The earth revolves around the sun in an orbit of radius $1.5\times10^{11}$ m with an orbital speed of 30 km/s. Find the quantum number that characterises its revolution using Bohr's model in this case (mass of earth $= 6.0\times10^{24}$ kg).

> [!TLDR] > Bohr’s angular momentum quantization condition $m v r = n \hbar$ applied to Earth’s orbit gives the quantum number $n \approx 2.6 \times 10^{74}$, an astronomically large number — showing why classical physics works for macroscopic objects. The Bohr model was originally proposed for the hydrogen atom, where the electron’s angular momentum is quantized in units of $\hbar = h / 2\pi$. The key idea is that for any object moving in a bound orbit under a central force, if we *force* the quantization condition $L = n\hbar$, we can solve for the quantum number $n$. For Earth, the angular momentum is enormous, so $n$ turns out to be huge — this is why quantum effects are invisible at our scale. Let’s work through it step by step. 1. **Write down the quantization condition.** Bohr’s postulate says that the angular momentum of a revolving body is an integer multiple of $\hbar$: $$L = n \hbar$$ For a planet in a circular orbit, $L = m v r$, where $m$ is Earth’s mass, $v$ its orbital speed, and $r$ the orbital radius. 2. **Plug in the known values.** Given: - $m = 6.0 \times 10^{24} \ \text{kg}$ - $v = 30 \ \text{km/s} = 3.0 \times 10^{4} \ \text{m/s}$ - $r = 1.5 \times 10^{11} \ \text{m}$ - $h = 6.63 \times 10^{-34} \ \text{J·s}$, so $\hbar = \frac{h}{2\pi} \approx 1.055 \times 10^{-34} \ \text{J·s}$ First compute the angular momentum: $$L = m v r = (6.0 \times 10^{24}) \times (3.0 \times 10^{4}) \times (1.5 \times 10^{11})$$ Multiply stepwise: - $6.0 \times 3.0 = 18.0$ - $10^{24} \times 10^{4} = 10^{28}$ - $18.0 \times 1.5 = 27.0$ - $10^{28} \times 10^{11} = 10^{39}$ So $L = 27.0 \times 10^{39} = 2.7 \times 10^{40} \ \text{kg·m}^2/\text{s}$. 3. **Solve for $n$.** From $L = n \hbar$: $$n = \frac{L}{\hbar} = \frac{2.7 \times 10^{40}}{1.055 \times 10^{-34}}$$ …

The earth revolves around the sun in an orbit of radius 1.5×10¹¹ m with an orbital speed of 30 km/s. Find the

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1.5\times10^{11}

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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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Bohr’s angular momentum quantization condition $m v r = n \hbar$ applied to Earth’s orbit gives the quantum number $n \approx 2.6 \times 10^{74}$ , an astronomically large number — showing why classical physics works for macroscopic objects.

The Bohr model was originally proposed for the hydrogen atom, where the electron’s angular momentum is quantized in units of $\hbar = h / 2\pi$ . The key idea is that for any object moving in a bound orbit under a central force, if we force the quantization condition $L = n\hbar$ , we can solve for the quantum number $n$ . For Earth, the angular momentum is enormous, so $n$ turns out to be huge — this is why quantum effects are invisible at our scale.

Let’s work through it step by step.

Write down the quantization condition. Bohr’s postulate says that the angular momentum of a revolving body is an integer multiple of $\hbar$ :

L = n \hbar

For a planet in a circular orbit, $L = m v r$ , where $m$ is Earth’s mass, $v$ its orbital speed, and $r$ the orbital radius.

Plug in the known values.

Given:
- $m = 6.0 \times 10^{24} \ \text{kg}$
- $v = 30 \ \text{km/s} = 3.0 \times 10^{4} \ \text{m/s}$
- $r = 1.5 \times 10^{11} \ \text{m}$
- $h = 6.63 \times 10^{-34} \ \text{J·s}$ , so $\hbar = \frac{h}{2\pi} \approx 1.055 \times 10^{-34} \ \text{J·s}$
First compute the angular momentum:

L = m v r = (6.0 \times 10^{24}) \times (3.0 \times 10^{4}) \times (1.5 \times 10^{11})

Multiply stepwise:

$6.0 \times 3.0 = 18.0$
$10^{24} \times 10^{4} = 10^{28}$
$18.0 \times 1.5 = 27.0$
$10^{28} \times 10^{11} = 10^{39}$ So $L = 27.0 \times 10^{39} = 2.7 \times 10^{40} \ \text{kg·m}^2/\text{s}$ .

