Show that the first few frequencies of light that is emitted when electrons fall to the $n$th level from levels higher than $n$, are approximate harmonics (i.e. in the ratio $1 : 2 : 3 \ldots$) when $n \gg 1$.

Show that the first few frequencies of light that is emitted when electrons fall to the nth level from levels

NCERT Exemplar · Q16

n

Gujarat GsebShort· 3mImportance★★★★★

73% · 37/51 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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}

Start your 14-day free trial to unlock the full solution →

For large $n$ , the energy difference between adjacent levels becomes nearly constant, so the emitted frequencies for transitions $n+p \to n$ ( $p=1,2,3,\dots$ ) are approximately in the ratio $1:2:3:\dots$ — i.e., they form a harmonic series.

The key insight comes from the Bohr model of the hydrogen atom. In this model, the energy of an electron in the $n$ th orbit is:

E_n = -\frac{13.6}{n^2} \ \text{eV}

When an electron jumps from a higher level $n+p$ (where $p$ is a positive integer) down to level $n$ , the energy of the emitted photon is:

\Delta E = E_{n+p} - E_n = 13.6 \left( \frac{1}{n^2} - \frac{1}{(n+p)^2} \right) \ \text{eV}

The frequency of the emitted light is proportional to this energy: $f \propto \Delta E$ .

Now, the question asks: what happens when $n$ is very large compared to $p$ ? That is, when the electron falls from a level just slightly above a very high $n$ ?

1. Rewrite the energy difference in a more revealing form

Start with:

\Delta E = 13.6 \left( \frac{1}{n^2} - \frac{1}{(n+p)^2} \right)

Combine the fractions:

\Delta E = 13.6 \left( \frac{(n+p)^2 - n^2}{n^2 (n+p)^2} \right)

The numerator simplifies:

(n+p)^2 - n^2 = n^2 + 2np + p^2 - n^2 = 2np + p^2

So:

\Delta E = 13.6 \left( \frac{2np + p^2}{n^2 (n+p)^2} \right)

2. Apply the condition $n \gg 1$ (and also $n \gg p$ )

When $n$ is very large compared to $p$ , two approximations become valid:

$n+p \approx n$
$p^2$ is negligible compared to $2np$ (since $p$ is small relative to $n$ )

Thus:

\Delta E \approx 13.6 \left( \frac{2np}{n^2 \cdot n^2} \right) = 13.6 \left( \frac{2p}{n^3} \right)

n \gg p

3. What does this tell us about frequencies?

Since $f \propto \Delta E$ , we have:

f_p \propto \frac{p}{n^3}

For a fixed large $n$ , the factor $1/n^3$ is constant. So the frequencies for $p = 1, 2, 3, \dots$ are:

f_1 : f_2 : f_3 : \dots = 1 : 2 : 3 : \dots

That is exactly the harmonic series. …

Start your 14-day free trial to unlock the full solution →

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 26 on this concept.

GUJCET 2026Set x1 markMCQ
$-1/2$
›Reveal solutionSolution

By the virial theorem for the Coulomb field, $KE=-\tfrac12 PE$ , giving the ratio $-\tfrac12$ .

For the electron in a hydrogen atom, with total energy $E$ :

$KE=-E,\qquad PE=2E$ …
Start your 14-day free trial to see the full solution →
GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$Q. In Hydrogen atom energy of electron in first excited state is ___.$
(a) -3.40 eV
(b) -1.51 eV
(c) -0.85 eV
(d) -13.6 eV
›Reveal solutionSolution

The Bohr model gives the energy of the nth stationary state of hydrogen as E_n = -13.6/n^2 eV; the first excited state corresponds to n = 2.

Ground state: n = 1, E_1 = -13.6 eV. …

Start your 14-day free trial to see the full solution →
GUJCET 2025Set 031 markMCQ
$5.3 \times 10^{-11}$
›Reveal solutionSolution
[!TLDR]

The orbital speed of the electron in the hydrogen atom is about $2.2\times10^{6}$ m/s.

Concept

For the hydrogen atom, the magnitude of the total (binding) energy equals the kinetic energy of the electron: $|E|=\tfrac12 mv^2 = 13.6$ eV in the ground state.

Solution

$KE=13.6\ \text{eV}=13.6\times1.6\times10^{-19}=2.176\times10^{-18}$ J.

From $\tfrac12 mv^2 = KE$ : …
Start your 14-day free trial to see the full solution →
GUJCET 2025Set 031 markMCQ
$-13.6$
›Reveal solutionSolution
[!TLDR] $PE = 2E = -27.2$ eV and $KE = -E = +13.6$ eV.

Concept

In Bohr's hydrogen atom the electron's kinetic and potential energies relate to the total energy $E$ by $KE = -E$ and $PE = 2E$ (a consequence of the virial theorem for the Coulomb potential: $PE = -2\,KE$ , and $E = KE + PE = -KE$ ).

Solution

Given total (ground-state) energy $E = -13.6$ eV: …
Start your 14-day free trial to see the full solution →
GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$Q. Potential energy of an electron in the first excited state in a hydrogen atom is ______ eV.$
(a) -3.4
(b) -6.8
(c) -10.2
(d) -13.6
›Reveal solutionSolution

For the Bohr hydrogen atom, total energy En = −13.6/n² eV and potential energy PE = 2En (kinetic energy KE = −En).

…

Start your 14-day free trial to see the full solution →
GUJCET 2024Set 131 markMCQ
$-13.6$
›Reveal solutionSolution

In the Bohr atom the potential energy is twice the total energy: $U = 2E = 2(-13.6) = -27.2$ eV.

