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Difficulty: MediumEnergy Levels and Atomic Spectra

An electron inside an excited gas atom undergoes a transition from an upper energy level of −2.40 eV-2.40\text{ eV} to a lower energy level of −5.15 eV-5.15\text{ eV}. What is the wavelength, in nanometers (nm\text{nm}), of the emitted photon? (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 s−1c = 3.0 \times 10^8\text{ m s}^{-1}, and 1 eV=1.6×10−19 J1\text{ eV} = 1.6 \times 10^{-19}\text{ J})

Answer: 450 nm

Answer

The wavelength of the emitted photon is 450 nm.
When an electron transitions from a higher energy level to a lower energy level, a photon is emitted with energy equal to the difference between the two energy states. Converting 2.75 eV to 4.40 x 10^-19 J and applying lambda = hc / E yields a wavelength of 4.50 x 10^-7 m, which equals 450 nm.

Step-by-Step Solution

1
Calculate the energy change of the transition in eV
\Delta E = -2.40\text{ eV} - (-5.15\text{ eV}) = 2.75\text{ eV}
The energy of the emitted photon equals the difference between the higher and lower electronic energy states.
2
Convert the transition energy into SI units (Joules)
\Delta E = 2.75 \times 1.6 \times 10^{-19}\text{ J} = 4.40 \times 10^{-19}\text{ J}
Standard physics constants h and c require energy to be expressed in Joules.
3
Calculate photon wavelength and convert to nanometers
\lambda = \frac{hc}{\Delta E} = \frac{1.98 \times 10^{-25}\text{ J m}}{4.40 \times 10^{-19}\text{ J}} = 4.50 \times 10^{-7}\text{ m} = 450\text{ nm}
Applying the de Broglie/Einstein relation connecting photon energy and wavelength.

Key Concept

Energy Level Transitions and Atomic Emission Spectra
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Topics

Energy Levels and Atomic Spectra

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