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Probing Deep Into Matter

Electron standing waves and the confined electron

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

Overview

Recall questions on why confinement gives discrete energies, the allowed wavelengths and resulting momentum and energy, nodes in higher states, why an atom has a definite size, and the model's limits.

What you need to know

Answer each one before you reveal it.

Your recallNot started
Key term

Why can a confined electron never have zero energy in this model?

Key term

Describe the simple model of an atom as an electron confined in a box.

Key term

Explain why confining an electron leads to a set of discrete energies.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Describe the simple model of an atom as an electron confined in a box.

    Model answer

    The electron is treated as a wave trapped between the walls of a small region, so only those wavelengths that fit a whole number of half-wavelengths into the box are allowed.

  2. 2

    Explain why confining an electron leads to a set of discrete energies.

    Model answer

    Only certain standing wave patterns fit inside the confining region, and each allowed wavelength corresponds through p = h/λ to one particular momentum and therefore one particular energy.

  3. 3

    Write the expression for the allowed wavelengths of an electron standing wave in a box of length L.

    Model answer

    The allowed wavelengths are λ = 2L/n, where n is a whole number counting the standing wave patterns.

  4. 4

    Calculate the wavelength of the lowest-energy standing wave of an electron confined in a box of length 1.0 × 10⁻¹⁰ m.

    Model answer

    With n = 1, λ = 2L = 2 × 1.0 × 10⁻¹⁰ = 2.0 × 10⁻¹⁰ m.

  5. 5

    Calculate the momentum of an electron whose de Broglie wavelength is 2.0 × 10⁻¹⁰ m.

    Model answer

    p = h/λ = 6.63 × 10⁻³⁴ ÷ 2.0 × 10⁻¹⁰ = 3.3 × 10⁻²⁴ N s.

  6. 6

    Calculate the kinetic energy of an electron of momentum 3.3 × 10⁻²⁴ N s, and express it in electronvolts.

    Model answer

    Ek = p²/2m = (3.3 × 10⁻²⁴)² ÷ (2 × 9.11 × 10⁻³¹) = 6.0 × 10⁻¹⁸ J, which is about 38 eV.

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