Balmer (nf=2), Paschen (nf=3), and so on. Within a series, lines are numbered by increasing initial level: the first Paschen line is the transition from the next level up (n=4→3), the second Paschen line is from one level further (n=5→3), etc. The Rydberg formula gives the frequency of any transition as f=Rc(nf21−ni21), so once we know which two transitions are meant, this is purely an algebra exercise in fractions.
Step-by-Step Solution
- First Paschen line (4→3): f1=Rc(91−161)=Rc⋅14416−9=1447Rc.
- Second Paschen line (5→3): f2=Rc(91−251)=Rc⋅22525−9=22516Rc.
- Find a common denominator for 22516 and 1447: since 225=9×25 and 144=16×9, the LCM is 3600.
- 22516=3600256, and 1447=3600175. …
- AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQQ.Of the following, Bohr's atomic model is applicable to (A) explain relative intensities of spectral lines emitted by hydrogen atoms (B) helium atom (C) lithium atom (D) hydrogenic atoms
›Reveal solutionSolution
Bohr's model is a single-electron model; it applies to hydrogen and hydrogen-like (hydrogenic) ions, not to multi-electron neutral atoms, and it does not predict spectral line intensities.
Concept and Intuition
Bohr's postulates were built specifically around one electron orbiting a nucleus of charge +Ze under a pure Coulomb force, with angular momentum quantized as nℏ. This works cleanly whenever there is exactly one electron in the system — hydrogen (Z=1) and "hydrogenic" ions like He+ (Z=2, one electron), Li2+ (Z=3, one electron), etc. As soon as more than one electron is present (neutral helium, neutral lithium), electron-electron repulsion and screening effects make the simple circular-orbit picture invalid, so Bohr's model does not extend to those. Also, Bohr's theory predicts energy levels and hence spectral line frequencies, but says nothing about intensities of the lines, which require quantum-mechanical transition probabilities beyond the scope of the model.
Step-by-Step Solution
- Recall Bohr's derivation assumes a single electron orbiting a central charge — the model's validity is tied to being "one-electron" in nature.
- Option (A), relative intensities of spectral lines, is outside Bohr's theory's predictive power (it only gives energy levels/frequencies, not transition probabilities) — reject. …
- AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQQ.When an electron beam of energy 10.2 eV is used to excite hydrogen gas, then the possible spectral line is (A) first Balmer line (B) first Lyman line (C) second Balmer line (D) second Lyman line
›Reveal solutionSolution
An electron beam of exactly 10.2eV can only excite a hydrogen atom from n=1 to n=2 (since higher-level excitations need more energy). The de-excitation n=2→n=1 produces the first Lyman line.
Concept and Intuition
Bohr's hydrogen energy levels are En=−n213.6eV. Electron-impact excitation can only promote the atom to a level if the beam energy equals (or in an inelastic collision, at least equals) the energy gap to that level — but if there's exactly enough energy for one specific transition and not the next, only that transition occurs. Here 10.2eV exactly matches E1→E2, so the atom is excited only up to n=2. When it relaxes back to the ground state (n=2→n=1), it emits a Lyman-series photon — specifically the first (lowest energy) line of that series, called Lyman-α.
Step-by-Step Solution
- E1=−13.6eV (ground state), E2=−413.6=−3.4eV.
- Energy gap E2−E1=−3.4−(−13.6)=10.2eV — exactly the beam energy given.
- Check E3=−913.6≈−1.51eV; gap E3−E1≈12.09eV, which is more than 10.2eV — so n=3 cannot be reached.
- So the atom is excited only to n=2. …
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