Concept. For a bound electron in the hydrogen atom the total energy $E$ , kinetic energy $K$ , and potential energy $U$ satisfy $K = -E$ and $U = 2E$ (virial theorem). …
Start your 14-day free trial to see the full solution →
GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$Q. Kinetic energy of electron in one of the orbit of hydrogen atom is x then its total energy is ___.$
(a) -x
(b) -x/2
(c) -2x
(d) -x/8
›Reveal solutionSolution

For an electron in a Bohr orbit, PE = −2KE, so Total Energy = KE + PE = KE − 2KE = −KE.

…

Start your 14-day free trial to see the full solution →
GUJCET 2023Set 091 markMCQ
$5438\,\text{\AA}$
›Reveal solutionSolution

Longest Balmer wavelength = smallest energy jump = $n=3\to2$ ( $H_\alpha$ ) $\approx 6563$ Å.

Concept — Balmer series. All Balmer transitions end at $n=2$ . Longest wavelength corresponds to the smallest energy difference, i.e. the transition from the nearest upper level $n=3\to n=2$ ( $H_\alpha$ ).

$\frac{1}{\lambda} = R\left(\frac{1}{2^2}-\frac{1}{3^2}\right) = R\left(\frac14-\frac19\right) = R\cdot\frac{5}{36}$ …
Start your 14-day free trial to see the full solution →
GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. What is the shortest wavelength present in the Balmer series of spectral lines? [Rydberg's constant R = 1.097 x 10^7 m^-1]$
(a) 365 nm
(b) 91 nm
(c) 26 nm
(d) 820 nm
›Reveal solutionSolution

The shortest Balmer wavelength (series limit, n -> infinity) satisfies 1/lambda = R(1/4 - 0) = R/4, giving lambda = 4/R = 365 nm.

Balmer series: 1/lambda = R(1/2^2 - 1/n^2). The shortest wavelength is the series limit (n -> infinity): …

Start your 14-day free trial to see the full solution →
GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. What is the ratio of total energy of an electron in hydrogen atom in first excited state and third excited state?$
(a) 4 : 1
(b) 3 : 1
(c) 1 : 1
(d) 1 : 4
›Reveal solutionSolution
First excited state = n=2, third excited state = n=4; since E_n proportional to -1/n^2, the ratio is (1/4)/(1/16) = 4:1.

Energy levels: E_n = -13.6/n^2 eV, so |E_n| proportional to 1/n^2.

First excited state: n = 2 (the state just above the ground state n = 1). …
Start your 14-day free trial to see the full solution →
GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQ
$Q. Which of the following spectral series lies in the ultraviolet region?$
(a) Paschen series
(b) Balmer series
(c) Lyman series
(d) Pfund series
›Reveal solutionSolution
Among the hydrogen spectral series, only the Lyman series lies in the ultraviolet.

The hydrogen spectral series are grouped by the final energy level (n_f) the electron falls to:

Lyman series (n_f = 1): all wavelengths in the ultraviolet.

Balmer series (n_f = 2): mostly in the visible region.

Paschen series (n_f = 3), Brackett (n_f = 4), Pfund (n_f = 5): all in the infrared. …
Start your 14-day free trial to see the full solution →
GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQ
$Q. The dimensional formula of m e^4 / (8 ε0^2 h^3 c) is ______.$
(a) M^0 L^-1 T^0
(b) M^0 L^1 T^0
(c) M^0 L^0 T^0
(d) M^-1 L^0 T^0
›Reveal solutionSolution

The quantity me⁴/(8ε0²h³c) is exactly the Rydberg constant from Bohr's theory, whose SI unit is m⁻¹.

Bohr's model gives the wave number of emitted radiation as:

$\bar{\nu} = \dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)$ , where $R = \dfrac{me^4}{8\varepsilon_0^2h^3c}$

…

Start your 14-day free trial to see the full solution →

←Previous questionQ15 — Using Bohr model, calculate the electric current created by the electr…→Next questionQ17 — What is the minimum energy that must be given to a H atom in ground st…

Q.Show that the first few frequencies of light that is emitted when electrons fall to the $n$ th level from levels higher than $n$ , are approximate harmonics (i.e. in the ratio $1 : 2 : 3 \ldots$ ) when $n \gg 1$ .

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

1. Rewrite the energy difference in a more revealing form

2. Apply the condition $n \gg 1$ (and also $n \gg p$ )

3. What does this tell us about frequencies?

Method: Continuum (Differential) Approximation for Large $n$

Steps

Concept

Solution

Concept

Solution

Q.Show that the first few frequencies of light that is emitted when electrons fall to the nnnth level from levels higher than nnn, are approximate harmonics (i.e. in the ratio 1:2:3…1 : 2 : 3 \ldots1:2:3…) when n≫1n \gg 1n≫1.

1. Rewrite the energy difference in a more revealing form

2. Apply the condition n≫1n \gg 1n≫1 (and also n≫pn \gg pn≫p)

3. What does this tell us about frequencies?

🎓Unlock everything free for 14 days

Q.Show that the first few frequencies of light that is emitted when electrons fall to the $n$ th level from levels higher than $n$ , are approximate harmonics (i.e. in the ratio $1 : 2 : 3 \ldots$ ) when $n \gg 1$ .

2. Apply the condition $n \gg 1$ (and also $n \gg p$ )

Unlock everything free for 14 days