Question

Difficulty: HardEnergy Levels and Atomic Spectra

An atom in a gas discharge tube has a ground state energy level of 10.4 eV-10.4\text{ eV} and an excited energy level of 3.8 eV-3.8\text{ eV}. An electron absorbs a single photon to undergo a transition directly from the ground state to this excited level. If the frequency of the absorbed photon is expressed as x×1015 Hzx \times 10^{15}\text{ Hz}, what is the numerical value of xx? (Take Planck's constant h=6.6×1034 Jsh = 6.6 \times 10^{-34}\text{ J}\cdot\text{s} and 1 eV=1.6×1019 J1\text{ eV} = 1.6 \times 10^{-19}\text{ J})

Answer: 1.6

Answer

The numerical value of xx is 1.6.
The energy of the absorbed photon is equal to the difference between the excited state and ground state energies: ΔE=3.8 eV(10.4 eV)=6.6 eV\Delta E = -3.8\text{ eV} - (-10.4\text{ eV}) = 6.6\text{ eV}. Converting this to Joules yields 6.6×1.6×1019 J=1.056×1018 J6.6 \times 1.6 \times 10^{-19}\text{ J} = 1.056 \times 10^{-18}\text{ J}. Using Einstein's photon relation E=hfE = hf, the frequency is f=1.056×10186.6×1034=1.6×1015 Hzf = \frac{1.056 \times 10^{-18}}{6.6 \times 10^{-34}} = 1.6 \times 10^{15}\text{ Hz}, which gives x=1.6x = 1.6.

Step-by-Step Solution

1
Determine the energy absorbed during transition
ΔE=6.6 eV\Delta E = 6.6\text{ eV}
The energy of the absorbed photon equals the energy difference between the initial and final states: ΔE=3.8 eV(10.4 eV)=6.6 eV\Delta E = -3.8\text{ eV} - (-10.4\text{ eV}) = 6.6\text{ eV}.
2
Convert energy from eV to Joules
ΔE=1.056×1018 J\Delta E = 1.056 \times 10^{-18}\text{ J}
Multiply by 1.6×1019 J/eV1.6 \times 10^{-19}\text{ J/eV} to convert energy to standard SI units.
3
Calculate photon frequency
f=1.6×1015 Hzf = 1.6 \times 10^{15}\text{ Hz}
Use Planck's energy equation f=ΔEh=1.056×1018 J6.6×1034 Js=1.6×1015 Hzf = \frac{\Delta E}{h} = \frac{1.056 \times 10^{-18}\text{ J}}{6.6 \times 10^{-34}\text{ J}\cdot\text{s}} = 1.6 \times 10^{15}\text{ Hz}.

Key Concept

Photon energy and atomic transition frequency
Estimated Time:2m 0s
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