# What is the ratio of energies of Hydrogen in its first excited state and second excited state?

Since Energy of an electron in a particular orbit is given by $$E= -13.6/n^2$$ eV So, I equated the energies in $$n=2$$ and $$n=3$$ which are the 1st and 2nd excited states respectively. The answer was 9/4. But since the electron in n=3 is in higher energy state and $$n= 2$$ in lower energy state, how do I make sense of this answer?

• Think about how negative numbers behave. Which energy is higher, -2 eV or -1 eV? What happens to signs if you take the ratio of negative numbers? – Bill N Jan 22 at 14:22

## 1 Answer

Your were calcualting the ratio of

$$\frac{9}{4} = \frac{E_\infty - E_2}{E_\infty - E_3}.$$

Where $$E_\infty$$ is the lowest energy of free electron as the limit of bound energies $$n=\infty$$ $$E_\infty = -\frac{13.6}{\infty^2} = 0 eV.$$

• A lot got cleared from here but a lot of questions arose too. Sorry if I ask dumb questions or something. So, the energy in the n= infinite is 0. All energies below this are decreasing till they reach the first energy level. where energy is -13.6 eV, which is the lowest possible energy state ( which can also be known intuitively as it can do the least possible work).If electrons were positively charged, the value of 9/4 would be intuitively understood because the charge closer to another positive charge could do greater work. What ratio do I have to take so that I would get the correct answer? – Bikash Adhikari Jan 25 at 10:11
• @BikashAdhikari if electron has a positive charge. Then there had no bound state for positron. – ytlu Jan 25 at 16:45