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Asked by Prasenjit Last Modified
The energy levels of a hydrogen atom are given by the formula for the energy levels of the hydrogen atom:
En=−RHn2En=−n2RH
Where:
The energy of a photon emitted or absorbed during a transition between two energy levels is given by the difference in energy between the initial and final energy levels:
ΔE=Efinal−EinitialΔE=Efinal−Einitial
For a transition from the second permitted energy level (n=2n=2) to the first permitted energy level (n=1n=1), the energy of the photon emitted (ΔE2→1ΔE2→1) is:
ΔE2→1=E1−E2ΔE2→1=E1−E2
For a transition from the highest permitted energy level (n=∞n=∞) to the first permitted energy level (n=1n=1), the energy of the photon emitted (ΔE∞→1ΔE∞→1) is:
ΔE∞→1=E1−E∞ΔE∞→1=E1−E∞
Substituting the expressions for the energy levels (EnEn) into these equations, we get:
ΔE2→1=−(−RH12)−(−RH22)ΔE2→1=−(−12RH)−(−22RH) ΔE2→1=−(−13.6 eV)+(−13.64 eV)ΔE2→1=−(−13.6eV)+(−413.6eV) ΔE2→1=13.6 eV−3.4 eVΔE2→1=13.6eV−3.4eV ΔE2→1=10.2 eVΔE2→1=10.2eV
ΔE∞→1=−(−RH12)−(−RH∞2)ΔE∞→1=−(−12RH)−(−∞2RH) ΔE∞→1=−(−13.6 eV)−(0 eV)ΔE∞→1=−(−13.6eV)−(0eV) ΔE∞→1=13.6 eVΔE∞→1=13.6eV
Now, to find the ratio of energies of photons:
ΔE2→1ΔE∞→1=10.2 eV13.6 eVΔE∞→1ΔE2→1=13.6eV10.2eV
ΔE2→1ΔE∞→1=10.213.6ΔE∞→1ΔE2→1=13.610.2
ΔE2→1ΔE∞→1≈0.75ΔE∞→1ΔE2→1≈0.75
So, the ratio of energies of photons produced due to the transition of an electron from the second permitted energy level to the first permitted level to the transition from the highest permitted energy level to the first permitted level is approximately 0.75.
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