An electron drops from the 4th energy level in an atom to the 3rd, then to the 1st. 2 frequencies of light are

Question

emitted. How does their combined energy compare with the energy of the single frequency that would have been emitted if the electron had dropped directly from the 4th level to the 1st?

...But is the combined frequency equal to, higher than, or lower then the single freq?

Answer 1

Their combined energy = the sum of the two separate energies.

That is because E (4-->3) = E4 - E3, ......(A)

where E4 is the fourth level's energy; and

E (3-->1) = E3 - E1. ......(B)

So, from (A) and (B), E (4-->3) + E (3-->1) = (E4 - E3) + (E3 - E1)

= E4 - E1. ......(C)

But E4 - E1 IS E (4-->1).

Therefore (E4 - E3) + (E3 - E1) = E (4-->3) + E (3-->1)

= E (4-->1) = E4 - E1, that is:

Their COMBINED ENERGY (= sum of the two separate energies of the photons emitted in the separate steps 4-->3
-->1) equals the energy of the single photon that would have been omitted in the more direct step 4-->1.

QED

Live long and prosper.

POSTSCRIPT: Under the heading "Additional details," you wrote "...But is the combined frequency equal to, higher than, or lower then the single freq?"

I really do not understand your psychology:

1. That is NOT an "Additional detail," it is ANOTHER QUESTION. If you will re-read your question, you will see that you only asked about ENERGIES; the word 'frequency,' as employed in your original question, had absolutely NO bearing on the answer that you requested.

2. Yet you obviously KNOW that there is a connection between frequency and energy. That being so, I must presume that you know that connection:

E = h nu, where h is Planck's constant, and ' nu ' is the frequency of the radiation,

since NO OTHER RELATIONSHIP whatsoever exists between the energy and frequency of radiation.

Therefore, ANY RELATIONSHIP WHATSOEVER that exists between successive and combined energies must also exist between the frequencies of successive transitions and the frequency of the "combined" transition from the initial to the final state.

How could it POSSIBLY be otherwise?

Mathematically, if

(E4 - E3) + (E3 - E1) = (E4 - E1), then

h nu(4,3) + h nu(3,1) = h nu(4,1),

where nu(i,j) means the frequency of radiation emitted when the electron drops from energy level Ei to level Ej.

So, dividing that last equation by h,

nu(4,3) + nu(3,1) = nu(4,1),

which is the answer to your additional QUESTION.

Answer 2

ok, it rather is worded weirdly... First, the place does it drop to n=2 from? If it drops from ionisation state (the main clever way from the help given, yet nevertheless especially illogcal in itself, yet however...) 13.6/2^2 = 3.4 ev divide by way of (a million.6x10^-19) to get its power in joules use E = hf and rearrange it to f = E/h, the place h is the Planck consistent so divide (3.4 x 16x10^-19) by way of 6.6x10^-34 and you get the respond. undemanding Sorry, i don't have a calculator, so do it your self lol