Atoms

AtomsNEET Physics · Class 12 · NCERT Chapter 12

Overview Notes NEET Questions Interactive Learning

8 interactive concept widgets for Atoms. Drag any slider, change any number, and watch the formula and the answer update live. Built so you understand how each NEET problem actually works, not just the final number.

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Distance of closest approach

Bohr orbit radius, velocity and energy

Hydrogen energy level diagram and transitions

Excitation and ionisation energy

Hydrogen spectrum: five series

Rydberg formula calculator

Spectral lines counter (n(n-1)/2)

de Broglie picture of Bohr orbits

Rutherford scattering

Distance of closest approach of an alpha to a target nucleus.

Rutherford scattering

Distance of closest approach in alpha scattering

Set the target Z and the alpha KE; see how close the alpha gets.

Rutherford fired alpha particles at a thin foil. Most went through; a few bounced back, proving a tiny dense nucleus. Closest head-on approach: r_0 = k · 2 Z e² / KE.

Atomic number Z (target): 79

Alpha KE: 5 MeV

Closest approach r_0

45.5 fm

r_{0} = \frac{1}{4 π ε _{0}} \frac{2 Z e ^{2}}{E _{k}}

Try this

Gold (Z = 79) and 5 MeV alphas: r_0 ≈ 45 fm. Nucleus is much smaller (~7 fm); alpha never touches.
Doubling the KE of the alpha halves the closest approach r_0.
Higher Z target gives larger r_0 (stronger repulsion).
About 1 in 8000 alphas at gold turn back, proving the dense central nucleus.

Bohr model: orbits, energy, transitions

Radius, velocity, energy and photon transitions in hydrogen-like atoms.

Bohr orbit

Bohr orbit radius, velocity and energy

Pick n and Z to see how each scales.

Bohr's n-th orbit: radius scales as n²/Z, velocity as Z/n, energy as -Z²/n². For ground state of hydrogen (n = 1, Z = 1), r = 0.529 Å, E = -13.6 eV.

Quantum number n: 1

Atomic number Z: 1

Radius r_n

0.529 Å

Velocity v_n

2.19 ×10⁶ m/s

Energy E_n

-13.60 eV

Frequency f_n

6.59e+15 Hz

r_{n} = \frac{n ^{2}}{Z} a_{0}, v_{n} = \frac{Z}{n} \frac{c}{137}, E_{n} = - \frac{13.6 Z ^{2}}{n ^{2}} eV

Try this

Hydrogen ground state (n = 1, Z = 1): r = 0.529 Å, E = -13.6 eV, v ≈ c/137.
Helium ion He+ (Z = 2): r = a_0 / 2, E = -54.4 eV (4× more bound).
For Li²+ (Z = 3): E_1 = -122.4 eV.
Higher n means larger orbit, slower electron, less bound.

Energy levels

Hydrogen energy level diagram and transitions

Click levels in the diagram to set a transition; see ΔE and λ.

Click two levels to see the photon emitted on the transition. ΔE = E_high - E_low → λ = 1240 / ΔE (nm).

From n = 3 → to n = 2

Energy released ΔE

1.89 eV

Wavelength λ

656.5 nm

Balmer (visible) series

From: E = -1.51 eV. To: E = -3.40 eV.

Try this

Balmer alpha (n = 3 → 2): λ = 656 nm (red). Visible.
Lyman alpha (n = 2 → 1): λ = 121.6 nm (UV). All Lyman lines are UV.
Brackett alpha (n = 5 → 4): far IR.
Click any two levels on the diagram to set the transition.

Excitation and ionisation

Excitation and ionisation energy

Pick Z and excited state n to read off the energies.

Excitation lifts the electron to a higher level. Ionisation removes it (n → ∞). For hydrogen, ionisation energy = 13.6 eV; first excitation (1 → 2) = 10.2 eV.

