[Bjc51192]

Hey everyone,

So I was working on this problem, and got the right answer I just dont understand why or even if the method I used was right in the first place. Here is the question, " which of the following transitions ( in a hydrogen atom) reoresent emission  of the longest wavelength photon?

A) n=3 to n=1
b) n=5 to n=4
c) n=4 to n=2
d) n=3 to n=4
e) n=1 to n=2


When I saw the choices the first thing that came to my mind is this formula deltaE= -2.18*10^-18 j(1/n^2f-1/n^2I) f meaning final, and I meaning initial. The correct answer is b and it was the largest negative of the others so it seemed like the right choice.

Thank you

[opti384]

The equation that you know might come in handy, but if you know the concepts working here, you can solve this problem more easily.

So let's start with the electron gaining energy. Do you know about the ground state and excited state?

[Bjc51192]

No I do not =/

[opti384]

You can look those up in wikipedia for more details, but in simply put, when an electron (let's say in a hydrogen atom) gains energy, it becomes in an excited state from ground state from n=1 to n=3 for example. The more energy an electron gains the further it goes. If an electron gains enough energy it will be ionized and the electron will reach n=∞. Anyway, from the excited state if the electron releases energy it will again become in the ground state. And the energy will be released as an electromagnetic wave. You can calculate the wavelength if you know the amount of energy from the equation E = hc/λ.