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 15 on this concept.
- KEAM 2026Set eng-2026-04174 marksMCQQ.The ratio of the longest wavelength of the Lyman series to that of Ballmer series of the hydrogen spectrum is (A) 3:29 (B) 7:27 (C) 5:27 (D) 4:29 (E) 5:29
›Reveal solutionSolution
The longest wavelength in each series comes from the smallest energy jump (2→1 for Lyman, 3→2 for Balmer). Their ratio is 5:27.
Rydberg formula: λ1=R(n121−n221).
Longest Lyman (n1=1, n2=2):
λL1=R(1−41)=43R⇒λL=3R4.
Longest Balmer (n1=2, n2=3): …
- KEAM 2026Set eng-2026-04184 marksMCQQ.In hydrogen spectral series, the wave numbers of the first Lyman line and the first Balmer line are in the ratio (A) 1 : 2 (B) 2 : 1 (C) 27 : 5 (D) 5 : 27 (E) 16 : 1
›Reveal solutionSolution
The first Lyman line has νˉ=3R/4 and the first Balmer line νˉ=5R/36; their ratio is 27:5.
Wave number νˉ=R(n121−n221).
First Lyman line (n1=1, n2=2):
νˉL=R(1−41)=43R
First Balmer line (n1=2, n2=3): …
- KEAM 2026Set pha-2026-0418F4 marksMCQQ.The angular momentum and the energy of the electron in the second Bohr's orbit are respectively (A) πh and -13.6 eV (B) π2h and -1.5 eV (C) πh and -3.4 eV (D) 2πh and -3.4 eV (E) π2h and -3.4 eV
›Reveal solutionSolution
For n=2: L=nh/2π=h/π and E=−13.6/n2=−3.4eV. …
- KEAM 2026Set pha-2026-0419F4 marksMCQQ.The energy of the electron in third excited state of the hydrogen atom is (A) -0.54 eV (B) -1.51 eV (C) -13.6 eV (D) -3.4 eV (E) -0.85 eV
›Reveal solutionSolution
Third excited state is n=4; E4=−13.6/16=−0.85 eV.
Energy levels of hydrogen are En=−n213.6 eV. The states are: ground n=1, first excited n=2, second excited n=3, third excited n=4. So …
- KEAM 2026Set pha-2026-0420F4 marksMCQQ.Pick out the wrong statement about Bohr atom model: (A) Orbit of the electron is circular (B) Model is applicable only for single electron systems (C) Orbits of the electron are non-radiating (D) Angular momentum of electron in an orbit is quantized. (E) Model is applicable for many electron systems also
›Reveal solutionSolution
Bohr's model succeeds only for single-electron (hydrogen-like) systems; claiming it applies to many-electron atoms is false.
The Bohr model correctly assumes circular, non-radiating (stationary) orbits with quantized angular momentum L=nℏ, and it works well for single-electron (hydrogen-like) systems. It fails for many-electron atoms because it ignores electron-electron interactions a …
- KEAM 2025Set eng-2025-04234 marksMCQQ.The greatest wavelength of the radiation that will ionize unexcited hydrogen atom is (A) 1820 Å (B) 450 Å (C) 910 Å (D) 700 Å (E) 1400 Å
›Reveal solutionSolution
The longest wavelength that can ionize ground-state hydrogen is about 910 Angstrom.
Concept and Intuition
Ionizing an unexcited hydrogen atom needs at least 13.6 eV. The greatest wavelength (least energy) corresponds exactly to this threshold energy through E = hc/lambda.
Step-by-Step Solution
- Ionization energy E = 13.6 eV.
- lambda = hc/E = 1240 eV*nm / 13.6 eV = 91.2 nm. …
- KEAM 2025Set eng-2025-04254 marksMCQQ.In a hydrogen atom, the transition of an electron from the energy level, n2=∞ and n1=3 is the (A) shortest wavelength of Paschen series (B) longest wavelength of Paschen series (C) shortest wavelength of Balmer series (D) longest wavelength of Balmer series (E) shortest wavelength of Bracket series
›Reveal solutionSolution
n1=3 marks the Paschen series (infrared). The transition from n2=∞ releases the most energy, i.e. the series limit — the shortest wavelength of the Paschen series.
The lower level n1=3 identifies the Paschen series. The wavelength satisfies λ1=R(n121−n221). …
- KEAM 2025Set eng-2025-04264 marksMCQQ.In hydrogen spectrum, the shortest wavelength of Bracket series is produced during the transition between the states (A) n2=5 and n1=4 (B) n2=4 and n1=1 (C) n2=4 and n1=3 (D) n2=∞ and n1=4 (E) n2=4 and n1=2
›Reveal solutionSolution
Shortest wavelength = largest energy jump = the series limit, so for the Brackett series (n1=4) it is the ∞→4 transition.
The wavenumber of a hydrogen line is
λ1=R(n121−n221)
Shortest wavelength corresponds to maximum 1/λ, i.e. maximum photon energy, obtained when n2→∞. …
- KEAM 2025Set eng-2025-04284 marksMCQQ.Light of wavelength 5R36m is emitted by a hydrogen atom during the transition of electrons from the state (A) n=3 to n=2 (B) n=4 to n=1 (C) n=4 to n=2 (D) n=4 to n=3 (E) n=3 to n=1
›Reveal solutionSolution
λ1=365R matches R(221−321), the n=3→n=2 (Balmer) transition.
Given λ=5R36, the wavenumber is λ1=365R. Using the Rydberg formula
λ1=R(n121−n221),
try n1=2,n2=3: …
- KEAM 2025Set pha-2025-0424F4 marksMCQQ.Ionization potential of hydrogen atom is 13.6 eV. Hydrogen atom in the ground state initially is excited by monochromatic radiation of photon energy 12.75 eV. The number of spectral lines emitted by the hydrogen atom, according to Bohr's theory will be (A) 2 (B) 4 (C) 3 (D) 6 (E) 5
›Reveal solutionSolution
−13.6+12.75=−0.85 eV corresponds to n=4. Number of spectral lines =2n(n−1)=6.
The energy of the nth level is En=−n213.6 eV. Absorbing 12.75 eV from the ground state:
En=−13.6+12.75=−0.85 eV.
Solving,
−n213.6=−0.85⇒n2=16⇒n=4. …
- KEAM 2024Set eng-2024-06064 marksMCQQ.The energy required to excite the hydrogen atom from its first excited state to second excited state is (A) 12.09 eV (B) 1.89 eV (C) 10.2 eV (D) 3.40 eV (E) 1.51 eV
›Reveal solutionSolution
From n=2 to n=3: ΔE=13.6(41−91)=1.89 eV.
The first excited state is n=2 with E2=−3.40 eV, and the second excited state is n=3 with E3=−1.51 eV. …
- KEAM 2024Set eng-2024-06084 marksMCQQ.Bohr atom model is invalid for (A) Hydrogen atom (B) doubly ionized helium atom (C) deuteron atom (D) singly ionized helium atom (E) doubly ionized lithium atom
›Reveal solutionSolution
The Bohr model works only for single-electron species (H, He+, Li2+, deuterium). Doubly ionized helium He2+ has zero electrons, so Bohr's orbit picture cannot apply.
The Bohr atomic model successfully describes hydrogen-like systems that contain exactly one electron: hydrogen, singly ionized helium (He+), doubly ionized lithium (Li2+), and deuterium (a hydrogen isotope, still one electron). Doubly ionized helium means both of helium's electrons have been removed, leaving He2+ — a b …
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