Q.Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n−1). For large n, show that this frequency equals the classical frequency of revolution of the electron in the orbit.
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.
The Bohr model works perfectly only for hydrogen and one-electron ions (like He+, Li2+). For multi-electron atoms, it fails — electron-electron repulsion changes the energy levels in ways Bohr's simple picture cannot capture. That's where quantum mechanics takes over.
Key Takeaways for Exams
- Energy levels are quantised — only specific values allowed, given by En=−13.6/n2 eV for hydrogen
- The ground state (n=1) is the most stable, lowest energy
- Excited states (n>1) are higher in energy (less negative)
- Transitions between levels produce line spectra — not continuous
- The ionisation energy of hydrogen (energy to remove the electron from ground state) is +13.6 eV
The negative sign in En is not optional — it tells you the electron is bound. A positive energy would mean a free electron (ionised atom).
Searches like "Bohr model energy levels formula" and "hydrogen spectrum series class 12 physics" are extremely common, since this concept anchors the Atoms chapter of the NCERT/CBSE Class 12 Physics curriculum. Energy-level transition and spectral-series questions built on this model are a staple of JEE Main and NEET.
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
That's the famous result. The 1/n2 dependence comes directly from the n2 dependence of the radius, which came from the angular momentum quantisation.
Why the Negative Sign Matters
The energy is negative because the electron is bound. To remove the electron from the atom (ionise it), you need to add +13.6 eV to get it to E=0 (free electron at rest). The ground state (n=1) is the most tightly bound; higher n states are less negative, meaning they're closer to being free.
A common mistake is to think the energy levels are equally spaced. They're not — the gap between n=1 and n=2 is about 10.2 eV, while between n=2 and n=3 is only 1.9 eV. The spacing shrinks as 1/n3 for large n.
The Physical Picture
The Bohr model gives you a ladder of energies because the electron can only exist in orbits whose angular momentum is an integer multiple of ℏ. Each orbit has a specific radius, and therefore a specific energy. When the electron jumps from a higher orbit to a lower one, the energy difference is emitted as a photon of frequency f=(Ei−Ef)/h — which exactly matches the hydrogen spectral lines.
The model fails for multi-electron atoms and doesn't explain why angular momentum is quantised in the first place. But for hydrogen, it's remarkably accurate — and the derivation shows that the 1/n2 energy law is a direct consequence of combining classical circular motion with a single quantisation postulate.
The photon frequency for a de-excitation from level n to n−1 follows directly from the hydrogen energy-level formula, and taking the large-n limit of that expression reproduces exactly the electron's own classical orbital (revolution) frequency -- a striking confirmation of Bohr's correspondence principle.
ν=4ε02h3me4[(n−1)21−n21]n largeν≈4ε02h3n3me4=νclassical
The de-excitation photon frequency works out to ν=4ε02h3me4[(n−1)21−n21]; for large n this reduces to 4ε02h3n3me4, which is exactly the classical orbital frequency of the electron computed independently from Bohr's orbit formulas -- confirming the correspondence principle.
Step 1 -- Energy of level n.
From the Bohr model,
En=−8ε02h2n2me4
Step 2 -- Photon frequency for the n→(n−1) transition.
The photon carries away ΔE=En−En−1 (the electron drops to the lower, more negative energy En−1, releasing the difference), so
ν=hΔE=8ε02h3me4[(n−1)21−n21]
Step 3 -- Large-n limit.
For large n, expand the bracket:
(n−1)21−n21=n2(n−1)2n2−(n−1)2=n2(n−1)22n−1≈n42n=n32(n≫1)
so
ν≈8ε02h3me4⋅n32=4ε02h3n3me4
Step 4 -- Compare with the classical orbital (revolution) frequency.
From Bohr's own orbit formulas, vn=2ε0hne2 and rn=πme2ε0h2n2, so the electron's classical frequency of revolution is
νclassical=2πrnvn=4ε02h3n3me4
This is identical to the large-n photon frequency derived in Step 3. This is the content of Bohr's correspondence principle: for large quantum numbers, quantum predictions (the discrete photon frequency between adjacent, closely-spaced levels) merge smoothly into classical predictions (the electron's own continuous orbital frequency).
ν=4ε02h3me4[(n−1)21−n21],n→∞limν=4ε02h3n3me4=νclassical
Showing the 12 most recent of 26 on this concept.
