Using Bohr model, calculate the electric current created by the electron when the H-atom is in the ground stat

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Q. Using Bohr model, calculate the electric current created by the electron when the H-atom is in the ground state.

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Concept understanding — Bohr Model Quantization

Why does an electron not spiral into the nucleus?

Imagine you're pushing a child on a swing. If you push at random moments, the swing jerks and slows down. But if you push exactly in rhythm with the swing's natural motion, each push adds energy smoothly and the swing goes higher and higher. The swing "prefers" to move at specific frequencies — its natural modes.

An electron orbiting a nucleus is similar, but with a crucial twist from the quantum world. Classical physics says an accelerating charge (like an electron going in a circle) must continuously radiate energy. If that were true, the electron would lose energy, spiral into the nucleus, and atoms would collapse in a flash of light. But atoms are stable. So something is fundamentally different.

The radical idea: allowed orbits only

Niels Bohr proposed in 1913 that the electron cannot occupy just any orbit. It can only exist in certain stationary states — orbits where it does not radiate energy. These are like the swing's natural frequencies, but for an electron.

The key condition that picks out these special orbits is called quantization of angular momentum.

L = n \frac{h}{2\pi}, \quad n = 1, 2, 3, \dots

Here $L$ is the orbital angular momentum of the electron, $h$ is Planck's constant, and $n$ is a positive integer called the principal quantum number.

What this means physically

Angular momentum for a circular orbit is $L = m v r$ , where $m$ is the electron mass, $v$ its speed, and $r$ the orbit radius. So the quantization condition becomes:

m v r = n \frac{h}{2\pi}

This is not a formula you derive — it is a postulate, a rule that nature follows. Bohr had no deeper explanation for why this rule works; he simply noticed it gave the right answers for hydrogen's spectrum.

Note

The quantity $\frac{h}{2\pi}$ appears so often that it has its own symbol: $\hbar$ (h-bar). So the condition is often written as $L = n\hbar$ .

What it predicts

Combining this quantization with Newton's law for circular motion (centripetal force = Coulomb attraction) gives:

Radius of the $n$ th orbit: $r_n = n^2 a_0$ , where $a_0 = 0.529 \, \text{Å}$ is the Bohr radius (the smallest orbit, $n=1$ ).
Energy of the $n$ th orbit: $E_n = -\frac{13.6 \, \text{eV}}{n^2}$

The negative sign means the electron is bound to the nucleus. As $n$ increases, the orbit gets larger and the energy becomes less negative (closer to zero).

The key insight for exams

Bohr quantization is not a derivation — it is a condition you apply. When you see a problem about hydrogen-like atoms (one electron), you:

Write $mvr = n\hbar$
Write the force balance: $\frac{mv^2}{r} = \frac{kZe^2}{r^2}$ (for nuclear charge $Ze$ )
Solve for $r$ and $v$ in terms of $n$ …

Why this formula?

Why Angular Momentum is Quantised in the Bohr Model

The Bohr model's most famous result — that angular momentum comes only in integer multiples of $\frac{h}{2\pi}$ — is not an arbitrary assumption. It follows directly from a single, elegant idea: the electron's wave must close on itself.

The core problem Bohr faced

By 1913, physicists knew two things that seemed contradictory:

Rutherford's nuclear model showed electrons orbiting the nucleus.
Maxwell's equations predicted that any accelerating charge (like an orbiting electron) must radiate energy, spiral inward, and collapse in about $10^{-11}$ seconds.

Atoms are stable. Something was missing.

Bohr's breakthrough was to combine the newly discovered quantum idea (Planck's constant $h$ ) with classical mechanics, but only for allowed orbits. The key constraint came from thinking of the electron not as a tiny planet, but as a standing wave.

The de Broglie wavelength argument (the cleanest derivation)

A few years after Bohr, de Broglie proposed that every moving particle has a wavelength:

\lambda = \frac{h}{p} = \frac{h}{mv}

For an electron in a circular orbit of radius $r$ , the circumference is $2\pi r$ . For the wave to be stable — not cancelling itself out — the circumference must contain an integer number of wavelengths:

2\pi r = n\lambda, \quad n = 1, 2, 3, \dots

Substitute $\lambda = h/(mv)$ :

2\pi r = n \cdot \frac{h}{mv}

Rearrange:

mvr = n \cdot \frac{h}{2\pi}

That's it. The left side $mvr$ is the angular momentum $L$ . So:

L = n\hbar, \quad \text{where } \hbar = \frac{h}{2\pi}

Note

This is not a separate postulate — it is a consequence of requiring the electron wave to be a standing wave. If the wave doesn't close on itself, it interferes destructively and the orbit cannot exist.

