In the Auger process an atom makes a transition to a lower state without emitting a photon. The excess energy is transferred to an outer electron which may be ejected by the atom. (This is called an Auger electron.) Assuming the nucleus to be massive, calculate the kinetic energy of an $n = 4$ Auger electron emitted by Chromium by absorbing the energy from a $n = 2$ to $n = 1$ transition.

In the Auger process an atom makes a transition to a lower state without emitting a photon. The excess energy

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n = 4

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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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The $n=2\to n=1$ transition in Cr releases $5875.2$ eV; ejecting the $n=4$ electron costs its binding energy $489.6$ eV, leaving $K=5875.2-489.6=5385.6$ eV $\approx 5.39$ keV.

Concept understanding. In the Auger process the energy released when an electron drops into an inner vacancy is not radiated as a photon but handed directly to another (outer) electron, which is then ejected. Energy conservation gives

K = \Delta E_{2\to1} - |E_4|,

where $\Delta E_{2\to1}$ is the energy freed by the filling transition and $|E_4|$ is the binding energy of the ejected $n=4$ electron.

Hydrogen-like levels. Treating the nucleus as massive (so no reduced-mass correction) and using the hydrogenic formula for Chromium, $Z=24$ :

E_n = -\frac{Z^2(13.6\ \text{eV})}{n^2} = -\frac{576\times 13.6}{n^2}\ \text{eV}.

$E_1 = -576\times13.6 = -7833.6$ eV
$E_2 = -\dfrac{576\times13.6}{4} = -1958.4$ eV …

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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 15 on this concept.

KEAM 2026Set eng-2026-04174 marksMCQ
$Q. The ratio of the longest wavelength of the Lyman series to that of Ballmer series of the hydrogen spectrum is (A) 3:29 (B) 7:27 (C) 5:27 (D) 4:29 (E) 5:29$
›Reveal solutionSolution

The longest wavelength in each series comes from the smallest energy jump ( $2\to1$ for Lyman, $3\to2$ for Balmer). Their ratio is $5:27$ .

Rydberg formula: $\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$ .

Longest Lyman ( $n_1=1,\ n_2=2$ ):
$\frac{1}{\lambda_L} = R\left(1 - \frac14\right) = \frac{3R}{4} \quad\Rightarrow\quad \lambda_L = \frac{4}{3R}.$
Longest Balmer ( $n_1=2,\ n_2=3$ ): …
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KEAM 2026Set eng-2026-04184 marksMCQ
$Q. In hydrogen spectral series, the wave numbers of the first Lyman line and the first Balmer line are in the ratio (A) 1 : 2 (B) 2 : 1 (C) 27 : 5 (D) 5 : 27 (E) 16 : 1$
›Reveal solutionSolution

The first Lyman line has $\bar\nu=3R/4$ and the first Balmer line $\bar\nu=5R/36$ ; their ratio is $27:5$ .

Wave number $\bar\nu=R\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)$ .

First Lyman line ( $n_1=1,\ n_2=2$ ):
$\bar\nu_L=R\left(1-\frac{1}{4}\right)=\frac{3R}{4}$
First Balmer line ( $n_1=2,\ n_2=3$ ): …
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KEAM 2026Set pha-2026-0418F4 marksMCQ
$\frac{h}{\pi}$
›Reveal solutionSolution

For $n=2$ : $L = nh/2\pi = h/\pi$ and $E = -13.6/n^2 = -3.4\,eV$ . …

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KEAM 2026Set pha-2026-0419F4 marksMCQ
$Q. The energy of the electron in third excited state of the hydrogen atom is (A) -0.54 eV (B) -1.51 eV (C) -13.6 eV (D) -3.4 eV (E) -0.85 eV$
›Reveal solutionSolution

Third excited state is $n=4$ ; $E_4 = -13.6/16 = -0.85\ \mathrm{eV}$ .

Energy levels of hydrogen are $E_n = -\dfrac{13.6}{n^2}$ eV. The states are: ground $n=1$ , first excited $n=2$ , second excited $n=3$ , third excited $n=4$ . So …
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KEAM 2026Set pha-2026-0420F4 marksMCQ
$Q. Pick out the wrong statement about Bohr atom model: (A) Orbit of the electron is circular (B) Model is applicable only for single electron systems (C) Orbits of the electron are non-radiating (D) Angular momentum of electron in an orbit is quantized. (E) Model is applicable for many electron systems also$
›Reveal solutionSolution

Bohr's model succeeds only for single-electron (hydrogen-like) systems; claiming it applies to many-electron atoms is false.