Solve for $n$ . From $L = n \hbar$ : $n = \frac{L}{\hbar} = \frac{2.7 \times 10^{40}}{1.055 \times 10^{-34}}$ …

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

Showing the 12 most recent of 70 on this concept.

CBSE 2026Set 55/1/11 markMCQ
$Q. For questions 13 to 16, two statements are given - one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) below: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Both Assertion (A) and Reason (R) are false. Assertion (A) : In Bohr model of hydrogen atom, the energy levels are discrete and quantised. Reason (R) : In a hydrogen atom, the electrostatic force on the electron provides the necessary centripetal force to it to revolve around the nucleus.$
›Reveal solutionSolution

The key idea is that Bohr’s model indeed has discrete energy levels (true), and the reason given — that electrostatic force provides centripetal force — is also true, but it is a classical condition that does not explain quantisation. So both are true, but the reason is not the correct explanation of the assertion. The answer is (B).

Let’s unpack this carefully. The question is from the Assertion-Reason format, so we need to check two things independently: whether each statement is true, and then whether the reason correctly explains the assertion.

1. Check the Assertion (A):

“In Bohr model of hydrogen atom, the energy levels are discrete and quantised.”

This is absolutely true. Bohr’s model was revolutionary precisely because it introduced quantised orbits — electrons can only occupy certain allowed energy levels, and they jump between these levels by absorbing or emitting photons of specific energies. The energy of the $n$ -th level is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$ , which is clearly discrete. So Assertion (A) is true.

2. Check the Reason (R):

“In a hydrogen atom, the electrostatic force on the electron provides the necessary centripetal force to it to revolve around the nucleus.”

This is also true — it’s a standard Newtonian condition for any circular orbit. For an electron of mass $m$ and speed $v$ at a distance $r$ from the nucleus, the Coulomb attraction $\frac{ke^2}{r^2}$ equals the centripetal force $m v^2 / r$ . So Reason (R) is true.

3. Now the crucial part: Does (R) correctly explain (A)?

The reason describes a classical force balance — it’s the same condition that would hold for a planet orbiting the Sun. But that classical condition alone does not lead to discrete, quantised energy levels. In fact, classical physics would allow any orbit radius, and hence any energy. Bohr had to add an extra postulate — quantisation of angular momentum ( $m v r = n \hbar$ ) — to get discrete levels. The electrostatic-centripetal balance is necessary for the model, but it is not the reason for quantisation. So (R) is true but does not explain (A). …
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CBSE 2026Set 55/2/11 markMCQ
$n = 5$
›Reveal solutionSolution
The energy difference between levels determines photon wavelength through $E = \frac{hc}{\lambda}$ . Since the $5 \to 2$ transition releases less energy than $5 \to 1$ , its photon has a longer wavelength. The answer is $\boxed{\frac{32}{7}\lambda}$ — option (D).

The Bohr model tells us that when an electron drops from a higher energy level to a lower one, it emits a photon whose energy exactly equals the energy difference between those levels. The key relationship is $E_{\text{photon}} = \frac{hc}{\lambda}$ , which shows that energy and wavelength are inversely related: a smaller energy gap produces a longer wavelength.

In hydrogen, the energy of the $n$ -th level is given by:
$E_n = -\frac{13.6 \text{ eV}}{n^2}$
The negative sign indicates that the electron is bound to the nucleus. When the electron transitions from level $n_i$ to $n_f$ , the energy released is:
$\Delta E = E_{n_i} - E_{n_f} = 13.6 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) \text{ eV}$
This energy becomes the photon's energy: $\Delta E = \frac{hc}{\lambda}$ .

For the first transition ( $5 \to 1$ ):

Calculate the energy difference:

$\Delta E_1 = 13.6 \left(\frac{1}{1^2} - \frac{1}{5^2}\right) = 13.6 \left(1 - \frac{1}{25}\right) = 13.6 \times \frac{24}{25}$

This energy corresponds to wavelength $\lambda$ :

$\frac{hc}{\lambda} = 13.6 \times \frac{24}{25}$
For the second transition ( $5 \to 2$ ):

Calculate the energy difference:

$\Delta E_2 = 13.6 \left(\frac{1}{2^2} - \frac{1}{5^2}\right) = 13.6 \left(\frac{1}{4} - \frac{1}{25}\right)$

Find a common denominator:

