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Difficulty: MediumEnergy Levels and Atomic Spectra

An atom has three stationary energy levels given by E1=−12.50 eVE_1 = -12.50\text{ eV}, E2=−6.80 eVE_2 = -6.80\text{ eV}, and E3=−3.50 eVE_3 = -3.50\text{ eV}. What is the wavelength, in nanometers (nm\text{nm}), of the photon emitted during the transition that produces the longest wavelength line in its emission spectrum? (Take Planck's constant h=6.6×10−34 J⋅sh = 6.6 \times 10^{-34}\text{ J}\cdot\text{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: 375 nm

Answer

The wavelength of the photon emitted for the longest wavelength spectral line is 375 nm.
Photon wavelength is related to transition energy by λ=hcΔE\lambda = \frac{hc}{\Delta E}. To find the longest wavelength spectral line, the transition with the smallest energy gap must be used. Evaluating all emission transitions between the levels gives ΔE3→2=3.30 eV\Delta E_{3 \to 2} = 3.30\text{ eV}, ΔE2→1=5.70 eV\Delta E_{2 \to 1} = 5.70\text{ eV}, and ΔE3→1=9.00 eV\Delta E_{3 \to 1} = 9.00\text{ eV}. The minimum energy difference is 3.30 eV3.30\text{ eV}. Converting 3.30 eV3.30\text{ eV} to Joules gives 3.30×1.6×10−19=5.28×10−19 J3.30 \times 1.6 \times 10^{-19} = 5.28 \times 10^{-19}\text{ J}. Substituting this into the wavelength formula yields λ=6.6×10−34×3.0×1085.28×10−19=3.75×10−7 m=375 nm\lambda = \frac{6.6 \times 10^{-34} \times 3.0 \times 10^8}{5.28 \times 10^{-19}} = 3.75 \times 10^{-7}\text{ m} = 375\text{ nm}.

Step-by-Step Solution

1
Determine which electronic transition yields the longest wavelength photon.
Transition from E3E_3 to E2E_2 yields the minimum energy difference of 3.30 eV3.30\text{ eV}.
Since λ=hcΔE\lambda = \frac{hc}{\Delta E}, the longest wavelength corresponds to the smallest energy transition.
2
Convert the transition energy from electron-volts to Joules.
ΔE=5.28×10−19 J\Delta E = 5.28 \times 10^{-19}\text{ J}.
SI units are required for calculations involving Planck's constant and the speed of light.
3
Calculate the wavelength λ\lambda using the photon energy formula λ=hcΔE\lambda = \frac{hc}{\Delta E}.
λ=375 nm\lambda = 375\text{ nm}.
Substituting h=6.6×10−34 J⋅sh = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}, c=3.0×108 m/sc = 3.0 \times 10^8\text{ m/s}, and ΔE=5.28×10−19 J\Delta E = 5.28 \times 10^{-19}\text{ J} gives 3.75×10−7 m3.75 \times 10^{-7}\text{ m}, which equals 375 nm375\text{ nm}.

Key Concept

Inverse relationship between transition energy and photon wavelength in atomic emission spectra
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Topics

Energy Levels and Atomic Spectra

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