- GUJCET 2026Set x1 markMCQQ.The ground state energy of hydrogen atom is -13.6 eV. What is the ratio of kinetic energy and potential energy of the electron of this state? (A) −1/2 (B) 1/2 (C) −1 (D) −2
›Reveal solutionSolution
By the virial theorem for the Coulomb field, KE=−21PE, giving the ratio −21.
For the electron in a hydrogen atom, with total energy E:
KE=−E,PE=2E
At the ground state E=−13.6 eV, so KE=+13.6 eV and PE=−27.2 eV.
PEKE=−27.213.6=−21
✓Final answerOption (A) −1/2
ANSWER: (A)
- GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQQ.In Hydrogen atom energy of electron in first excited state is ___.(a) -3.40 eV(b) -1.51 eV(c) -0.85 eV(d) -13.6 eV
›Reveal solutionSolution
The Bohr model gives the energy of the nth stationary state of hydrogen as E_n = -13.6/n^2 eV; the first excited state corresponds to n = 2.
Ground state: n = 1, E_1 = -13.6 eV.
First excited state: n = 2.
E_2 = -13.6 / (2)^2 = -13.6/4 = -3.40 eV.
✓Final answer(a) -3.40 eV.
- GUJCET 2025Set 031 markMCQQ.13.6 eV energy is required to separate a hydrogen atom into proton and an electron. If the orbital radius of an electron in hydrogen atom is 5.3×10−11 m, then velocity of electron is ______. (A) 6.25×107 ms−1 (B) 1.36×105 ms−1 (C) 2.4×108 ms−1 (D) 2.2×106 ms−1
›Reveal solutionSolution
[!TLDR]
The orbital speed of the electron in the hydrogen atom is about 2.2×106 m/s.
Concept
For the hydrogen atom, the magnitude of the total (binding) energy equals the kinetic energy of the electron: ∣E∣=21mv2=13.6 eV in the ground state.
Solution
- KE=13.6 eV=13.6×1.6×10−19=2.176×10−18 J.
- From 21mv2=KE:
v=m2KE=9.1×10−312×2.176×10−18=4.78×1012≈2.2×106 m/s.
(The same value follows from v=4πε0mre using r=5.3×10−11 m.)
[!ANSWER]
(D) 2.2×106 ms−1
- GUJCET 2025Set 031 markMCQQ.The ground state energy of hydrogen atom is −13.6 eV. The potential and kinetic energies of the electron in this state ______. (A) −13.6 eV, −27.2 eV (B) −27.2 eV, −13.6 eV (C) −27.2 eV, +13.6 eV (D) −13.6 eV, +27.2 eV
›Reveal solutionSolution
[!TLDR] PE=2E=−27.2 eV and KE=−E=+13.6 eV.
Concept
In Bohr's hydrogen atom the electron's kinetic and potential energies relate to the total energy E by KE=−E and PE=2E (a consequence of the virial theorem for the Coulomb potential: PE=−2KE, and E=KE+PE=−KE).
Solution
Given total (ground-state) energy E=−13.6 eV:
- Potential energy: PE=2E=2(−13.6)=−27.2 eV.
- Kinetic energy: KE=−E=−(−13.6)=+13.6 eV.
The question asks for potential and kinetic energies in that order: −27.2 eV and +13.6 eV.
[!ANSWER] (C)
- GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQQ.Potential energy of an electron in the first excited state in a hydrogen atom is ______ eV.(a) -3.4(b) -6.8(c) -10.2(d) -13.6
›Reveal solutionSolution
For the Bohr hydrogen atom, total energy En = −13.6/n² eV and potential energy PE = 2En (kinetic energy KE = −En).
First excited state: n = 2. En = −13.6/4 = −3.4 eV.
PE = 2En = 2(−3.4) = −6.8 eV.
✓Final answer(b) −6.8 eV.
- GUJCET 2024Set 131 markMCQQ.The ground state energy of hydrogen atom is −13.6 eV, then the potential energy of the electron in this state will be ________. (A) −6.8 eV (B) −27.2 eV (C) 13.6 eV (D) 27.2 eV
›Reveal solutionSolution
In the Bohr atom the potential energy is twice the total energy: U=2E=2(−13.6)=−27.2 eV.
Concept. For a bound electron in the hydrogen atom the total energy E, kinetic energy K, and potential energy U satisfy K=−E and U=2E (virial theorem).