$\Delta E_2 = 13.6 \left(\frac{25 - 4}{100}\right) = 13.6 \times \frac{21}{100}$

This energy corresponds to wavelength $\lambda'$ : …
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CBSE 2026Set V11 markMCQ
$Q. Let K be the kinetic energy, U be the potential energy and E be the total energy of an electron revolving around the nucleus in a hydrogen atom, then which of the following is correct?$
(a) $K > 0,\ U > 0,\ E > 0$
(b) $K > 0,\ U < 0,\ E < 0$
(c) $K > 0,\ U > 0,\ E < 0$
(d) $K < 0,\ U < 0,\ E < 0$
›Reveal solutionSolution

Option (b) $K > 0,\ U < 0,\ E < 0$ . …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Formula for total energy of the electron in the nth stationary state of the hydrogen atom is -$
(i) $-\dfrac{11.2}{n^2}$ eV
(ii) $+\dfrac{11.2}{n^2}$ eV
(iii) $-\dfrac{13.6}{n^2}$ eV
(iv) $+\dfrac{13.6}{n^2}$ eV
›Reveal solutionSolution

Total energy of the electron in the nth Bohr orbit of hydrogen is $-13.6/n^2$ eV.

…

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CBSE 2026Set A1 markMCQ
$Q. The solar spectrum is (A) continuous (B) line spectrum (C) spectrum of black lines (D) spectrum of black bands$
›Reveal solutionSolution

The solar spectrum is a line-absorption spectrum: a continuous background crossed by dark (Fraunhofer) lines.

The hot, dense interior of the Sun emits a continuous spectrum. As this light passes through the cooler gases of the Sun's outer atmosphere, atoms there absorb their characteristic wavelengths, leaving dark lines (Fraunhofer lines) superimposed on the …

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CBSE 2026Set A1 markMCQ
$Q. The kinetic energy (K) of an electron in a Bohr orbit is related to its potential energy (U) by (A) K = U (B) K = -U (C) K = -U/2 (D) K = -2U$
›Reveal solutionSolution

For an electron in a Bohr orbit, K = -U/2 (with U taken negative), which also gives total energy E = -K.

In a hydrogen-like Bohr orbit the Coulomb attraction provides the centripetal force:

$\frac{k e^2}{r^2} = \frac{m v^2}{r} \;\Rightarrow\; m v^2 = \frac{k e^2}{r}.$

So the kinetic energy is

$K = \tfrac{1}{2} m v^2 = \frac{k e^2}{2r}.$ …

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CBSE 2026Set ANNUAL1 mark
$Q. Write the definition of emission line spectrum.$
›Reveal solutionSolution

When atoms of a rarefied gas are excited (e.g. by heating or an electric discharge) and their electrons fall back to lower energy levels, they emit light only at specific wavelengths, producing a spectrum of separated bright lines rather than a continuous band.

An emission line spectrum is obtained when the light emitted directly by a source of excited atoms (such as a gas discharge tube) is passed through a spectrometer/prism. Because each element's electrons can only occupy discrete (quantised) energy levels, transitions between these levels emit photons of only certain specific energies (and hence specific wavelengths/frequencies). The resulting spectrum consists of a series of bright, sharp coloured lines at these part …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Case study: Bohr's model addressed the instability of the Rutherford model by introducing quantization. The model is based on three postulates, which successfully explained the discrete line spectrum of hydrogen. According to the model, the radius of the nth stationary orbit is r_n \propto n², and the total energy is E_n = -13.6 eV / n², when an electron jumps from a higher energy level (E_i) to a lower one (E_f), a photon of energy hν = E_i - E_f is emitted. Transitions ending at the n = 1 level form the Lyman series. The energy of an electron in the first excited state (n = 2) of hydrogen is:$
(a) –13.6 eV
(b) –6.8 eV
(c) –3.4 eV
(d) –1.7 eV
›Reveal solutionSolution

Bohr energy levels of hydrogen follow $E_n = -13.6/n^2$ eV; for $n=2$ , $E_2 = -3.4$ eV.

$E_n = \dfrac{-13.6}{n^2}\,\text{eV}$

For the first excited state, $n=2$ : …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Case study: Bohr's model addressed the instability of the Rutherford model by introducing quantization. The model is based on three postulates, which successfully explained the discrete line spectrum of hydrogen. According to the model, the radius of the nth stationary orbit is r_n \propto n², and the total energy is E_n = -13.6 eV / n², when an electron jumps from a higher energy level (E_i) to a lower one (E_f), a photon of energy hν = E_i - E_f is emitted. Transitions ending at the n = 1 level form the Lyman series. Transitions ending at n = 3 belong to which series?$
(a) Lyman series
(b) Balmer series
(c) Paschen series
(d) Brackett series
›Reveal solutionSolution

Transitions of hydrogen electrons ending on the $n=3$ level constitute the Paschen series (infrared).

