JEE Mains Physics · Atoms

Transition Energies, Excitation and X-rays

A jump between levels gives a photon carrying the energy gap; a photon is absorbed only if it matches a gap, a sample in level n gives n(n − 1)/2 lines, and the photon's momentum makes the atom recoil.

All Atoms notes JEE Mains Physics strategy

Why this matters

Twenty-seven PYQs, seventeen of them multiple choice, and one from 2026. Nine turn a jump into a photon's energy, wavelength or frequency, two of them read off an energy-level diagram. Twelve are about excitation: six count the spectral lines a sample emits, and six find a level, an energy or an atomic number from the energy taken in or given out. Six are about photon momentum: three find the recoil of the emitting atom, and three are about X-rays.

Concept 1 of 3: Photon energy and wavelength of a transition

When the electron falls from one level to another, the photon carries away exactly the gap between them. Work out the gap in electron-volts, then turn it into a wavelength with hc = 1240 eV nm. The gaps shrink as you go up the ladder, so the biggest photon energies, and the highest frequencies, come from jumps that end on the ground state.

Definition

Gap: $\Delta E = 13.6\,Z^2\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right)$ eV.
Wavelength: $\lambda\,(\text{nm}) = \dfrac{1240}{\Delta E\,(\text{eV})}$ . If a question gives its own h or hc (1242, 1245, or $h = 4 \times 10^{-15}$ eV s, which with $c = 3 \times 10^{8}$ m/s makes hc = 1200 eV nm), use that value.
Frequency: $\nu = \dfrac{\Delta E}{h}$ .
Hydrogen's gaps shrink upward: 2 → 1 is 10.2 eV, 3 → 2 is 1.89 eV, 4 → 3 is 0.66 eV. Among single jumps, the one nearest the ground state has the largest frequency.
Which level did a photon come from? Set $\dfrac{1}{\lambda} = RZ^2\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right)$ and solve for $n_i$ .
Ratio of orbit radii given: $r \propto n^2$ , so take square roots to find the two n values before working out the gap.

Energy and wavelength of a transition

\Delta E = 13.6\,Z^2\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)\ \text{eV}, \qquad \lambda\,(\text{nm}) = \frac{1240}{\Delta E\,(\text{eV})}

Worked example

The electron in He⁺ falls from n = 3 to n = 2. Find the photon's energy, wavelength and frequency. (hc = 1240 eV nm,

h = 4.14 \times 10^{-15}

eV s)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 24 Jan 2023 · Q97Moderate

Example 1 · Atoms · Transition Energies, Excitation and X-rays

A photon is emitted in transition from

n = 4

n = 1

level in hydrogen atom. The corresponding wavelength for this transition is (given,

h = 4 \times10^{- 15}eVs

Use the constants the question gives

hc = 1240 eV nm is the default, but a question that states h = 4 × 10⁻¹⁵ eV s means hc = 1200 eV nm, and its answer is built on that. Options are often 2–3% apart.

The gap scales as Z²

The same jump in He⁺ releases 4 times the energy it does in hydrogen, and in Li²⁺ 9 times. Leaving Z out treats every ion as hydrogen.

Big frequency means a jump near the ground state

The 2 → 1 gap is larger than any jump between higher levels, such as 5 → 4. The higher the levels, the closer they sit, the smaller the photon.

Drill 8 more on photon energy and wavelength of a transition

Concept 2 of 3: Excitation, absorption and the number of spectral lines

A photon is all or nothing: it is absorbed only if its energy matches a gap exactly. An electron hitting the atom can hand over just part of its energy, so it lifts the atom to the highest level it can afford. Once a sample of atoms is excited to level n, different atoms fall by different routes, and every pair of levels gives its own line.

Definition

Lines from a sample excited to level n: $N = \dfrac{n(n - 1)}{2}$ . A single atom gives at most n − 1 photons on its way down.
Hydrogen's gaps from the ground state: n = 2: 10.2 eV; n = 3: 12.09 eV; n = 4: 12.75 eV; n = 5: 13.06 eV; ionisation: 13.6 eV.
Photon: absorbed only if hν equals one of these gaps exactly, or exceeds 13.6 eV (then it ionises).
Electron of energy K: lifts the atom to the highest level whose gap is at most K.
To see any Balmer line, an atom must first reach n = 3 or higher, because a Balmer line ends on n = 2.
Franck–Hertz: the current dips when the accelerating voltage reaches the first excitation energy, and the excited atoms then emit the 2 → 1 line.
Capture: a free electron of kinetic energy K caught into level n gives a photon of $K + |E_n|$ .
Atomic number from a gap: set $13.6\,Z^2\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right)$ equal to the given gap and solve for Z.

Lines from level n and the gap from the ground state

N = \frac{n(n-1)}{2}, \qquad E_n - E_1 = 13.6\,Z^2\left(1 - \frac{1}{n^2}\right)\ \text{eV}

Worked example

Hydrogen gas in its ground state is bombarded with electrons of energy 13.0 eV. Which is the highest level reached, and how many spectral lines are emitted? What changes if photons of 13.0 eV are used instead?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 12 Apr 2023 · Q18Moderate

Example 2 · Atoms · Transition Energies, Excitation and X-rays

12.5eV

electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be:

A photon must match; an electron need not

A 12.5 eV photon passes through ground-state hydrogen untouched, because no gap equals 12.5 eV. A 12.5 eV electron excites it to n = 3 and keeps the rest of its energy.

