What would be the photon energy of light emitted from a transition between the = 5 state and the ground state? Show all your working and give your final answer in eV.
Question
What would be the photon energy of light emitted from a transition between the = 5 state and the ground state? Show all your working and give your final answer in eV.
Solution
The energy of a photon emitted during a transition between two energy levels in a hydrogen atom can be calculated using the Rydberg formula:
E = E1 - E2 = RH * (1/n1² - 1/n2²)
where:
- E is the energy of the photon,
- E1 and E2 are the energies of the initial and final states,
- RH is the Rydberg constant for hydrogen (approximately 13.6 eV),
- n1 and n2 are the principal quantum numbers of the initial and final states.
In this case, the transition is from n1 = 5 (the initial state) to n2 = 1 (the ground state). Substituting these values into the formula gives:
E = 13.6 eV * (1/1² - 1/5²) E = 13.6 eV * (1 - 0.04) E = 13.6 eV * 0.96 E = 13.056 eV
So, the energy of the photon emitted during this transition is approximately 13.056 eV.
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