The Hydrogen Atom

[Phoebe Chen 4I]

What is supposed to be the right answer?

[David S]

Since the answers of this question all have to do with energy gained/absorbed as a result of electron movement between energy level shells, we will use the following formula to determine the right answer:

, where h, c, and -R_h are all constants that we can ignore for this problem.

For answer choice A, note that E is directly proportional to the magnitude of the energy difference () between an electron's initial energy level shell n_i and its final energy level shell n_f. When comparing the transition between n=4 to n=2 and n=5 to n=3, we get that |1/16 - 1/4| > |1/25 - 1/9|. Thus, the transition between n=4 to n=2 corresponds with a larger amount of energy released. A is wrong.

For answer choice B, note that lambda is inversely proportional to the magnitude of the energy difference () between an electron's initial energy level shell n_i and its final energy level shell n_f. When comparing the transition between n=4 to n=2 and n=5 to n=1, we get that |1/16 - 1/4| < |1/25 - 1/1|. The transition between n=4 to n=2 has a smaller energy difference magnitude, which corresponds with a larger (or longer) wavelength. Since this is what B says, B is the correct answer.

As for C, when an electron goes from a higher energy level shell to a lower energy level shell ( n_i > n_f), it emits a photon. Photon absorption corresponds with cases when an electron goes from lower energy level shell to a higher one ( n_i < n_f). C is wrong.

As for D, we must consider that the typical Hydrogen atom has 1 electron which resides in the lowest subshell possible, 1s, which has principal quantum number n=1. Thus, a transition from n=1 to n=infinity would give the ionization energy. Since choice D states n=2 to n="some large value", it is also incorrect.

I hope this helps!

[Phoebe Chen 4I]

It does! Thank you!