Why this fixes the energy levels

Once angular momentum is quantised, the rest follows from classical physics. For a circular orbit, the centripetal force is provided by the Coulomb attraction:

\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}

Combine this with $mvr = n\hbar$ and solve for $r$ and $E$ :

r_n = \frac{4\pi\epsilon_0 \hbar^2}{m e^2} \cdot n^2 = a_0 n^2

$E_n = -\frac{m e^4}{8\epsilon_0^2 h^2} \cdot \frac{1}{n^2} = -\frac{13.6 \text{ eV}}{n^2}$ …

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In the Bohr model, the orbiting electron constitutes a tiny current loop. For hydrogen in the ground state ( $n=1$ ), the current is $I = e/T$ , where $T$ is the orbital period. Using the Bohr radius $r_1 = 0.529 \, \text{Å}$ and velocity $v_1 = 2.18 \times 10^6 \, \text{m/s}$ , the current comes out to about $1.05 \times 10^{-3} \, \text{A}$ .

Why this works: the electron as a current loop

The Bohr model treats the electron as moving in a circular orbit around the proton. A moving charge is an electric current. If you watch a single point on the orbit, one electron passes that point once every orbital period $T$ . That’s exactly what current is: charge per unit time. So the average current is simply $I = e/T$ , where $e = 1.6 \times 10^{-19} \, \text{C}$ .

The trick is to find $T$ from the Bohr model’s ground state parameters. We don’t need to memorise a special formula — we can derive $T$ from the orbital radius and speed.

Step-by-step calculation

1. Recall the Bohr radius for $n=1$

The radius of the $n$ -th orbit is:

r_n = n^2 a_0, \quad a_0 = \frac{4\pi \epsilon_0 \hbar^2}{m_e e^2} = 0.529 \times 10^{-10} \, \text{m}

For the ground state, $n=1$ , so $r_1 = a_0$ .

2. Find the electron’s speed in the ground state

In the Bohr model, the centripetal force is provided by the Coulomb attraction:

\frac{m_e v^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}

Cancel one $r$ :

m_e v^2 = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r}

So:

v = \sqrt{ \frac{e^2}{4\pi\epsilon_0 m_e r} }

Plug in $r = a_0$ :

v_1 = \sqrt{ \frac{e^2}{4\pi\epsilon_0 m_e a_0} }

This evaluates to $v_1 \approx 2.18 \times 10^6 \, \text{m/s}$ (about 1/137 of the speed of light).

Tip

You can also get $v_1$ directly from the fine-structure constant: $v_1 = \alpha c$ , where $\alpha \approx 1/137$ . That’s a neat shortcut if you remember it.

3. Calculate the orbital period $T$

The circumference of the orbit is $2\pi r_1$ . Time for one revolution:

T = \frac{\text{distance}}{\text{speed}} = \frac{2\pi r_1}{v_1}

Substitute numbers: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

JKBOSE Class 12 Annual Regular Examination 2024Set SZ3 marks
$Q. On the basis of Bohr's atomic model, find an expression for radius of nth orbit of a hydrogen atom.$
›Reveal solutionSolution

Equating the Coulomb attraction to the required centripetal force, and combining it with Bohr's quantization of angular momentum, gives the radius of the nth orbit as r_n = n²h²ε0/(πme²) — increasing as the square of the orbit number n.

Step 1 — Centripetal force condition: In Bohr's model, an electron of mass m and charge −e moves in a circular orbit of radius r around the nucleus (charge +e for hydrogen), held in orbit by the electrostatic (Coulomb) force acting as the centripetal force:

(1/4πε0) × e²/r² = mv²/r

⟹ mv² = e² / (4πε0 r) ... (1)

Step 2 — Bohr's quantization postulate: The angular momentum of the electron is quantized in integral multiples of h/2π:

mvr = nh/2π ⟹ v = nh / (2πmr) ... (2)

Step 3 — Combine (1) and (2): Substitute v from (2) into (1):

m × [nh/(2πmr)]² = e²/(4πε0 r)

n²h² / (4π²mr²) = e² / (4πε0 r)

Multiplying both sides by r and rearranging for r:

…

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JKBOSE Class 12 Annual Regular Examination 2023Set ANNUAL3 marks
$Q. Write the postulates of Bohr's modal of hydrogen atom.$
›Reveal solutionSolution
Bohr's model of the hydrogen atom rests on three postulates: stable non-radiating orbits, quantized angular momentum (mvr = nh/2pi), and photon emission/absorption only during transitions between orbits (h*nu = delta-E).

