The binding energy of a H-atom, considering an electron moving around a fixed nucleus (proton), is $B = -\dfrac{m e^4}{8 n^2 \varepsilon_0^2 h^2}$ ($m$ = electron mass). If one decides to work in a frame of reference where the electron is at rest, the proton would be moving around it. By similar arguments, the binding energy would be $B = -\dfrac{M e^4}{8 n^2 \varepsilon_0^2 h^2}$ ($M$ = proton mass). This last expression is not correct because (a) $n$ would not be integral. (b) Bohr-quantisation applies only to electron. (c) the frame in which the electron is at rest is not inertial. (d) the motion of the proton would not be in circular orbits, even approximately.

> [!TLDR] > The binding energy formula depends on the reduced mass of the system, not on either mass alone. The frame where the electron is at rest is non-inertial because the electron accelerates, so the proton’s motion in that frame is not governed by the same simple Bohr quantisation — making option **(C)** the correct reason. The problem highlights a subtle but crucial point about reference frames in Bohr’s model. The original derivation assumes the nucleus is fixed — an excellent approximation because the proton is nearly 2000 times heavier than the electron. But if you flip the frame and treat the electron as fixed, you are no longer in an inertial frame. The electron accelerates (it is in orbit around the proton in the lab frame), so a frame attached to it is accelerating. Newton’s laws — and therefore Bohr’s quantisation conditions — do not hold in their simple form there. Let’s walk through the reasoning step by step. 1. **Why the two expressions differ** In the standard Bohr model for hydrogen, the electron orbits a fixed proton. The centripetal force is provided by the Coulomb attraction: \[ \frac{m v^2}{r} = \frac{e^2}{4\pi\varepsilon_0 r^2} \] Combined with Bohr’s quantisation $m v r = n \hbar$, you derive the energy levels: \[ E_n = -\frac{m e^4}{8 n^2 \varepsilon_0^2 h^2} \] The binding energy $B$ is just $E_n$ (since the zero of energy is at infinite separation). This uses the electron mass $m$ because the proton is taken as stationary. If you naively swap roles and treat the electron as fixed, you would write the same equations with the proton mass $M$: \[ \frac{M v_p^2}{r} = \frac{e^2}{4\pi\varepsilon_0 r^2}, \quad M v_p r = n \hbar \] giving $B = -\dfrac{M e^4}{8 n^2 \varepsilon_0^2 h^2}$. This is not correct — but why? 2. **The real motion: both bodies move** In truth, both the electron and proton revolve around their common centre of mass. The correct treatment replaces the electron mass $m$

The binding energy of a H-atom, considering an electron moving around a fixed nucleus (proton), is B = -(m e⁴)

NCERT Exemplar · Q2

B = -\dfrac{m e^4}{8 n^2 \varepsilon_0^2 h^2}

(a)

n

would not be integral.

(b) Bohr-quantisation applies only to electron.

(d) the motion of the proton would not be in circular orbits, even approximately.

Mizoram MbseMCQ· 1mImportance★★★★★

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✓ Free question

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 $\frac{h}{2\pi}$ (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:

E_n = -\frac{13.6 \text{ eV}}{n^2}

Here:

$n$ is the principal quantum number — a positive integer ( $n = 1, 2, 3, \dots$ )
$E_n$ 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 $-13.6 \text{ eV}$ . The higher states ( $n=2, 3, 4, \dots$ ) are excited states — they have less negative (higher) energies.

Note

As $n$ increases, the energy levels get closer together. At $n = \infty$ , the energy becomes $0 \text{ eV}$ — the electron is completely free from the atom (ionisation).

What This Explains

When an electron jumps from a higher level ( $n_i$ ) to a lower level ( $n_f$ ), it emits a photon of energy:

\Delta E = E_{n_f} - E_{n_i} = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ 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.

Watch out

The Bohr model works perfectly only for hydrogen and one-electron ions (like $\text{He}^+$ , $\text{Li}^{2+}$ ). 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 $E_n = -13.6/n^2$ 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

Important

The negative sign in $E_n$ 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.

\frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2} = \frac{mv^2}{r}

This gives you a relation between speed $v$ and radius $r$ :

v^2 = \frac{e^2}{4\pi\varepsilon_0 m r}

Second, the total energy of the electron is the sum of its kinetic and potential energies. Potential energy for a Coulomb force is $-\frac{1}{4\pi\varepsilon_0}\frac{e^2}{r}$ (negative because the force is attractive, and we set zero at infinity).

E = \frac12 mv^2 - \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r}

Substitute $v^2$ from above:

E = \frac12 \left( \frac{e^2}{4\pi\varepsilon_0 r} \right) - \frac{e^2}{4\pi\varepsilon_0 r} = -\frac{1}{2} \frac{e^2}{4\pi\varepsilon_0 r}

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 $\hbar = h/2\pi$ :

m v r = n \hbar, \quad n = 1, 2, 3, \dots

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 $2\pi r = n \lambda$ gives $m v r = n \hbar$ directly. So the quantisation condition is really a wave condition imposed on a particle picture.

Note

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 $m v r = n \hbar$ , we get $v = n\hbar / (mr)$ . Substitute this into the centripetal force equation:

\frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2} = \frac{m}{r} \left( \frac{n\hbar}{mr} \right)^2 = \frac{n^2 \hbar^2}{m r^3}

Solve for $r$ :

r_n = \frac{4\pi\varepsilon_0 \hbar^2}{m e^2} \, n^2

The constant in front is the Bohr radius $a_0 \approx 0.529 \, \text{Å}$ . So the radii are $r_n = a_0 n^2$ .

Now plug $r_n$ back into the energy expression $E = -\frac12 \frac{e^2}{4\pi\varepsilon_0 r}$ :

E_n = -\frac12 \frac{e^2}{4\pi\varepsilon_0} \cdot \frac{1}{a_0 n^2} = -\frac{1}{2} \frac{e^2}{4\pi\varepsilon_0 a_0} \cdot \frac{1}{n^2}

Substitute $a_0 = \frac{4\pi\varepsilon_0 \hbar^2}{m e^2}$ :

E_n = -\frac{1}{2} \frac{e^2}{4\pi\varepsilon_0} \cdot \frac{m e^2}{4\pi\varepsilon_0 \hbar^2} \cdot \frac{1}{n^2} = -\frac{m e^4}{8 \varepsilon_0^2 h^2} \cdot \frac{1}{n^2}

E_n = -\frac{13.6 \text{ eV}}{n^2}

That's the famous result. The $1/n^2$ dependence comes directly from the $n^2$ 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 $+13.6 \text{ eV}$ 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.

Watch out

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 $1/n^3$ 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 $\hbar$ . 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 $f = (E_i - E_f)/h$ — 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 $1/n^2$ energy law is a direct consequence of combining classical circular motion with a single quantisation postulate.

The binding energy formula depends on the reduced mass of the system, not on either mass alone. The frame where the electron is at rest is non-inertial because the electron accelerates, so the proton’s motion in that frame is not governed by the same simple Bohr quantisation — making option (C) the correct reason.

The problem highlights a subtle but crucial point about reference frames in Bohr’s model. The original derivation assumes the nucleus is fixed — an excellent approximation because the proton is nearly 2000 times heavier than the electron. But if you flip the frame and treat the electron as fixed, you are no longer in an inertial frame. The electron accelerates (it is in orbit around the proton in the lab frame), so a frame attached to it is accelerating. Newton’s laws — and therefore Bohr’s quantisation conditions — do not hold in their simple form there.

Let’s walk through the reasoning step by step.

Why the two expressions differ In the standard Bohr model for hydrogen, the electron orbits a fixed proton. The centripetal force is provided by the Coulomb attraction:

\frac{m v^2}{r} = \frac{e^2}{4\pi\varepsilon_0 r^2}

Combined with Bohr’s quantisation $m v r = n \hbar$ , you derive the energy levels:

E_n = -\frac{m e^4}{8 n^2 \varepsilon_0^2 h^2}

The binding energy $B$ is just $E_n$ (since the zero of energy is at infinite separation). This uses the electron mass $m$ because the proton is taken as stationary.

