A hydrogen atom consists of an electron revolving in a circular orbit of radius $r$ with a certain velocity $v$ around a proton located at the nucleus. The electrostatic force of attraction between the revolving electron and the proton provides the requisite centripetal force to keep it in certain stable orbits. The angular momentum of the electron in these orbits is some integral multiple of $\frac{h}{2\pi}$. When an electron makes a transition from one orbit of higher energy to that of lower energy, a photon is emitted having energy equal to the difference between the energies of the initial and final states. Assuming the mass and charge of the electron as $m$ and $-e$ respectively, answer the following questions. (Take $K = \frac{1}{4\pi\varepsilon_0}$.) (i) The expression for the speed of the electron $v$ in terms of the radius of the orbit $(r)$ and the physical constant $K$ is: (ii) The total energy of the atom in terms of $r$ and the physical constant $K$ is: (iii) A photon of wavelength $500$ nm is emitted when an electron makes a transition from one state to another state in an atom. The change in the total energy of the electron and the change in its kinetic energy (in eV) are: (A) $2.48,\ -2.48$ (B) $1.24,\ 1.24$ (C) $-2.48,\ 2.48$ (D) $1.24,\ -1.24$ (iv)(a) The frequency of revolution of the electron in its $n^{th}$ orbit is proportional to: (A) $n$ (B) $\frac{1}{n}$ (C) $\frac{1}{n^2}$ (D) $\frac{1}{n^3}$ OR (iv)(b) An electron makes a transition from the $-3.4$ eV state to the ground state in a hydrogen atom. Its radius of orbit changes by: (radius of orbit of electron in the ground state $= 0.53$ Å) (A) $0.53$ Å (B) $1.06$ Å (C) $1.59$ Å (D) $2.12$ Å

> [!TLDR] > (i) $v=\sqrt{\dfrac{Ke^2}{mr}}$; (ii) $E=-\dfrac{Ke^2}{2r}$; (iii) **(C)** $-2.48,\ 2.48$; (iv)(a) **(D)** $f\propto 1/n^3$; (iv)(b) **(C)** $\Delta r=1.59\ \text{Å}$. ## Part (a) **(i) Speed of the electron.** The Coulomb attraction supplies the centripetal force: $$\frac{Ke^2}{r^2}=\frac{mv^2}{r}\Rightarrow v^2=\frac{Ke^2}{mr}\Rightarrow v=\sqrt{\frac{Ke^2}{mr}}.$$ **(ii) Total energy.** $$KE=\tfrac12 mv^2=\frac{Ke^2}{2r},\qquad PE=-\frac{Ke^2}{r},$$ $$E=KE+PE=\frac{Ke^2}{2r}-\frac{Ke^2}{r}=-\frac{Ke^2}{2r}.$$ The negative sign shows the electron is bound; note $KE=-E$. **(iii) Photon of $500\ \text{nm}$.** Using $hc=1240\ \text{eV nm}$, $$E_{\text{photon}}=\frac{1240}{500}=2.48\ \text{eV}.$$ On emission the atom's total energy drops: $\Delta E_{\text{total}}=-2.48\ \text{eV}$. Since $KE=-E$, a fall in total energy means a rise in kinetic energy of the same magnitude: $\Delta KE=+2.48\ \text{eV}$. **Option (C).** **(iv)(a) Frequency of revolution.** From Bohr's model $r\propto n^2$ and $v\propto 1/n$, so $$f=\frac{v}{2\pi r}\propto\frac{1/n}{n^2}=\frac{1}{n^3}.$$ …

A hydrogen atom consists of an electron revolving in a circular orbit of radius r with a certain velocity v ar

Question

r

(i) The expression for the speed of the electron

v

in terms of the radius of the orbit

(r)

and the physical constant

K

is:

(ii) The total energy of the atom in terms of

r

and the physical constant

K

is:

(iii) A photon of wavelength

500

nm is emitted when an electron makes a transition from one state to another state in an atom. The change in the total energy of the electron and the change in its kinetic energy (in eV) are: (A)

2.48,\ -2.48

(B)

1.24,\ 1.24

(C)

-2.48,\ 2.48

(D)

1.24,\ -1.24

(iv)(a) The frequency of revolution of the electron in its

n^{th}

orbit is proportional to: (A)

n

(B)

\frac{1}{n}

(C)

\frac{1}{n^2}

(D)

\frac{1}{n^3}

(OR)

(iv)(b) An electron makes a transition from the

-3.4

eV state to the ground state in a hydrogen atom. Its radius of orbit changes by: (radius of orbit of electron in the ground state

= 0.53

Å) (A)

0.53

Å (B)

1.06

Å (C)

1.59

Å (D)

2.12

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Part (a)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}

Part (b)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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(i) $v=\sqrt{\dfrac{Ke^2}{mr}}$ ; (ii) $E=-\dfrac{Ke^2}{2r}$ ; (iii) (C) $-2.48,\ 2.48$ ; (iv)(a) (D) $f\propto 1/n^3$ ; (iv)(b) (C) $\Delta r=1.59\ \text{Å}$ .