Atomic number Z: 1

Excite to n: 3

Excitation energy (1 → 3)

12.09 eV

Ionisation energy (1 → ∞)

13.60 eV

E_1 = -13.60 eV, E_3 = -1.51 eV

Δ E = - 13.6 Z^{2} (\frac{1}{n ^{2}} - 1) eV

Try this

Hydrogen first excitation: 10.2 eV (1 → 2). Second: 12.09 eV (1 → 3). Third: 12.75 eV (1 → 4).
Hydrogen ionisation energy: 13.6 eV. He+ ionisation: 54.4 eV (= 13.6 × 4).
Most NEET problems on this fall in the 1 → 2 or 1 → 3 transitions.
Ionisation energy = -E_1. Always positive (energy must be added to free the electron).

Hydrogen spectrum and Rydberg formula

Five named series and the universal formula linking n_1, n_2 to wavelength.

Hydrogen spectrum

Hydrogen spectrum: five series

Click each series to see its transitions and wavelengths.

Click any series to see its lines. Lyman is UV, Balmer is visible (the only series we can see directly with our eyes), Paschen and onwards are IR.

Lyman (UV)

Balmer (Visible)

Paschen (IR)

Brackett (IR)

Pfund (Far IR)

Balmer series (Visible)

All transitions ending at n_1 = 2.

Transition

λ (nm)

Label

3→2

656.3

H-α (red)

4→2

486.1

H-β (blue-green)

5→2

434.0

H-γ (violet)

6→2

410.2

H-δ (violet)

∞→2

364.6

Series limit

\frac{1}{λ} = R (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})

R = 1.097 × 10⁷ m⁻¹ for hydrogen.

Try this

NEET favourites: H-α at 656.3 nm and H-β at 486.1 nm.
Series limit: n_2 → ∞. Shortest λ in the series.
Lyman series limit at 91.2 nm corresponds to ionising photon (13.6 eV).
Each series has infinitely many lines bunching towards the limit.

Rydberg formula

Rydberg formula calculator

Compute the wavelength for any (n_1, n_2) transition in hydrogen-like atoms.

Pick n_1 (the lower level) and n_2 (the higher level). The Rydberg formula gives the wavelength of the photon emitted when the electron jumps from n_2 down to n_1.

n_1 (lower): 2

n_2 (higher): 3

Atomic number Z: 1

Wavelength λ

656.34 nm

Frequency f

4.57e+14 Hz

Photon E

1.89 eV

\frac{1}{λ} = R Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})

Try this

n_1 = 2, n_2 = 3 (H-α): λ = 656.3 nm (red).
For He+ (Z = 2), all wavelengths shrink by Z² = 4 compared to H.
n_2 = ∞ gives the series limit: 1/λ = R Z² / n_1².
The biggest jump in 1/λ is Lyman α (n=2→1); largest λ is Pfund α onwards.

Spectral lines counter

How many lines from level n?

Useful NEET shortcut: total possible transitions back to ground from level n.

When an electron de-excites from level n, it can take any path back. Total possible spectral lines: n(n - 1) / 2.

Initially excited to n: 4

Total spectral lines emitted

Lines = \frac{n ( n - 1 )}{2}

Transitions

4→3

4→2

4→1

3→2

3→1

2→1

Try this

n = 2: 1 line (2→1, Lyman α).
n = 3: 3 lines (3→2, 3→1, 2→1).
n = 4: 6 lines.
NEET shortcut: lines emitted from level n is just C(n, 2) = n(n-1)/2.

de Broglie standing wave picture

How matter waves explain Bohr's quantisation as a standing-wave condition.

de Broglie standing wave

de Broglie picture of Bohr orbits

See why Bohr's quantisation is just a standing-wave condition.

de Broglie said: an electron in orbit is a standing wave. For the wave to fit, the orbit circumference must be a whole number of wavelengths: 2 π r = n λ. Plug λ = h / m v: m v r = n ℏ. That is exactly Bohr's postulate.

n: 3

2 π r = nλ \Rightarrow m v r = n ℏ

Try this

n = 1: ground state, single wavelength fits around the orbit.
n = 6: six wavelengths fit. The orbit must be larger.
A non-integer n would mean the wave does not match itself: destructive interference, electron not stable.
This is why orbits are quantised: only standing-wave conditions survive.

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