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Probing Deep Into Matter

Discrete energy levels in atoms

  • 3 things to know
  • Components 01, 02 and 03

Overview

Recall questions on energy levels and the evidence of line spectra, ΔE = hf, frequency and wavelength calculations for a transition, absorption against emission spectra, electron collisions as evidence, electrons below the excitation energy, ionisation energy, negative energy values, ground and excited states, excitation against ionisation, and exciting an atom with a 10.4 V electron.

What you need to know

Answer each one before you reveal it.

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Key term

How do line spectra provide evidence for discrete energy levels?

Key term

Why is the emission spectrum of a gas a set of lines rather than a continuous band of colour?

Key term

Write the relationship between the energy difference of two levels and the frequency of the photon emitted.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    What is meant by an energy level in an atom?

    Model answer

    An energy level is one of the set of definite energies that an electron in an atom is allowed to have, with no values in between.

  2. 2

    How do line spectra provide evidence for discrete energy levels?

    Model answer

    A hot gas emits only certain sharply defined wavelengths, and each corresponds to a fixed photon energy, which can only happen if the electron energies before and after emission are themselves fixed.

  3. 3

    Why is the emission spectrum of a gas a set of lines rather than a continuous band of colour?

    Model answer

    Only the differences between allowed energy levels are available as photon energies, so only a limited set of frequencies can be emitted.

  4. 4

    Write the relationship between the energy difference of two levels and the frequency of the photon emitted.

    Model answer

    The photon energy equals the difference between the levels, so ΔE = hf.

  5. 5

    Calculate the frequency of the photon emitted in a transition of energy 2.1 eV.

    Model answer

    The energy is 2.1 × 1.60 × 10⁻¹⁹ = 3.36 × 10⁻¹⁹ J, so f = ΔE/h = 3.36 × 10⁻¹⁹ ÷ 6.63 × 10⁻³⁴ = 5.1 × 10¹⁴ Hz.

  6. 6

    Calculate the wavelength of a photon of frequency 5.1 × 10¹⁴ Hz.

    Model answer

    λ = c/f = 3.00 × 10⁸ ÷ 5.1 × 10¹⁴ = 5.9 × 10⁻⁷ m, which is 590 nm.

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