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NEET UG Β· Physics

Atoms

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NEET UG Β· Physics

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Read photon gaps, distinguish cascades and check hydrogenic or scattering assumptions. Atoms: read energy gaps and state the model

About 6 min Β· NEET UG

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Atoms: read energy gaps and state the model

Read photon gaps, distinguish cascades and check hydrogenic or scattering assumptions.

What you will understand

For NEET UG Physics students: turn an energy-level difference into photon energy, distinguish a cascade from one transition, calculate a head-on Coulomb turning distance and state when a hydrogen-like formula applies.

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Quick answer: compare states before choosing a formula

An electron level energy is measured relative to a chosen reference. In the elementary hydrogen model, zero is a separated electron at rest infinitely far away; bound levels are negative. A downward transition lowers the electron energy and emits a positive-energy photon equal to the gap. A head-on scattering question instead compares incoming kinetic energy with electric potential energy at a turning point. These are different energy balances, and neither level energy nor a scattering separation is automatically a photon wavelength or a nuclear radius.

Hydrogen levels: a cascade contains separate photon energies

Original energy-level diagram for hydrogen in the fixed-nucleus nonrelativistic Bohr approximation. A common energy scale places the ionisation continuum at zero, n=3 near -1.511 eV, n=2 at -3.400 eV and n=1 at -13.600 eV; rounded labels are shown. Downward transitions 3 to 2 and 2 to 1 indicate a possible two-step cascade. The first releases about 1.889 eV and corresponds to about 656 nm using supplied hc=1240 eV nm. Vertical spacing depicts electron energy, not orbital radius or a trajectory. Different cascade paths need not occur with equal probability. Photon recoil, fine structure and reduced-mass corrections are neglected.
The level positions follow the fixed-nucleus Bohr approximation on a common energy scale. A possible n=3 to n=2 to n=1 cascade emits two photons, with each energy set by its own step. Vertical spacing represents energy rather than orbital radius. The rounded labels do not imply equal probability for every possible path.

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From n=3 to n=2: subtract the levels, then find wavelength

For hydrogen use En=βˆ’13.6/n2 eVE_n=-13.6/n^2\,\mathrm{eV} in the supplied fixed-nucleus nonrelativistic approximation. Thus E3β‰ˆβˆ’1.511 eV and E2=βˆ’3.400 eV. The electron change is

Ξ”Eelectron=E2βˆ’E3β‰ˆβˆ’1.889 eV\Delta E_\mathrm{electron}=E_2-E_3\approx-1.889\,\mathrm{eV}
, while emitted photon energy is +1.889 eV. With supplied hc=1240 eV nmhc=1240\,\mathrm{eV\,nm}, the wavelength is about 1240/1.889=656 nm. A subsequent 2 to 1 photon has 10.20 eV. Their sum is about 12.089 eV, the gap for a direct 3 to 1 transition; their wavelengths must not be added or averaged to represent that direct photon. Photon recoil and fine-structure corrections are neglected.

Absorption, excitation and ionisation need distinct endpoints

To excite a hydrogen electron from n=2 to n=3 requires about 1.889 eV in this model. To ionise it from n=2 to the zero-energy continuum requires 3.40 eV. A photon supplies its entire energy in a simple absorption event; it is not an electron projectile that can give only part of an arbitrary kinetic energy to excitation. A collision problem may therefore have a different threshold/energy-sharing interpretation. Specify the initial level and whether the requested final state is bound or ionised.

Rutherford head-on approach: distance is a centre separation

Original head-on alpha-particle and gold-nucleus calculation diagram. Both rows share the same target-centre reference. For a fixed target with charge +79e and an alpha charge +2e, supplied k e squared=1.44 MeV fm gives rmin=1.44 times 2 times 79 divided by K in MeV. At 5.00 MeV the turn separation is 45.504 fm; at 10.00 MeV it is 22.752 fm. The two drawn centre separations have a 2:1 ratio, but particle symbols are not to physical size. The incoming direction is toward the target. At a head-on turn radial kinetic energy vanishes in this fixed-target Coulomb model. These distances are not the nuclear radius. Finite target recoil, electron screening and nuclear-force deviations are neglected.
The alpha particle approaches a fixed gold nucleus along the line joining their centres. In the pure repulsive Coulomb model, its kinetic energy becomes electric potential energy at the turn. Doubling incoming energy halves the head-on turn separation for unchanged charges. Particle symbols are not drawn to nuclear size, and the separation is not the gold nucleus radius.