Steps. With ground-state total energy E=−13.6 eV,
U=2E=2(−13.6)=−27.2 eV.
✓Final answer(B) −27.2 eV
ANSWER: (B)
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQQ.Kinetic energy of electron in one of the orbit of hydrogen atom is x then its total energy is ___.(a) -x(b) -x/2(c) -2x(d) -x/8
›Reveal solutionSolution
For an electron in a Bohr orbit, PE = −2KE, so Total Energy = KE + PE = KE − 2KE = −KE.
If KE = x, then Total Energy E = −x.
✓Final answer(a) −x.
- GUJCET 2023Set 091 markMCQQ.The longest wavelength present in the Balmer series of spectral line is ______. (A) 5438A˚ (B) 6563A˚ (C) 7369A˚ (D) 3646A˚
›Reveal solutionSolution
Longest Balmer wavelength = smallest energy jump = n=3→2 (Hα) ≈6563 Å.
Concept — Balmer series. All Balmer transitions end at n=2. Longest wavelength corresponds to the smallest energy difference, i.e. the transition from the nearest upper level n=3→n=2 (Hα).
λ1=R(221−321)=R(41−91)=R⋅365
With R=1.097×107m−1:
λ=5R36≈6563 A˚
(The value 3646 Å is the series limit — the shortest wavelength — not the longest.)
✓Final answerOption (B) 6563A˚
ANSWER: (B)
- GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQQ.What is the shortest wavelength present in the Balmer series of spectral lines? [Rydberg's constant R = 1.097 x 10^7 m^-1](a) 365 nm(b) 91 nm(c) 26 nm(d) 820 nm
›Reveal solutionSolution
The shortest Balmer wavelength (series limit, n -> infinity) satisfies 1/lambda = R(1/4 - 0) = R/4, giving lambda = 4/R = 365 nm.
Balmer series: 1/lambda = R(1/2^2 - 1/n^2). The shortest wavelength is the series limit (n -> infinity):
1/lambda = R(1/4 - 0) = R/4.
lambda = 4/R = 4/(1.097 x 10^7) = 3.646 x 10^-7 m = 364.6 nm approximately 365 nm.
✓Final answer(a) 365 nm.
- GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQQ.What is the ratio of total energy of an electron in hydrogen atom in first excited state and third excited state?(a) 4 : 1(b) 3 : 1(c) 1 : 1(d) 1 : 4
›Reveal solutionSolution
First excited state = n=2, third excited state = n=4; since E_n proportional to -1/n^2, the ratio is (1/4)/(1/16) = 4:1.
Energy levels: E_n = -13.6/n^2 eV, so |E_n| proportional to 1/n^2.
- First excited state: n = 2 (the state just above the ground state n = 1).
- Third excited state: n = 4.
Ratio of total energies:
E_2 : E_4 = (1/2^2) : (1/4^2) = (1/4) : (1/16) = 4 : 1.
✓Final answer(a) 4 : 1.
- GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQQ.Which of the following spectral series lies in the ultraviolet region?(a) Paschen series(b) Balmer series(c) Lyman series(d) Pfund series
›Reveal solutionSolution
Among the hydrogen spectral series, only the Lyman series lies in the ultraviolet.
The hydrogen spectral series are grouped by the final energy level (n_f) the electron falls to:
- Lyman series (n_f = 1): all wavelengths in the ultraviolet.
- Balmer series (n_f = 2): mostly in the visible region.
- Paschen series (n_f = 3), Brackett (n_f = 4), Pfund (n_f = 5): all in the infrared.
Since transitions ending at n=1 release the largest energy photons (shortest wavelength), only the Lyman series lies in the ultraviolet.
✓Final answer(c) Lyman series.
- GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQQ.The dimensional formula of m e^4 / (8 ε0^2 h^3 c) is ______.(a) M^0 L^-1 T^0(b) M^0 L^1 T^0(c) M^0 L^0 T^0(d) M^-1 L^0 T^0
›Reveal solutionSolution
The quantity me⁴/(8ε0²h³c) is exactly the Rydberg constant from Bohr's theory, whose SI unit is m⁻¹.
Bohr's model gives the wave number of emitted radiation as:
νˉ=λ1=R(n121−n221), where R=8ε02h3cme4
Since νˉ=1/λ has units of inverse length, R (the Rydberg constant) must also have dimensions of inverse length:
[R]=M0L−1T0
✓Final answer(a) M^0 L^-1 T^0.
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