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Difficulty: MediumEnergy Levels and Atomic Spectra

An electron in an atom transitions from an excited state with energy −1.7×10−19 J-1.7 \times 10^{-19}\text{ J} to a lower state with energy −5.0×10−19 J-5.0 \times 10^{-19}\text{ J}. What is the wavelength of the emitted photon in nanometers (nm\text{nm})? (Take Planck's constant h=6.6×10−34 J⋅sh = 6.6 \times 10^{-34}\text{ J}\cdot\text{s} and the speed of light c=3.0×108 m/sc = 3.0 \times 10^8\text{ m/s})

Answer: 600 nm

Answer

The wavelength of the emitted photon is 600 nm.
The energy released during the atomic transition is ΔE=(−1.7×10−19 J)−(−5.0×10−19 J)=3.3×10−19 J\Delta E = (-1.7 \times 10^{-19}\text{ J}) - (-5.0 \times 10^{-19}\text{ J}) = 3.3 \times 10^{-19}\text{ J}. Substituting this into λ=hcΔE\lambda = \frac{hc}{\Delta E} yields λ=6.6×10−34×3.0×1083.3×10−19=6.0×10−7 m\lambda = \frac{6.6 \times 10^{-34} \times 3.0 \times 10^8}{3.3 \times 10^{-19}} = 6.0 \times 10^{-7}\text{ m}. Converting to nanometers (1 m=109 nm1\text{ m} = 10^9\text{ nm}) gives 600 nm600\text{ nm}.

Step-by-Step Solution

1
Calculate energy of the emitted photon
ΔE=3.3×10−19 J\Delta E = 3.3 \times 10^{-19}\text{ J}
The energy of the photon equals the difference between the initial higher energy level and the final lower energy level.
2
Calculate wavelength in meters using Planck's relation
λ=6.0×10−7 m\lambda = 6.0 \times 10^{-7}\text{ m}
Rearranging ΔE=hcλ\Delta E = \frac{hc}{\lambda} gives λ=hcΔE\lambda = \frac{hc}{\Delta E}.
3
Convert wavelength to nanometers
600 nm600\text{ nm}
Multiply by 109 nm/m10^9\text{ nm/m} to obtain the final value in nanometers.

Key Concept

Energy level transitions and photon emission wavelength
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Topics

Energy Levels and Atomic Spectra

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