Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its second permitted energy level to the first permitted level and the highest permitted energy level to the first permitted level.

The energy levels of a hydrogen atom are given by the formula for the energy levels of the hydrogen atom:

En=−RHn2En=−n2RH

Where:

• EnEn is the energy of the nth energy level,
• RHRH is the Rydberg constant (RH=13.6RH=13.6 eV),
• nn is the principal quantum number.

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.