An electron de-excites from the fourth quantum level to the third and then directly to the first. Two frequencies of light are emitted. How do their combined energies compare to the energy of the single frequency that would be emitted by de-excitation from the fourth level directly to the first level?

The answer is explained below.

Explanation:

The energy emitted during the de-excitation of an electron from a higher energy level to a lower energy level is directly proportional to the frequency of the emitted light.

Here, the total sum of the energies of 2 frequencies of light emitted in different stages is equal to the energy of a single frequency of light during the de-excitation of fourth level to ground level directly.

Hence the total sum of of the frequencies of 2 lights emitted in different stages is equal to the frequency of single frequency of light emitted during the de-excitation  from fourth level to ground level directly.

The some of the energies of 2 frequencies emitted by one electron is equal to the energy of a single frequency when electron jumps directly.

snell's law

refractve index = sin incidence / sin refraction

answer is slate

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