…

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CBSE 2026Set ANNUAL1 mark
$Q. Case study: Bohr's model addressed the instability of the Rutherford model by introducing quantization. The model is based on three postulates, which successfully explained the discrete line spectrum of hydrogen. According to the model, the radius of the nth stationary orbit is r_n \propto n², and the total energy is E_n = -13.6 eV / n², when an electron jumps from a higher energy level (E_i) to a lower one (E_f), a photon of energy hν = E_i - E_f is emitted. Transitions ending at the n = 1 level form the Lyman series. Explain the physical significance of the negative sign in the expression for the total energy E_n of an electron in the Bohr model. OR What is the ratio of the radius of the third Bohr orbit (n = 3) to the radius of the first Bohr orbit (n = 1)?$
›Reveal solutionSolution

A negative total energy means the electron is in a bound state — energy must be added to remove it to infinity (where E=0).

The total energy of an electron in the $n^{th}$ Bohr orbit is $E_n = -\dfrac{13.6}{n^2}\,\text{eV}$ . The negative sign signifies that the electron is bound to the nucleus by the attractive electrostatic force — its potential energy (taken as zero at infinite separation) is more negative than its kinetic energy is positive, so the net energy is negative. This means external energy (at least $|E_n|$ ) must be supplied …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. In which part of the electromagnetic spectrum, is the Lyman series of hydrogen found?$
(a) Ultraviolet
(b) Scene
(c) Infrared
(d) X-rays
›Reveal solutionSolution

The Lyman series corresponds to electron transitions from higher levels $n=2,3,4,\ldots$ down to $n=1$ ; these are the highest-energy (shortest-wavelength) hydrogen transitions and fall in the ultraviolet.

Using the Rydberg formula $\dfrac{1}{\lambda}=R\left(\dfrac{1}{1^2}-\dfrac{1}{n^2}\right)$ for $n=2,3,4,\ldots$ , the Lyman series wavelengths work out to roughly 91–122 nm, which lies well below the visible range (about 400–700 nm) and squarely in the ultraviolet band. (By contrast, the Ba …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following statements is true?$
(a) Its kinetic energy increases and its potential and total energies decrease.
(b) Its kinetic energy decreases, potential energy increases and its total energy remains the same.
(c) Its kinetic and total energies decrease and its potential energy increases.
(d) Its kinetic energy, potential energy and total energy all decrease.
›Reveal solutionSolution

In the Bohr model $KE=+\dfrac{ke^2}{2r}$ , $PE=-\dfrac{ke^2}{r}=-2KE$ , and $E=KE+PE=-\dfrac{ke^2}{2r}$ . As the electron moves to a lower (ground) state, $r$ decreases, so $KE$ increases while $PE$ and $E$ both decrease (become more negative).

Energies in the Bohr model

For an electron in a circular orbit of radius $r$ about the nucleus (charge $+e$ ), the Coulomb force supplies the centripetal force:

$\frac{mv^2}{r}=\frac{ke^2}{r^2}\ \implies\ mv^2=\frac{ke^2}{r}$

Kinetic energy:

$KE=\frac12 mv^2=\frac{ke^2}{2r}\quad(\text{always positive})$

Potential energy (electrostatic PE between electron and nucleus, taking $PE=0$ at $r=\infty$ ):

$PE=-\frac{ke^2}{r}\quad(\text{always negative})$

Total energy:

$E=KE+PE=\frac{ke^2}{2r}-\frac{ke^2}{r}=-\frac{ke^2}{2r}\quad(\text{always negative})$

Note that $PE=-2\,KE$ and $E=-KE$ — standard virial-theorem relations for the Coulomb (inverse-square) force.

Effect of a transition from an excited state to the ground state

Going from an excited state to the ground state means the electron moves to a smaller orbit radius $r$ (ground state, $n=1$ , has the smallest Bohr radius).

…

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Q.The earth revolves around the sun in an orbit of radius 1.5×10111.5\times10^{11}1.5×1011 m with an orbital speed of 30 km/s. Find the quantum number that characterises its revolution using Bohr's model in this case (mass of earth =6.0×1024= 6.0\times10^{24}=6.0×1024 kg).

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Q.The earth revolves around the sun in an orbit of radius $1.5\times10^{11}$ m with an orbital speed of 30 km/s. Find the quantum number that characterises its revolution using Bohr's model in this case (mass of earth $= 6.0\times10^{24}$ kg).

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