Part (a)

Speed of the electron. The Coulomb attraction supplies the centripetal force: $\frac{Ke^2}{r^2}=\frac{mv^2}{r}\Rightarrow v^2=\frac{Ke^2}{mr}\Rightarrow v=\sqrt{\frac{Ke^2}{mr}}.$
Total energy. $KE=\tfrac12 mv^2=\frac{Ke^2}{2r},\qquad PE=-\frac{Ke^2}{r},$ $E=KE+PE=\frac{Ke^2}{2r}-\frac{Ke^2}{r}=-\frac{Ke^2}{2r}.$ The negative sign shows the electron is bound; note $KE=-E$ . (iii) Photon of $500\ \text{nm}$ . Using $hc=1240\ \text{eV nm}$ , $E_{\text{photon}}=\frac{1240}{500}=2.48\ \text{eV}.$ On emission the atom's total energy drops: $\Delta E_{\text{total}}=-2.48\ \text{eV}$ . Since $KE=-E$ , a fall in total energy means a rise in kinetic energy of the same magnitude: $\Delta KE=+2.48\ \text{eV}$ . Option (C). (iv)(a) Frequency of revolution. From Bohr's model $r\propto n^2$ and $v\propto 1/n$ , so $f=\frac{v}{2\pi r}\propto\frac{1/n}{n^2}=\frac{1}{n^3}.$ …

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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 99 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
$n$
›Reveal solutionSolution
In the Bohr model, the orbital radius scales as $r_n \propto n^2$ . For large $n$ , the gap between consecutive orbits $\Delta r = r_{n+1} - r_n$ behaves like $2n + 1$ , which is proportional to $n$ — so the answer is (B).

The Bohr model gives a beautifully simple picture of the hydrogen atom: electrons orbit the nucleus in fixed circular paths, with quantised angular momentum. The radius of the $n$ -th orbit is
$r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2} \quad \text{or, more compactly,} \quad r_n = a_0 n^2,$
where $a_0 \approx 0.529 \, \text{Å}$ is the Bohr radius. So the radius grows as $n^2$ .

The question asks about the distance between consecutive orbits — that is, the difference $r_{n+1} - r_n$ — for large $n$ . Many students instinctively think this difference is constant, or that it grows like $n^2$ because the radii themselves do. But a difference between two quadratic terms behaves differently from either term alone. Let’s work it out.

Write the radii for two neighbouring orbits:

$r_n = a_0 n^2, \qquad r_{n+1} = a_0 (n+1)^2.$

The gap between them is

$\Delta r = r_{n+1} - r_n = a_0 \left[ (n+1)^2 - n^2 \right].$

Expand $(n+1)^2 = n^2 + 2n + 1$ . Then

$\Delta r = a_0 \left( n^2 + 2n + 1 - n^2 \right) = a_0 (2n + 1).$

For large $n$ , the constant $1$ becomes negligible compared to $2n$ . So

$\Delta r \approx 2 a_0 n.$
Thus the spacing between consecutive orbits is proportional to $n$ itself — not $n^2$ , not $\sqrt{n}$ , not $n^3$ . …
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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
$Q. The radius of the lowest Bohr's orbit in hydrogen atom is r0. The radius of Bohr's second orbit is (A) r0 (B) 2r0 (C) 4r0 (D) r0/2$
›Reveal solutionSolution

Bohr radii scale as n², so r₂ = 4 r₀.

In Bohr's model of hydrogen the radius of the $n$ -th orbit is:

$r_n=n^2 r_0$

…

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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 markMCQ
$Q. If the first Bohr radius of hydrogen atom be R, then the radius of the third orbit is$
(a) 9R
(b) R/3
(c) 3R
(d) R/9
›Reveal solutionSolution

Bohr radius of the nth orbit scales as $n^2$ , so the third orbit's radius is $9R$ .

In the Bohr model, the radius of the $n$ th orbit of the hydrogen atom is

$r_n = n^2 r_1$

…

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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
$Q. Case study: Bohr's model addressed the instability of the Rutherford model by introducing quantization. The model is based on three postulates, which successfully explained the discrete line spectrum of hydrogen. According to the model, the radius of the nth stationary orbit is r_n \propto n², and the total energy is E_n = -13.6 eV / n², when an electron jumps from a higher energy level (E_i) to a lower one (E_f), a photon of energy hν = E_i - E_f is emitted. Transitions ending at the n = 1 level form the Lyman series. According to Bohr's second postulate, which quantity is quantized?$
(a) Energy of electron
(b) Orbital angular momentum
(c) Linear momentum
(d) Frequency of revolution
›Reveal solutionSolution

Bohr's second postulate quantizes the electron's orbital angular momentum in integer multiples of $h/2\pi$ .

Bohr's second postulate states that an electron can revolve only in those orbits for which its orbital angular momentum is an integral multiple of $\hbar = h/2\pi$ : …

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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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Part (a)

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