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Why do atoms emit and absorb only certain wavelengths of light, and what does this reveal about energy levels?

Describe the Bohr model with quantised electron energy levels, explain how photons are emitted or absorbed when electrons change levels, and apply the energy-level relationship Ephoton=Eiβˆ’EfE_{photon} = E_i - E_f for hydrogen.

A Regents Physics answer on the Bohr model and atomic spectra: quantised electron energy levels, the emission and absorption of photons when electrons jump between levels, and the energy-level relationship for hydrogen, using the Reference-Table equation and energy-level diagram, with worked examples.

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  1. What this topic is asking
  2. Quantised energy levels
  3. Emission and absorption of photons
  4. Applying the energy-level relationship
  5. Reference Tables note
  6. Try this

What this topic is asking

This dot point explains why atoms emit and absorb light only at specific wavelengths, using the Bohr model of quantised energy levels. The Physical Setting/Physics course asks you to describe the model (electrons occupy only certain allowed energy levels), to explain how a photon is emitted or absorbed when an electron changes level, and to apply the energy-level relationship Ephoton=Eiβˆ’EfE_{photon} = E_i - E_f for hydrogen, reading the levels from the energy-level diagram on the Reference Tables. The Regents tests both the concept of line spectra and the energy calculation.

Quantised energy levels

This was a radical departure from the idea that an electron could have any energy. The quantisation means an atom is like a staircase, not a ramp: an electron can sit on a step but not between steps. The allowed levels are specific to each element, which is why every element has its own characteristic spectrum, in effect an atomic fingerprint.

Emission and absorption of photons

Because only certain level differences exist, only photons of certain energies (and so certain frequencies and wavelengths) can be absorbed or emitted. This is why a heated gas emits a line spectrum, a set of discrete bright lines rather than a continuous range, and why the same gas absorbs exactly those wavelengths from white light passing through it. The pattern of lines identifies the element.

Applying the energy-level relationship

The energy levels are quoted as negative numbers, measured relative to a completely free electron at zero energy; the ground state is the most negative. When an electron drops to a lower level, Eiβˆ’EfE_i - E_f comes out positive (an emitted photon); when it absorbs a photon to rise, you supply the energy. Careful handling of the negative signs is the main calculation skill, and you read the level values from the hydrogen diagram on the tables.

Reference Tables note

The Reference Tables print Ephoton=Eiβˆ’EfE_{photon} = E_i - E_f in the Modern Physics section and provide an energy-level diagram for hydrogen giving the energy of each level in electronvolts. The constant 11 electronvolt =1.60Γ—10βˆ’19= 1.60 \times 10^{-19} J is in the constants list for converting to joules. You recall that energy levels are quantised, that emission and absorption involve photons matching the gap, and that this produces line spectra; the photon energy itself connects to Ephoton=hfE_{photon} = hf from the dual nature of light.

Try this

Q1. State why an atom emits light of only certain wavelengths. [2 points]

  • Cue. Its electron energy levels are quantised, so transitions release photons of only certain energies (and so wavelengths), giving a line spectrum.

Q2. An electron drops from a βˆ’0.85-0.85 eV level to a βˆ’3.40-3.40 eV level. Calculate the energy of the emitted photon. [2 points]

  • Cue. Ephoton=Eiβˆ’Ef=(βˆ’0.85)βˆ’(βˆ’3.40)=2.55E_{photon} = E_i - E_f = (-0.85) - (-3.40) = 2.55 eV.

Exam-style practice questions

Practice questions written in the style of NYSED exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Regents (style)2 marksPart B-2 (constructed response). An electron in a hydrogen atom drops from an energy level of βˆ’1.51-1.51 eV to a level of βˆ’3.40-3.40 eV. Calculate the energy of the emitted photon. Show the equation, substitution and answer.
Show worked answer β†’

A 2-point constructed-response calculation using the Reference-Table energy-level relationship Ephoton=Eiβˆ’EfE_{photon} = E_i - E_f.

Equation: Ephoton=Eiβˆ’EfE_{photon} = E_i - E_f.
Substitution: Ephoton=(βˆ’1.51)βˆ’(βˆ’3.40)=βˆ’1.51+3.40E_{photon} = (-1.51) - (-3.40) = -1.51 + 3.40.
Answer: Ephoton=1.89E_{photon} = 1.89 eV (an emitted photon, since the electron dropped to a lower level).

Markers reward the equation from the tables, careful handling of the negative energy-level values, and a positive emitted-photon energy. Reading the levels from the hydrogen energy-level diagram on the tables is expected.

Regents (style)1 marksPart A (multiple choice). An atom emits light of only certain specific wavelengths because (1) its electrons can have any energy (2) its electron energy levels are quantised (3) light is a continuous wave (4) the nucleus emits the light. Justify your choice.
Show worked answer β†’

A 1-point Part A item on quantised energy levels. The answer is (2).

An atom's electrons can occupy only specific, quantised energy levels. When an electron drops from a higher to a lower level, it emits a photon whose energy equals the difference between the levels, Ephoton=Eiβˆ’EfE_{photon} = E_i - E_f. Because only certain level differences exist, only certain photon energies (and so wavelengths) are emitted, producing a line spectrum. If electrons could have any energy, the spectrum would be continuous.

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