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Difficulty: MediumEnergy Levels and Atomic Spectra

An electron inside an excited atom drops from an energy state of −2.10 eV-2.10\text{ eV} to a lower energy state of −4.575 eV-4.575\text{ eV}. 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 s}, speed of light c=3.0×108 m/sc = 3.0 \times 10^8\text{ m/s}, and 1 eV=1.6×10−19 J1\text{ eV} = 1.6 \times 10^{-19}\text{ J}).

Answer: 500 nm

Answer

The wavelength of the emitted photon is 500 nm500\text{ nm}.
The energy of the emitted photon is calculated from the energy change of the electron, ΔE=−2.10 eV−(−4.575 eV)=2.475 eV\Delta E = -2.10\text{ eV} - (-4.575\text{ eV}) = 2.475\text{ eV}. Converting this to Joules gives 2.475×1.6×10−19 J=3.96×10−19 J2.475 \times 1.6 \times 10^{-19}\text{ J} = 3.96 \times 10^{-19}\text{ J}. Substituting into the wavelength formula λ=hcΔE=6.6×10−34×3.0×1083.96×10−19=5.0×10−7 m=500 nm\lambda = \frac{hc}{\Delta E} = \frac{6.6 \times 10^{-34} \times 3.0 \times 10^8}{3.96 \times 10^{-19}} = 5.0 \times 10^{-7}\text{ m} = 500\text{ nm}.

Step-by-Step Solution

1
Calculate the energy difference between the initial excited state and final state
ΔE=−2.10 eV−(−4.575 eV)=2.475 eV\Delta E = -2.10\text{ eV} - (-4.575\text{ eV}) = 2.475\text{ eV}
The energy of the emitted photon equals the difference between the two energy levels.
2
Convert the photon energy from electron-volts (eV) to Joules (J)
ΔE=2.475×1.6×10−19 J=3.96×10−19 J\Delta E = 2.475 \times 1.6 \times 10^{-19}\text{ J} = 3.96 \times 10^{-19}\text{ J}
Standard SI units must be used to calculate wavelength in meters.
3
Apply the Planck-Einstein relation λ=hcΔE\lambda = \frac{hc}{\Delta E} to calculate the wavelength in meters and convert to nanometers
λ=(6.6×10−34 J s)(3.0×108 m/s)3.96×10−19 J=5.0×10−7 m=500 nm\lambda = \frac{(6.6 \times 10^{-34}\text{ J s})(3.0 \times 10^8\text{ m/s})}{3.96 \times 10^{-19}\text{ J}} = 5.0 \times 10^{-7}\text{ m} = 500\text{ nm}
Converting meters to nanometers requires multiplying by 10910^9.

Key Concept

Photon emission wavelength during atomic energy level transitions
Estimated Time:1m 30s
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Topics

Energy Levels and Atomic Spectra

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