Let $E_n = -\dfrac{m e^4}{8 n^2 \varepsilon_0^2 h^2}$ be the energy of the $n$th level of H-atom. If all the H-atoms are in the ground state and radiation of frequency $(E_2 - E_1)/h$ falls on it, (a) it will not be absorbed at all. (b) some of atoms will move to the first excited state. (c) all atoms will be excited to the $n = 2$ state. (d) no atoms will make a transition to the $n = 3$ state.

Let Eₙ = -(m e⁴)/(8 n² varepsilon₀² h²) be the energy of the nth level of H-atom. If all the H-atoms are in th

NCERT Exemplar · Q28

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

(a) it will not be absorbed at all.

(b) some of atoms will move to the first excited state.

n = 2

state.

(d) no atoms will make a transition to the

n = 3

state.

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Concept understanding — Photon Energy

Photon Energy: The Energy Carried by Light

Light does not carry its energy in a continuous stream. It comes in discrete packets — like individual drops instead of a steady hose. Each packet is called a photon: the smallest possible unit of light of a given frequency. You cannot have half a photon; it is all or nothing.

The Intuition: Colour Determines Energy

Red light and blue light are both light, but they behave differently — blue light causes sunburn and drives chemical reactions more readily than red. The reason is that the colour (frequency) of light directly fixes how much energy each photon carries, and blue photons carry more energy than red photons.

The Planck–Einstein Relation

The energy $E$ of a single photon is directly proportional to its frequency $f$ , and inversely proportional to its wavelength $\lambda$ :

E = h f = \frac{h c}{\lambda}

Where:

$E$ = energy of one photon (joules, J)
$f$ = frequency (hertz, Hz)
$\lambda$ = wavelength (metres, m)
$c$ = speed of light in vacuum ( $3.00 \times 10^8$ m/s)
$h$ = Planck's constant ( $6.626 \times 10^{-34}$ J·s)

Important

Planck's constant $h$ is extremely tiny, so an individual photon carries a very small amount of energy — which is why we never feel single photons striking our skin.

What This Tells You

Higher frequency = higher energy. Gamma rays have extremely energetic photons; radio waves have very low-energy photons.
Shorter wavelength = higher energy. Ultraviolet photons are more energetic than visible-light photons; infrared photons are less.
Energy is quantised. Light energy comes in fixed packets — you can have 1, 2, or 1000 photons, but never 0.5. This was Max Planck's revolutionary idea of 1900.

This is why UV light causes sunburn (UV photons carry enough energy to damage DNA, while visible photons do not), why X-ray photons are energetic enough to pass through soft tissue but get absorbed by bone, and why a photon needs enough energy to actually break a bond or drive a photochemical/biological reaction, rather than merely wavelength-dependent intensity.

A Concrete Example

Note

Question: Which has more energy — a photon of red light ( $\lambda = 700$ nm) or a photon of blue light ( $\lambda = 450$ nm)?

Since $E = \dfrac{hc}{\lambda}$ , energy is inversely proportional to wavelength, so the shorter-wavelength blue photon carries more energy:

Red: $E_{\text{red}} = \dfrac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{700 \times 10^{-9}} \approx 2.84 \times 10^{-19}$ J …

Why this formula?

Why Photon Energy is $E = hf$ — The Reasoning Behind the Formula

The formula $E = hf$ is not something you memorise and plug numbers into. It comes from a deep shift in how physicists understood light.

The problem that forced the formula

By the late 1800s, classical physics said light was a wave — continuous, carrying energy in proportion to its intensity (brightness). But blackbody radiation and the photoelectric effect showed something bizarre: when you shine light on a metal, electrons are ejected only if the light's frequency is above a certain threshold, no matter how bright the light is. Below that frequency, even the brightest lamp won't kick out a single electron. Classically this made no sense — a wave's energy depends on amplitude, not frequency.

Einstein's radical idea (1905)

Einstein proposed that light comes in discrete packets — photons — each carrying a fixed energy that depends only on its frequency, with Planck's constant $h$ as the proportionality:

A single photon's energy is $E = hf$ , where $f$ is the frequency of the light.

This directly explains the photoelectric effect: an electron needs a minimum energy (the work function $\phi$ ) to escape. If a photon's energy $hf < \phi$ , no electron is ejected — regardless of how many photons you send, because each photon interacts with one electron individually.

Important

$E = hf$ is not derived from deeper principles — it is a postulate justified by the experiments it explains, and it unified optics with quantum mechanics.

The equivalent forms

Since $c = f\lambda$ for light in vacuum, the same idea in terms of wavelength is:

E = \frac{hc}{\lambda}

Use this when you're given $\lambda$ instead of $f$ . A photon also carries momentum (despite having no mass): …

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The photon energy exactly matches the $n=1 \to n=2$ transition energy, so some atoms absorb it and jump to the first excited state. Not all atoms absorb because absorption is probabilistic, and the $n=3$ state is not reachable with this single photon energy.

