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A sample of excited atoms lie 3.314×10−19 J above the ground state. Determine the emission wavelength of these atoms.

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The emission wavelength of the atoms excited above the ground state is 599.8 nm.

Emission wavelength of the atoms

The emission wavelength of the atom is determined by using the following formulas as shown below;

E = hf

E = hc/λ

where;

  • E is the energy of the atom
  • h is Planck's constant
  • c is speed of light
  • λ is the emission wavelength

λ = hc/E

λ = (6.626 x 10⁻³⁴ x 3 x 10⁸) / (3.314 x 10⁻¹⁹)

λ = 5.998 x 10⁻⁷ m

λ = 599.8 x 10⁻⁹ m

λ = 599.8 nm

Thus, the emission wavelength of the atoms is 599.8 nm.

Learn more about wavelength here: https://brainly.com/question/10728818

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