A sample of excited atoms lie 3.314×10−19 J above the ground state. Determine the emission wavelength of these atoms.
1 answer:
The emission wavelength of the atoms excited above the ground state is 599.8 nm.
<h3>Emission wavelength of the atoms</h3>
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: brainly.com/question/10728818
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