The first four spectral lines in the Lyman series of a H-atom are $\lambda = 1218\ \text{\AA}, 1028\ \text{\AA}, 974.3\ \text{\AA}$ and $951.4\ \text{\AA}$. If instead of Hydrogen, we consider Deuterium, calculate the shift in the wavelength of these lines.

The first four spectral lines in the Lyman series of a H-atom are λ = 1218AA, 1028AA, 974.3AA and 951.4AA. If

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\lambda = 1218\ \text{\AA}, 1028\ \text{\AA}, 974.3\ \text{\AA}

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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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Because deuterium's nucleus is about twice as heavy, its reduced mass (and Rydberg constant) is slightly larger, so each Lyman line shifts to a shorter wavelength by the fixed fraction $m_e/(2m_p) \approx 2.72\times10^{-4}\;(0.0272\%)$ . The shift is linear in $\lambda$ , giving $\approx 0.33,\ 0.28,\ 0.27,\ 0.26\,\text{Å}$ .

1. Why the shift happens.

The electron and nucleus both orbit their common centre of mass, so the effective mass is the reduced mass $\mu = \dfrac{m_e M}{m_e + M}$ . The Rydberg constant of a one-electron atom is

R = R_\infty\,\frac{\mu}{m_e} = R_\infty\,\frac{M}{m_e + M}.

For hydrogen $M = m_p$ ; for deuterium $M \approx 2m_p$ . The heavier deuteron gives a larger $\mu$ , hence a larger $R$ , hence a shorter wavelength for the same transition.

2. Wavelength is inversely proportional to $R$ .

For any line, $\dfrac{1}{\lambda} \propto R$ , so $\lambda \propto 1/\mu$ and

\frac{\Delta\lambda}{\lambda} = \frac{\lambda_D-\lambda_H}{\lambda_H} = \frac{\mu_H}{\mu_D} - 1.

3. Evaluate the fractional shift.

With $\mu \approx m_e\left(1 - \dfrac{m_e}{M}\right)$ ,

$\left|\frac{\Delta\lambda}{\lambda}\right| \approx \frac{m_e}{M_H} - \frac{m_e}{M_D} = \frac{m_e}{m_p} - \frac{m_e}{2m_p} = \frac{m_e}{2m_p} \approx \frac{1}{2\times1836} \approx 2.72\times10^{-4}.$ …

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

Higher Secondary (+2 Stage) Examination 2026Set ANNUAL1 markMCQ
$Q. The energy of an electron in the ground state of a hydrogen atom is -13.6 eV. What will be the energy of an electron in the third orbit of the same atom?$
(a) -1.51 eV
(b) -3.40 eV
(c) -4.53 eV
(d) -6.80 eV
›Reveal solutionSolution

The energy of the nth Bohr orbit of hydrogen is En = -13.6/n² eV; putting n = 3 gives -1.51 eV.

For a hydrogen atom, the energy of the electron in the nth orbit is

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

For the third orbit, n = 3: …

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Higher Secondary (+2 Stage) Examination 2026Set ANNUAL1 mark
$Q. A hydrogen atom is in its third excited state. What is the maximum number of spectral lines that can be emitted by this atom?$
›Reveal solutionSolution

The third excited state of hydrogen is n = 4 (ground = n1, then 1st, 2nd, 3rd excited states are n = 2, 3, 4); the number of possible spectral lines from n down to 1 is n(n−1)/2 = 6.

For hydrogen, the ground state is n = 1. The 1st excited state is n = 2, the 2nd excited state is n = 3, and the 3rd excited state is n = 4. From the n = 4 state, an electron can drop to any of the lower levels via all possible transit …

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Higher Secondary (+2 Stage) Examination 2025Set ANNUAL1 markMCQ
$Q. If the total energy of an electron in the first excited state of a hydrogen-like atom is -3.4 eV, then the kinetic energy and potential energy of the electron in that energy level are respectively —$
(a) -1.7 eV, -1.7 eV
(b) -3.4 eV, 0 eV
(c) 3.4 eV, -6.8 eV
(d) -6.8 eV, 3.4 eV
›Reveal solutionSolution

For an electron bound by a Coulomb (1/r) force, PE = -2xKE and Total Energy = -KE (virial theorem), so from E = -3.4 eV we get KE = +3.4 eV and PE = -6.8 eV.

For an electron orbiting under an inverse-square (Coulomb) force, the virial theorem gives:

$PE = -2\times KE$

Total energy: $E = KE + PE = KE - 2KE = -KE$ , so $KE = -E$ .

…

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Higher Secondary (+2 Stage) Examination 2025Set ANNUAL1 mark
$Q. Which series of the hydrogen spectrum lies in the visible region?$
›Reveal solutionSolution

Of all the hydrogen spectral series, only the Balmer series (transitions ending at n=2) gives lines in the visible range (~400-700 nm); the others fall in the UV or IR.

The hydrogen spectrum consists of several series depending on which energy level the electron falls to: Lyman (falls to n=1, ultraviolet), Balmer (falls to n=2), and Paschen, Brackett, Pfund (fall to n=3,4,5 -- all in the infrared). Only the Balmer series's wavele …

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Higher Secondary (+2 Stage) Examination 2023Set ANNUAL1 mark
$Q. The total energy of an electron in the first Bohr orbit of a hydrogen atom is -13.6 eV. What is the kinetic energy of this electron?$
›Reveal solutionSolution

For an electron bound by the Coulomb force in a Bohr orbit, total energy = −(kinetic energy), so if total energy is −13.6 eV, kinetic energy is +13.6 eV.

For an electron revolving in a Bohr orbit under the Coulomb attraction of the nucleus, the potential energy is twice the negative of the kinetic energy (a consequence of the virial theorem for an inverse-square force):

PE = −2 KE

…

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Q.The first four spectral lines in the Lyman series of a H-atom are $\lambda = 1218\ \text{\AA}, 1028\ \text{\AA}, 974.3\ \text{\AA}$ and $951.4\ \text{\AA}$ . If instead of Hydrogen, we consider Deuterium, calculate the shift in the wavelength of these lines.

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

Method: The Universal Isotope-Shift Constant (Memorised Shortcut)

Steps

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