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Double the alpha energy and halve the head-on turn distance

Take an alpha charge +2e and fixed gold-nucleus charge +79e. Assume an initially distant projectile, negligible screening and a pure Coulomb interaction. With supplied ke2=1.44 MeV fmke^2=1.44\,\mathrm{MeV\,fm}, energy conservation gives rmin⁑=k(2e)(79e)/Kr_{\min}=k(2e)(79e)/K. At K=5.00 MeV this is 1.44(2)(79)/5.00=45.504 fm1.44(2)(79)/5.00=45.504\,\mathrm{fm}, about 45.5 fm. At 10.00 MeV it is about 22.8 fm. These are centre-to-centre turning distances within the model. A nonzero impact parameter retains tangential motion at closest approach, so blindly setting all kinetic energy to potential energy there would be wrong. Finite target recoil and nuclear-force deviations require a different treatment.

Bohr scaling applies to one-electron systems, with stated approximations

For a one-electron ion of nuclear charge +Ze, the elementary fixed-nucleus model gives En=βˆ’13.6Z2/n2 eVE_n=-13.6Z^2/n^2\,\mathrm{eV} and rn=a0n2/Zr_n=a_0n^2/Z, with supplied a0=0.0529 nm. A hydrogen atom and He+ fit this one-electron idealisation; neutral helium has two electrons and does not obey it directly. Bohr radii are model orbit radii, not general quantum trajectories or the nucleus size. Reduced mass, relativistic effects and detailed spectral structure are outside this elementary scaling.

Same energy does not imply the same Bohr radius

Compare hydrogen at n=1 with He+ at n=2. Hydrogen has E=βˆ’13.6 eVE=-13.6\,\mathrm{eV} and radius a0. He+ has Z=2, so E=βˆ’13.6(22)/(22)=βˆ’13.6 eVE=-13.6(2^2)/(2^2)=-13.6\,\mathrm{eV} as well, but its radius is a0(22)/2=2a0=0.1058 nma_0(2^2)/2=2a_0=0.1058\,\mathrm{nm}. The energy depends on Z2/n2, while radius depends on n2/Z. Equal model energy therefore does not force equal radius. The equality uses the stated fixed-nucleus approximation; it is not an exact high-precision identity between real hydrogen and helium-ion spectral lines.

Series labels depend on the final level

Hydrogen transitions ending at n=1 belong to the Lyman series; those ending at n=2 belong to the Balmer series. The illustrated 3 to 2 line is the approximately 656 nm Balmer line. Classify by the final level, then calculate the energy gap if a wavelength is requested. The name of a series does not mean every member has the same wavelength, and energy levels are not equally spaced.

Common mistakes: keep sign, gap and model separate

A negative bound-level energy does not mean a negative emitted photon energy. Do not use the magnitude of the initial energy when a question asks for a transition between two bound levels. Add photon energies across a cascade, not wavelengths. A head-on Coulomb turn distance is not a nuclear radius. Check that the system has one electron before applying hydrogenic scaling, and do not infer equal radius from equal energy.

Try it: ionisation from an excited level

A hydrogen electron starts at n=2, with model energy βˆ’3.40 eV. What minimum energy takes it to the zero-energy continuum? Would using 13.6 eV answer this initial-state question?

Reveal the answer

The minimum is 0βˆ’(βˆ’3.40)=3.40 eV in the stated model. The 13.6 eV ground-state ionisation value uses n=1 as the initial state. Read the starting level before recalling a familiar number.

Try it: a gap can be shared by different cascade paths

Hydrogen can fall directly from n=3 to n=1, or through n=2. Ignoring recoil, do both routes release the same total photon energy? Must they produce the same number of photons or the same wavelength?

Reveal the answer

The total released energy is the same initial-to-final gap, about 12.089 eV in the stated model. The direct route emits one photon; the specified cascade emits two, about 1.889 eV and 10.20 eV. Their individual wavelengths differ from the direct photon wavelength. Equality of summed energy does not require equal photon count or a common wavelength.

Come back tomorrow: name the states before the gap

Draw three hydrogen levels from memory, label one photon gap and explain why its energy is positive. Then compare two hydrogen-like radii. Revisit NEET Electric Potential and Capacitance for the Coulomb energy reference, or continue to NEET Nuclei to distinguish electronic energy from nuclear binding.

Chapter practice opens the broad Atoms pool.

Sources and review

Technical review: Codex (automated). Primary textbook sources checked on 23 October 2026. Diagrams, worked examples and self-checks are original. Numerical calculations use stated approximations and supplied rounded constants. Human subject review is not claimed. Chapter catalogues and these explanations do not establish complete coverage of every current examination format or syllabus.

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Before the next practice set, write the initial state, final state and quantity requested. After a mistake, correct the model or sign first, then redo the calculation without viewing the solution.

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Sources and reading
  • NCERT XII Physics Part II: Atoms β€” Reprint 2026–27, Chapter 12: Rutherford head-on approach, Bohr hydrogen energy and radius, transitions and model limits. Concept evidence, not a claim of complete current exam coverage.
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