The Bohr model correctly assumes circular, non-radiating (stationary) orbits with quantized angular momentum $L = n\hbar$ , and it works well for single-electron (hydrogen-like) systems. It fails for many-electron atoms because it ignores electron-electron interactions a …
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KEAM 2025Set eng-2025-04234 marksMCQ
$Q. The greatest wavelength of the radiation that will ionize unexcited hydrogen atom is (A) 1820 Å (B) 450 Å (C) 910 Å (D) 700 Å (E) 1400 Å$
›Reveal solutionSolution
The longest wavelength that can ionize ground-state hydrogen is about 910 Angstrom.

Concept and Intuition

Ionizing an unexcited hydrogen atom needs at least 13.6 eV. The greatest wavelength (least energy) corresponds exactly to this threshold energy through E = hc/lambda.

Step-by-Step Solution

Ionization energy E = 13.6 eV.

lambda = hc/E = 1240 eV*nm / 13.6 eV = 91.2 nm. …
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KEAM 2025Set eng-2025-04254 marksMCQ
$n_2 = \infty$
›Reveal solutionSolution

$n_1=3$ marks the Paschen series (infrared). The transition from $n_2=\infty$ releases the most energy, i.e. the series limit — the shortest wavelength of the Paschen series.

The lower level $n_1=3$ identifies the Paschen series. The wavelength satisfies $\dfrac{1}{\lambda}=R\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)$ . …
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KEAM 2025Set eng-2025-04264 marksMCQ
$n_2 = 5$
›Reveal solutionSolution

Shortest wavelength = largest energy jump = the series limit, so for the Brackett series ( $n_1 = 4$ ) it is the $\infty \to 4$ transition.

The wavenumber of a hydrogen line is
$\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$
Shortest wavelength corresponds to maximum $1/\lambda$ , i.e. maximum photon energy, obtained when $n_2 \to \infty$ . …
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KEAM 2025Set eng-2025-04284 marksMCQ
$\frac{36}{5R}\,m$
›Reveal solutionSolution

$\dfrac{1}{\lambda}=\dfrac{5R}{36}$ matches $R\left(\dfrac{1}{2^2}-\dfrac{1}{3^2}\right)$ , the $n=3\to n=2$ (Balmer) transition.

Given $\lambda=\dfrac{36}{5R}$ , the wavenumber is $\dfrac{1}{\lambda}=\dfrac{5R}{36}$ . Using the Rydberg formula
$\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right),$
try $n_1=2,\;n_2=3$ : …
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KEAM 2025Set pha-2025-0424F4 marksMCQ
$eV$
›Reveal solutionSolution

$-13.6+12.75=-0.85$ eV corresponds to $n=4$ . Number of spectral lines $=\dfrac{n(n-1)}{2}=6$ .

The energy of the $n$ th level is $E_n=-\dfrac{13.6}{n^2}$ eV. Absorbing $12.75$ eV from the ground state:
$E_n=-13.6+12.75=-0.85\ \text{eV}.$
Solving,

$-\frac{13.6}{n^2}=-0.85\;\Rightarrow\; n^2=16\;\Rightarrow\; n=4.$ …
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KEAM 2024Set eng-2024-06064 marksMCQ
$Q. The energy required to excite the hydrogen atom from its first excited state to second excited state is (A) 12.09 eV (B) 1.89 eV (C) 10.2 eV (D) 3.40 eV (E) 1.51 eV$
›Reveal solutionSolution

From $n=2$ to $n=3$ : $\Delta E=13.6\left(\dfrac{1}{4}-\dfrac{1}{9}\right)=1.89$ eV.

The first excited state is $n=2$ with $E_2=-3.40$ eV, and the second excited state is $n=3$ with $E_3=-1.51$ eV. …
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KEAM 2024Set eng-2024-06084 marksMCQ
$Q. Bohr atom model is invalid for (A) Hydrogen atom (B) doubly ionized helium atom (C) deuteron atom (D) singly ionized helium atom (E) doubly ionized lithium atom$
›Reveal solutionSolution

The Bohr model works only for single-electron species (H, He $^+$ , Li $^{2+}$ , deuterium). Doubly ionized helium He $^{2+}$ has zero electrons, so Bohr's orbit picture cannot apply.

The Bohr atomic model successfully describes hydrogen-like systems that contain exactly one electron: hydrogen, singly ionized helium (He $^+$ ), doubly ionized lithium (Li $^{2+}$ ), and deuterium (a hydrogen isotope, still one electron). Doubly ionized helium means both of helium's electrons have been removed, leaving He $^{2+}$ — a b …
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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: Combine the Three Energy Terms into One Algebraic Expression First

Steps

Concept and Intuition

Step-by-Step Solution

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