The key here is to understand what happens when light of a specific frequency shines on a collection of hydrogen atoms. The energy of a photon is $E = h\nu$ . The problem gives the frequency as $\nu = (E_2 - E_1)/h$ , so the photon energy is exactly $E_2 - E_1$ .

This is the textbook condition for resonant absorption. An electron in the ground state ( $n=1$ ) can absorb this photon only if the photon’s energy precisely matches the gap to some higher energy level. Here, the gap to $n=2$ is exactly $E_2 - E_1$ , so absorption is possible.

But does every atom absorb? No. Absorption is a quantum mechanical process — it’s probabilistic. When a beam of photons passes through a gas of atoms, each atom has a certain probability of absorbing a photon. Some atoms will absorb and get excited; others will not. So we never get “all atoms” excited unless the light is extremely intense and the interaction time is long enough — and even then, stimulated emission competes. The question’s phrasing “all atoms” is a trap.

Also, can an atom in the ground state absorb this photon and jump to $n=3$ ? No — the energy required for $n=1 \to n=3$ is $E_3 - E_1$ , which is larger than $E_2 - E_1$ . A single photon cannot supply that energy unless its frequency is higher. So option (D) is actually correct: no atom can go to $n=3$ with this photon.

Let’s go through the options step by step.

Photon energy matches exactly one transition.

The photon energy is $h\nu = E_2 - E_1$ . This is the precise energy difference between the ground state ( $n=1$ ) and the first excited state ( $n=2$ ). So absorption is energetically allowed.
Not all atoms absorb — absorption is probabilistic.

When a beam of photons passes through a sample, each atom has a certain cross-section for absorption. Some atoms will absorb a photon and excite to $n=2$ ; others will not. The fraction that absorbs depends on the intensity and duration of exposure, but it is never “all” in a typical scenario unless specified otherwise. Option (C) says “all atoms will be excited” — that is false.
The $n=3$ state is unreachable with this photon.

The energy needed to go from $n=1$ to $n=3$ is $E_3 - E_1$ . Since $E_3$ is higher than $E_2$ , this gap is larger than $E_2 - E_1$ . A single photon of energy $E_2 - E_1$ cannot bridge that gap. So no atom can transition to $n=3$ via this photon. Option (D) is correct. …

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

KCET 2024Set D-21 markMCQ
$1:16$
›Reveal solutionSolution
$r\propto n^2 \Rightarrow$ area $\propto n^4$ ; for $n=2$ vs $n=1$ that is $2^4:1^4 = 16:1$ .

Step 1 — Bohr's radius formula

For a hydrogen-like atom, Bohr's quantisation of angular momentum ( $mvr = n\hbar$ ) combined with the Coulomb-force–centripetal-force balance gives
$r_n=\frac{n^{2}h^{2}\varepsilon_{0}}{\pi m e^{2}Z}\qquad\Longrightarrow\qquad r_n\propto n^{2}\ \ (\text{for fixed }Z)$
For hydrogen ( $Z=1$ ), $r_n = n^2 a_0$ with $a_0 = 0.529\ \text{Å}$ .

Step 2 — From radius to area

The orbit is a circle, so the area enclosed is
$A_n=\pi r_n^{2}$
Since $r_n \propto n^2$ ,
$A_n \propto \left(n^{2}\right)^{2} = n^{4}$
This $n^4$ (not $n^2$ ) is the crux of the question — the squaring of an already-squared quantity.

Step 3 — Identify the two states

Ground state: $n=1$ .

First excited state: the next level up, $n=2$ . (Not $n=3$ — the "first excited" state is the first level above the ground state.)

Step 4 — Take the ratio …
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COMEDK 2024Set 2024-M1 markMCQ
$13.75 \mathrm{~eV}$
›Reveal solutionSolution
Using $E=hc/\lambda$ , the 13.75 eV energy difference corresponds to a wavelength of 90 nm, i.e. 900 Å — matching option (D).

Step-by-step reasoning

Convert the energy to joules.

$E = 13.75\ \text{eV} \times 1.6\times10^{-19}\ \text{J/eV} = 2.2\times10^{-18}\ \text{J}$

Solve for wavelength.

$\lambda = \frac{hc}{E} = \frac{(6.6\times10^{-34})(3\times10^8)}{2.2\times10^{-18}} = \frac{19.8\times10^{-26}}{2.2\times10^{-18}} = 9.0\times10^{-8}\ \text{m}$

Convert to convenient units.

$9.0\times10^{-8}\ \text{m} = 90\ \text{nm} = 900\ \text{Å} \quad (\text{since } 1\ \text{nm}=10\ \text{Å})$
This matches the option listing 900 Å. …
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KCET 2022Set B-31 markMCQ
$0.53~\overset{\circ}{A}$
›Reveal solutionSolution

Bohr radii go as $n^2$ ; the radius has grown by a factor of 4, so $n^2 = 4$ and the atom is excited to $n = 2$ .

