Q.The ground state energy of hydrogen atom is −13.6 eV. What are the kinetic and potential energies of the electron in this state?
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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
In the Bohr model, the total energy E=K+U for a hydrogen atom is E=−13.6 eV in the ground state. The virial theorem for a Coulomb potential gives K=−E and U=2E.
Reasoning
- For an electron in a stable circular orbit around a proton, the centripetal force is provided by the electrostatic attraction: rmv2=r2ke2.
- Kinetic energy K=21mv2=2rke2. Potential energy U=−rke2. …
In the Bohr model, the total energy E=−13.6 eV is split such that kinetic energy K=−E and potential energy U=2E. For the ground state, K=13.6 eV and U=−27.2 eV.
The Bohr model of the hydrogen atom gives us a beautifully simple relationship between the electron's total, kinetic, and potential energies. The key insight is that the electron is held in a circular orbit by the electrostatic attraction to the proton, and the centripetal force condition leads to a direct proportionality between these energies.
For any stable circular orbit in an inverse-square force law (like Coulomb's law), the kinetic energy is exactly half the magnitude of the potential energy, but with opposite sign. This is a special case of the virial theorem for a 1/r potential.
For a Coulomb potential U=−rke2, the virial theorem gives:
K=−21U
Since total energy E=K+U, we get:
K=−EandU=2E
Let's verify this from first principles.
- Write the total energy. The electron has kinetic energy K=21mv2 and potential energy U=−rke2 (taking U=0 at infinity). So:
E=K+U=21mv2−rke2
- Apply the centripetal force condition. For a stable circular orbit, the electrostatic force provides the necessary centripetal acceleration:
r2ke2=rmv2
Multiply both sides by r:
rke2=mv2
- Relate kinetic energy to potential. Notice that mv2=2K. So:
rke2=2K
But rke2=−U (since U=−rke2). Therefore:
−U=2K⇒K=−21U
- Express K and U in terms of E. From E=K+U and U=−2K, substitute: E=K−2K=−K⇒K=−E …
Method: Using the Bohr model relations for a hydrogen-like atom.
In the Bohr model, the electron moves in a circular orbit around the nucleus. The total energy E is the sum of kinetic energy K and potential energy U:
E=K+U
For a Coulomb force, the potential energy is U=−rke2 and the kinetic energy is K=21mv2. From the centripetal force condition rmv2=r2ke2, we get:
K=2rke2
Comparing with U=−rke2, we see:
K=−21U
So the total energy becomes:
E=K+U=−21U+U=21U
This gives U=2E and K=−E.
Step-by-step for the given problem:
-
The ground state energy is given: E=−13.6 eV.
-
Using K=−E:
K=−(−13.6 eV)=+13.6 eV.
-
Using U=2E: …
Students often know the formula En=−n213.6 eV but then fumble when asked for kinetic or potential energy separately. Here are the most common mistakes and how to avoid each.
Mistake 1: Confusing total energy with kinetic energy
Many students see −13.6 eV and think that's the kinetic energy, or they assume kinetic energy is also negative. In the Bohr model, kinetic energy is always positive — the electron is moving, so K>0.
How to avoid: Remember the key relationship derived from the Bohr model:
- Total energy: E=K+U
- For a circular orbit, the virial theorem for a Coulomb force gives K=−E and U=2E.
Since E=−13.6 eV, you get:
K=−E=13.6 eV
U=2E=−27.2 eV
A quick check: kinetic energy is always positive and equal in magnitude to the total energy (for hydrogen-like atoms). Potential energy is twice the total energy and negative.
Mistake 2: Forgetting that potential energy is negative and twice the total energy
Students sometimes write U=E−K and then plug K=13.6 eV, getting U=−13.6−13.6=−27.2 eV — which is correct, but they don't understand why it's double. Without the reasoning, they might misapply it in other contexts.
How to avoid: Derive it once from force balance:
- Centripetal force: rmv2=r2ke2
- Kinetic energy: K=21mv2=2rke2
- Potential energy: U=−rke2
- So U=−2K, and E=K+U=K−2K=−K.
From E=−K, you get K=−E and U=2E. This is a clean, exam-ready chain.
Mistake 3: Using n=1 but forgetting the sign convention
Some students write E1=−1213.6=−13.6 eV, then say K=1213.6=13.6 eV — which works here, but they don't generalise. For n=2, they might incorrectly write K=2213.6=3.4 eV, when actually K=−E=3.4 eV (same number, but the reasoning is fragile).
How to avoid: Always use the relationships K=−E and U=2E, not a separate formula for each n. That way, for any state:
Kn=−En=n213.6 eV
Un=2En=−n227.2 eV …
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 …
- 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. …
- 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: …
- 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: …
- 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).
…
- 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). …
- 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.
…
- 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 …
- 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): …
- 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). …
- 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. …
- 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
…
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