A sample and a single atom count differently

A sample excited to level n shows n(n − 1)/2 lines, because different atoms fall by different routes. One atom falling step by step gives at most n − 1 photons.

Balmer lines need n = 3 first

Lifting the atom to n = 2 is not enough: from there it can only fall to n = 1, a Lyman line. The least energy for any Balmer line is the gap to n = 3.

Drill 11 more on excitation, absorption and the number of spectral lines

Concept 3 of 3: Recoil of the emitting atom and X-ray photons

A photon of energy E carries momentum E/c. When an atom emits one, the atom is pushed back with the same momentum, but because the atom is heavy its recoil speed is only a few metres per second. X-rays are the same physics at thousands of electron-volts: an electron accelerated through V can give at most eV to one photon, which fixes the shortest wavelength, and a vacancy in an inner shell gives the target's own lines.

Definition

Photon momentum: $p = \dfrac{E}{c} = \dfrac{h}{\lambda}$ .
Recoil speed: $v = \dfrac{E}{Mc}$ , where M is the mass of the whole atom, not of the electron.
The recoil takes a tiny share of the gap: the photon's fractional loss of energy, and so its fractional increase of wavelength, is about $\dfrac{E}{2Mc^2}$ .
X-ray cut-off: an electron accelerated through V gives at most eV to one photon, so $\lambda_{\min} = \dfrac{hc}{eV}$ , or $\lambda_{\min}\,(\text{nm}) = \dfrac{1240}{V\,(\text{volts})}$ . It depends only on V, not on the target.
Electrons given by their de Broglie wavelength: find their kinetic energy from $\lambda = \dfrac{h}{\sqrt{2mK}}$ , then $\lambda_{\min} = \dfrac{hc}{K}$ .
Characteristic Kα line: an L electron fills a hole in the K shell. Photon energy = energy of the atom with a K hole − energy with an L hole.

Photon momentum, recoil and the X-ray cut-off

p = \frac{E}{c} = \frac{h}{\lambda}, \qquad v_{\text{recoil}} = \frac{E}{Mc}, \qquad \lambda_{\min} = \frac{hc}{eV}

Worked example

A hydrogen atom of mass

1.67 \times 10^{-27}

kg emits the photon of the 4 → 1 jump (12.75 eV). Find the photon's momentum and the atom's recoil speed. (

1\ \text{eV} = 1.6 \times 10^{-19}

c = 3 \times 10^{8}

m/s)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 29 January 2024 · Q21Moderate

Example 3 · Atoms · Transition Energies, Excitation and X-rays

When a hydrogen atom going from

n= 2

n= 1

emits a photon, its recoil speed is

\frac{X}{5}\text{ }m/s

. Where

x =

______ . (Use: mass of hydrogen atom

= 1.6 \times10^{- 27}\text{ }kg

)

Recoil uses the atom's mass

The whole atom recoils, so divide the photon's momentum by the atom's mass, about 1.67 × 10⁻²⁷ kg for hydrogen. Dividing by the electron's mass gives a speed nearly 2000 times too large.

The cut-off wavelength ignores the target

λ_min = hc/eV depends only on the tube voltage. The target metal sets the characteristic lines such as Kα, not the cut-off.

Kα energy is a difference of two hole energies

The photon carries the energy of the K-hole state minus the energy of the L-hole state. Taking the K-hole energy alone as the photon energy is the usual slip.

Drill 5 more on recoil of the emitting atom and x-ray photons

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Photon energy and wavelength of a transition
Energy and wavelength of a transition
$\Delta E = 13.6\,Z^2\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)\ \text{eV}, \qquad \lambda\,(\text{nm}) = \frac{1240}{\Delta E\,(\text{eV})}$
Excitation, absorption and the number of spectral lines
Lines from level n and the gap from the ground state
$N = \frac{n(n-1)}{2}, \qquad E_n - E_1 = 13.6\,Z^2\left(1 - \frac{1}{n^2}\right)\ \text{eV}$
Recoil of the emitting atom and X-ray photons
Photon momentum, recoil and the X-ray cut-off
$p = \frac{E}{c} = \frac{h}{\lambda}, \qquad v_{\text{recoil}} = \frac{E}{Mc}, \qquad \lambda_{\min} = \frac{hc}{eV}$

Watch out for (9)

Use the constants the question gives
→ Photon energy and wavelength of a transition
The gap scales as Z²
→ Photon energy and wavelength of a transition
Big frequency means a jump near the ground state
→ Photon energy and wavelength of a transition
A photon must match; an electron need not
→ Excitation, absorption and the number of spectral lines
A sample and a single atom count differently
→ Excitation, absorption and the number of spectral lines
Balmer lines need n = 3 first
→ Excitation, absorption and the number of spectral lines
Recoil uses the atom's mass
→ Recoil of the emitting atom and X-ray photons
The cut-off wavelength ignores the target
→ Recoil of the emitting atom and X-ray photons
Kα energy is a difference of two hole energies
→ Recoil of the emitting atom and X-ray photons

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