1. The Bohr radius law

Quantising the angular momentum ( $mvr = nh/2\pi$ ) and balancing the Coulomb force against the centripetal requirement gives, for a hydrogen-like atom,
$r_n = \frac{n^{2}h^{2}\varepsilon_0}{\pi m e^{2} Z} \;=\; \frac{n^{2}}{Z}\,a_0$
where $a_0 = 0.53\ \text{Å}$ is the Bohr radius. For hydrogen $Z = 1$ , so simply
$\boxed{\,r_n = n^{2}\,a_0\,}$
The radius grows as the square of the principal quantum number — that quadratic dependence is the whole content of the problem.

2. Set up the ratio

Ground state: $r_1 = 0.53\ \text{Å}$ (with $n=1$ , consistent with the formula).

Final state after the collision: $r_n = 2.12\ \text{Å}$ .

Taking the ratio kills $a_0$ entirely:
$\frac{r_n}{r_1} = \frac{n^{2}a_0}{1^{2}a_0} = n^{2}$
3. Solve
$n^{2} = \frac{2.12}{0.53} = 4$ $n = \sqrt{4} = 2$
4. Physical reading …
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KCET 2019Set A-11 markMCQ
$n^{th}$
›Reveal solutionSolution
The frequency of revolution of an electron in the $n^{th}$ orbit of a hydrogen atom is proportional to $\frac{1}{n^3}$ , making option (D) correct.

The key here is to connect the frequency of revolution — how many times per second the electron circles the nucleus — to the orbital radius and velocity. In Bohr's model, the electron moves in a circular orbit under electrostatic attraction, and its angular momentum is quantized. Frequency is simply $v/(2\pi r)$ , so if we find how $v$ and $r$ depend on $n$ , we can combine them.

Let's work through it step by step.

Write the force balance for a stable orbit. The centripetal force is provided by the Coulomb attraction between the electron and the proton:

$\frac{mv^2}{r} = \frac{ke^2}{r^2}$
where $m$ is the electron mass, $v$ its speed, $r$ the orbit radius, $k = 1/(4\pi\epsilon_0)$ , and $e$ the elementary charge.

Apply Bohr's quantization condition. Angular momentum is an integer multiple of $\hbar = h/(2\pi)$ :

$mvr = n\hbar$

Solve for $r$ in terms of $n$ . From the quantization condition, $v = n\hbar/(mr)$ . Substitute into the force equation:

$\frac{m}{r} \left( \frac{n\hbar}{mr} \right)^2 = \frac{ke^2}{r^2}$
Simplify:
$\frac{n^2 \hbar^2}{m r^3} = \frac{ke^2}{r^2}$
Multiply both sides by $r^3$ :
$\frac{n^2 \hbar^2}{m} = ke^2 r$
So:
$r = \frac{n^2 \hbar^2}{m k e^2}$
This shows $r \propto n^2$ .

Find $v$ in terms of $n$ . From $mvr = n\hbar$ , we have $v = n\hbar/(mr)$ . Substitute $r \propto n^2$ :

$v \propto \frac{n}{n^2} = \frac{1}{n}$
So $v \propto 1/n$ .

Now compute the frequency of revolution. Frequency $f$ is the number of orbits per second: $f = \frac{v}{2\pi r}$ …
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←Previous questionQ27 — The Balmer series for the H-atom can be observed (a) if we measure the…→Next questionQ29 — The simple Bohr model is not applicable to He⁴ atom because (a) He⁴ is…

Q.Let $E_n = -\dfrac{m e^4}{8 n^2 \varepsilon_0^2 h^2}$ be the energy of the $n$ th level of H-atom. If all the H-atoms are in the ground state and radiation of frequency $(E_2 - E_1)/h$ falls on it,

Concept understanding — Photon Energy

Photon Energy: The Energy Carried by Light

The Intuition: Colour Determines Energy

The Planck–Einstein Relation

What This Tells You

A Concrete Example

Why this formula?

Why Photon Energy is $E = hf$ — The Reasoning Behind the Formula

The problem that forced the formula

Einstein's radical idea (1905)

The equivalent forms

Method: The Discrete-Resonance Decision Procedure

Steps

Step 1 — Bohr's radius formula

Step 2 — From radius to area

Step 3 — Identify the two states

Step 4 — Take the ratio …

1. The Bohr radius law

2. Set up the ratio

3. Solve

4. Physical reading …

Q.Let En=−me48n2ε02h2E_n = -\dfrac{m e^4}{8 n^2 \varepsilon_0^2 h^2}En​=−8n2ε02​h2me4​ be the energy of the nnnth level of H-atom. If all the H-atoms are in the ground state and radiation of frequency (E2−E1)/h(E_2 - E_1)/h(E2​−E1​)/h falls on it,

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Q.Let $E_n = -\dfrac{m e^4}{8 n^2 \varepsilon_0^2 h^2}$ be the energy of the $n$ th level of H-atom. If all the H-atoms are in the ground state and radiation of frequency $(E_2 - E_1)/h$ falls on it,

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