Niels Bohr proposed the following postulates to explain the stability of atoms and the discrete (line) spectrum of hydrogen, combining classical mechanics with early quantum ideas:

Postulate of stationary (stable) orbits: An electron in an atom revolves around the nucleus only in certain specific, permitted circular orbits, called stationary orbits, without radiating energy - even though it is accelerating (contrary to classical electromagnetic theory, which would predict continuous energy loss and the electron spiralling into the nucleus). In these orbits, the necessary centripetal force is provided by the electrostatic (Coulomb) attraction between the electron and the nucleus.

Postulate of quantization of angular momentum: Only those orbits are permitted (stable) for which the angular momentum of the electron is an integral multiple of h/(2pi): L = mvr = nh/(2*pi), where n = 1, 2, 3, ... is the principal quantum number, m is electron mass, v its orbital speed, r the orbit radius, and h is Planck's constant.

…
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JKBOSE Class 12 Annual Regular Examination 2022Set SZ3 marks
$Q. Write down the postulates of Bohr's model of hydrogen atom.$
›Reveal solutionSolution
Bohr postulated stationary orbits with quantised angular momentum, and that spectral lines arise from photon emission/absorption during transitions between these orbits.

Niels Bohr proposed the following postulates for the hydrogen atom, combining classical mechanics with quantum ideas:

Stationary (stable) orbits: An electron in an atom revolves around the nucleus in certain fixed, stable circular orbits without radiating energy, contrary to what classical electromagnetic theory (an accelerating charge should continuously radiate energy) would predict. These allowed orbits are called stationary states.

Quantisation of angular momentum: The electron can revolve only in those orbits for which its orbital angular momentum is an integral multiple of $h/2\pi$ :

$mvr = \frac{nh}{2\pi}, \quad n = 1,2,3,\dots$

where $n$ is called the principal quantum number.

…
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JKBOSE Class 12 Annual Regular Examination 2021Set SZ3 marks
$Q. State the basic postulates of Bohr's model of atom.$
›Reveal solutionSolution

Bohr combined classical mechanics with quantum ideas: electrons orbit without radiating, only in orbits with quantised angular momentum, and photons are emitted/absorbed only during a jump between orbits.

Niels Bohr proposed three basic postulates to explain the stability of atoms and the discrete line spectra observed (overcoming the problem that a classically accelerating orbiting electron should radiate energy continuously and spiral into the nucleus):

1. Postulate of stationary orbits: An electron in an atom can revolve only in certain special, discrete circular orbits, called stationary states or orbits, without radiating energy, even though it is undergoing centripetal acceleration. In these orbits, the electrostatic force of attraction between the nucleus and electron provides the necessary centripetal force:

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

2. Postulate of quantisation of angular momentum: Only those orbits are allowed for which the angular momentum of the electron is an integral multiple of $h/2\pi$ :

$L = mvr = \frac{nh}{2\pi}, \qquad n = 1,2,3,\ldots$

where $n$ is called the principal quantum number.

…

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JKBOSE Class 12 Annual Regular Examination 2020Set SZ3 marks
$Q. State postulates of Bohr's theory of Hydrogen atom.$
›Reveal solutionSolution
Bohr's three postulates fix stable non-radiating orbits with quantised angular momentum, and explain spectral lines as photon emission/absorption during orbit jumps.

Stationary orbits postulate: An electron in an atom revolves around the nucleus in certain fixed circular orbits, called stationary orbits, without radiating any energy, even though it is accelerating (which classically should cause continuous radiation and spiral collapse into the nucleus).

Quantisation of angular momentum: Only those orbits are permitted for which the angular momentum of the electron is an integral multiple of h/2π (h being Planck's constant): mvr = nh/2π, n = 1, 2, 3, … where n is called the principal quantum number. …
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Q.Using Bohr model, calculate the electric current created by the electron when the H-atom is in the ground state.

Concept understanding — Bohr Model Quantization

Why does an electron not spiral into the nucleus?

The radical idea: allowed orbits only

What this means physically

What it predicts

The key insight for exams

Why this formula?

Why Angular Momentum is Quantised in the Bohr Model

The core problem Bohr faced

The de Broglie wavelength argument (the cleanest derivation)

Why this fixes the energy levels

Why this works: the electron as a current loop

Step-by-step calculation

Method: Get the Orbital Current via Quantised Angular Momentum, Not Force Balance

Steps

Q.Using Bohr model, calculate the electric current created by the electron when the H-atom is in the ground state.

Why this works: the electron as a current loop

Step-by-step calculation

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