Would the Bohr formula for the H-atom remain unchanged if proton had a charge $(+4/3)e$ and electron a charge $(-3/4)e$, where $e = 1.6 \times 10^{-19}\ \text{C}$? Give reasons for your answer.

Would the Bohr formula for the H-atom remain unchanged if proton had a charge (+4/3)e and electron a charge (-

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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. …

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}

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The Bohr formula depends on the product of the charges, not their individual values. Since $(+4/3)e \times (-3/4)e = -e^2$ , the product is unchanged, so the Bohr formula remains exactly the same.

The Bohr model for the hydrogen atom rests on three key ideas: the electron moves in circular orbits under the Coulomb attraction to the proton, only certain orbits are allowed where the angular momentum is quantized, and the electron does not radiate energy in these stationary states. The energy levels and radii that come out of this model depend directly on the Coulomb force between the proton and the electron.

The Coulomb force between two charges $q_1$ and $q_2$ separated by distance $r$ is

F = \frac{1}{4\pi\varepsilon_0} \frac{|q_1 q_2|}{r^2}.

In the standard hydrogen atom, $q_1 = +e$ (proton) and $q_2 = -e$ (electron), so the product is $q_1 q_2 = -e^2$ . The magnitude of the force is proportional to $e^2$ .

Now the question asks: what if the proton had charge $+\frac{4}{3}e$ and the electron had charge $-\frac{3}{4}e$ ? The product becomes

\left(+\frac{4}{3}e\right) \times \left(-\frac{3}{4}e\right) = -\frac{4}{3} \cdot \frac{3}{4} \, e^2 = -e^2.

The product is exactly the same as before. Since every formula in the Bohr model — the radius of the $n$ th orbit, the energy of the $n$ th level, the Rydberg constant — depends on $e^2$ (or equivalently on $|q_1 q_2|$ ), nothing changes.

Let’s verify this step by step.

Quantization of angular momentum

Bohr’s postulate says $m v r = n \hbar$ , where $m$ is the electron mass, $v$ its speed, $r$ the orbit radius, and $n$ a positive integer. This condition involves only the electron’s mass and velocity — it does not involve the charges at all. So this step is unaffected.
Coulomb force provides centripetal force

The equation of motion is

\frac{1}{4\pi\varepsilon_0} \frac{|q_1 q_2|}{r^2} = \frac{m v^2}{r}.

Here $|q_1 q_2|$ appears. In both the standard and modified cases, $|q_1 q_2| = e^2$ . So the equation is identical.

Solving for radius and velocity From step 1, $v = n\hbar/(mr)$ . Substitute into step 2:

\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}.

Rearranging gives the Bohr radius:

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

This depends only on $e^2$ , not on the individual charges. So $r_n$ is unchanged.

Energy of the $n$ th orbit Total energy $E = \text{K.E.} + \text{P.E.}$ . Kinetic energy is $\frac12 m v^2$ , and potential energy for an attractive Coulomb force is $-\frac{1}{4\pi\varepsilon_0} \frac{e^2}{r}$ . Using $v$ from step 1 and $r$ from step 3, you get $E_n = -\frac{m e^4}{2 (4\pi\varepsilon_0)^2 \hbar^2} \frac{1}{n^2}.$ …

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Q.Would the Bohr formula for the H-atom remain unchanged if proton had a charge $(+4/3)e$ and electron a charge $(-3/4)e$ , where $e = 1.6 \times 10^{-19}\ \text{C}$ ? Give reasons for your answer.

Concept understanding — Bohr Model Energy Levels

The Intuition: Why Can't an Electron Just Sit Anywhere?

The Precise Statement

What This Explains

Why this formula?

Why the Bohr Model Gives Those Energy Levels

The Two Non-Negotiable Pieces

The Quantisation Condition

Deriving the Allowed Radii and Energies

Method: Spot the Invariant Combination Instead of Re-Deriving the Whole Formula

Steps

Q.Would the Bohr formula for the H-atom remain unchanged if proton had a charge (+4/3)e(+4/3)e(+4/3)e and electron a charge (−3/4)e(-3/4)e(−3/4)e, where e=1.6×10−19 Ce = 1.6 \times 10^{-19}\ \text{C}e=1.6×10−19 C? Give reasons for your answer.

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Q.Would the Bohr formula for the H-atom remain unchanged if proton had a charge $(+4/3)e$ and electron a charge $(-3/4)e$ , where $e = 1.6 \times 10^{-19}\ \text{C}$ ? Give reasons for your answer.

Unlock everything free for 14 days