If you naively swap roles and treat the electron as fixed, you would write the same equations with the proton mass $M$ :

\frac{M v_p^2}{r} = \frac{e^2}{4\pi\varepsilon_0 r^2}, \quad M v_p r = n \hbar

giving $B = -\dfrac{M e^4}{8 n^2 \varepsilon_0^2 h^2}$ . This is not correct — but why?

The real motion: both bodies move In truth, both the electron and proton revolve around their common centre of mass. The correct treatment replaces the electron mass $m$ with the reduced mass $\mu = \frac{m M}{m+M}$ . The binding energy becomes:

B = -\frac{\mu e^4}{8 n^2 \varepsilon_0^2 h^2}

Since $M \gg m$ , $\mu \approx m$ , so the original formula is an excellent approximation. But neither $m$ nor $M$ alone is exact.

Why the “electron-at-rest” frame fails

The electron in a hydrogen atom is accelerating (it has centripetal acceleration). A frame where the electron is at rest is therefore non-inertial. In such a frame, fictitious forces appear — the proton experiences not just the Coulomb force but also a centrifugal or Coriolis-type force. The simple Bohr quantisation condition $M v_p r = n \hbar$ no longer applies because it was derived assuming an inertial frame with only the Coulomb force.

Watch out

A common mistake is to think that Bohr quantisation “applies only to the electron” (option B). That is not the core issue — you can quantise the proton’s angular momentum in principle, but only in an inertial frame. The real problem is that the electron’s rest frame is accelerating, so the proton’s dynamics there are not governed by the simple Coulomb-plus-centripetal balance.
Checking the other options
- (A) $n$ not being integral: No, $n$ remains an integer in the correct reduced-mass treatment.
- (B) Bohr quantisation applies only to the electron: This is misleading. Bohr quantisation applies to any bound particle’s angular momentum in an inertial frame. The proton can be quantised too — but not in a non-inertial frame.
- (D) The proton’s orbit not being circular: In the centre-of-mass frame, both orbits are circular (or elliptical in the Sommerfeld extension). In the electron’s rest frame, the proton’s path is indeed circular — but the dynamics are wrong because of fictitious forces. So this option is not the fundamental reason.
Only option (C) correctly identifies the root cause: the electron’s rest frame is non-inertial.

Tip

A quick way to see the flaw: if you set the electron fixed, the proton’s mass $M$ appears in the energy formula. But the actual binding energy must be symmetric in $m$ and $M$ (it depends on the reduced mass). Swapping roles should give the same physical result — it doesn’t here, signalling that the frame is invalid.

✓Final answer

The correct option is (C) — the frame in which the electron is at rest is not inertial.

Method: Checking Whether a Reference-Frame Switch Is Valid in a Derivation

Many exam questions dress up a physics result in a different frame ("what if we work in the frame where X is at rest?") and ask you to spot what breaks. This method gives a general checklist for that class of question, rather than analysing this one scenario from scratch.

Steps

Step 1: Identify what was held FIXED in the original, correct derivation

Every standard derivation (Bohr's model here) picks a frame and typically treats one body as stationary to simplify the dynamics -- that choice is baked into which mass appears in the final formula and which body's angular momentum gets quantised.

Step 2: Ask whether the "flipped" frame is inertial

Newton's laws (and any quantisation rule built on them, like $m v r = n\hbar$ ) are only valid, unmodified, in an inertial frame. If the body you are now calling "at rest" was itself accelerating in the original (correct) description, then a frame attached to it is non-inertial, and fictitious forces would have to be added to make the dynamics work -- the original equations do NOT simply carry over by swapping labels.

Step 3: Check for a hidden extra physical ingredient the naive swap ignores

Often the naive "just swap the masses" substitution silently drops a real effect -- here, the correct two-body treatment needs the reduced mass $\mu = \dfrac{mM}{m+M}$ , not either mass alone. Spotting that a swap ignores such a term is a strong sign the swapped expression is not simply the mirror image of the original.

Step 4: Applying to this problem

The electron accelerates around the nucleus in the correct picture, so a frame in which the electron is "at rest" is non-inertial -- Bohr's quantisation cannot be applied there in its simple form. This directly identifies option (C) as the flaw, and explains (without needing to compute anything) why option (B) is a distractor: quantisation isn't restricted "only to the electron" by rule, it fails here because of the frame, not because of who is being quantised.

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