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N 2 to N 1 Calculate Energy Ph Photon

Reviewed by Calculator Editorial Team

This calculator determines the energy of a photon emitted when an electron transitions from the n=2 energy level to the n=1 ground state in a hydrogen atom. The calculation uses quantum mechanics principles and the Rydberg formula.

Introduction

When an electron in a hydrogen atom moves from a higher energy level to a lower one, it emits a photon with energy equal to the difference between the two levels. The most common transition is from n=2 to n=1, which produces a photon in the ultraviolet spectrum.

This calculation is fundamental in atomic physics and spectroscopy. Understanding these energy transitions helps in analyzing atomic spectra, designing lasers, and studying quantum mechanical systems.

Formula

The energy of the photon (E) emitted during a transition from level n2 to n1 in a hydrogen atom is given by the Rydberg formula:

E = R∞hc (1/n12 - 1/n22)

Where:

  • E = Energy of the photon (in joules)
  • R∞ = Rydberg constant (1.0973731568539(55) × 107 m-1)
  • h = Planck's constant (6.62607015 × 10-34 J·s)
  • c = Speed of light (2.99792458 × 108 m/s)
  • n1 = Principal quantum number of the lower level (1 for ground state)
  • n2 = Principal quantum number of the higher level (2 for this calculation)

For the n=2 to n=1 transition, the formula simplifies to:

E = 1.8897 × 10-18 J

Calculation

The calculation involves plugging the known values into the Rydberg formula. The Rydberg constant, Planck's constant, and the speed of light are all fundamental physical constants. The principal quantum numbers n1 and n2 are specific to the transition being analyzed.

The result is the energy of the photon emitted during the transition, which can be converted to other units such as electron volts (eV) or wavenumbers (cm-1).

Example

Let's calculate the energy of a photon emitted when an electron transitions from n=2 to n=1 in a hydrogen atom.

  1. Identify the values:
    • R∞ = 1.0973731568539 × 107 m-1
    • h = 6.62607015 × 10-34 J·s
    • c = 2.99792458 × 108 m/s
    • n1 = 1
    • n2 = 2
  2. Plug the values into the formula:

    E = (1.0973731568539 × 107)(6.62607015 × 10-34)(2.99792458 × 108) × (1/12 - 1/22)

  3. Calculate the difference in energy levels:

    1/12 - 1/22 = 1 - 0.25 = 0.75

  4. Multiply by the constants:

    E = (1.0973731568539 × 107)(6.62607015 × 10-34)(2.99792458 × 108) × 0.75

    E ≈ 1.8897 × 10-18 J

The energy of the photon emitted during this transition is approximately 1.8897 × 10-18 joules, which corresponds to a wavelength of about 121.57 nanometers in the ultraviolet spectrum.

FAQ

What is the energy of a photon emitted when an electron transitions from n=2 to n=1 in a hydrogen atom?
The energy is approximately 1.8897 × 10-18 joules, which is the difference in energy between the n=2 and n=1 energy levels.
What is the wavelength of the photon emitted in this transition?
The wavelength is approximately 121.57 nanometers, which is in the ultraviolet spectrum.
What is the Rydberg formula used for?
The Rydberg formula is used to calculate the wavelengths or frequencies of light emitted or absorbed when electrons transition between energy levels in atoms.
What are the units for the energy of a photon?
The energy of a photon can be expressed in joules (J), electron volts (eV), or wavenumbers (cm-1).