# Finding excited stage of electron from its potential energy

The potential energy of an electron in the hydrogen atom is $$\pu{-6.8 eV}.$$ Indicate the excited stage in which electron is present.

Total energy would be equal to $$\pu{-3.4 eV}.$$ I used the formula

$$-13.6\left(\frac{1}{1^2} - \frac{1}{n^2}\right) = \pu{-3.4 eV};$$

$$n^2 - 1/ n^2 = 4.$$

I got answer as $$3n^2 = 4,$$ which is obviously not correct. Where am I going wrong? I need help either with the formula or calculation I did wrong.

• Replace n by infinity and right side by 0 eV for a free electron and you will see the error. Or, use n=1 and right side -13.6 eV. You would probably avoid the error,if you used algebraic symbolic equation first, as error in thinking would be easier to spot. In fact, if you spent the time for writing the question by additional thinking, you would have the answer much sooner. Commented Dec 15, 2020 at 19:13
• Why is n = infinity. If n is equal to infinity , then we don’t have any variable . Do you mean to say that we don’t know whether electron is in first orbit or 2nd orbit ? I didn’t understand why n = infinity@Poutnik Commented Dec 15, 2020 at 20:01
• If n is from 1 to 2 , it is 1st excited change.Why does n has to be = infinity ? Commented Dec 15, 2020 at 20:07
• n=infinity to make your error more obvious. :-) Commented Dec 15, 2020 at 20:07
• Big one. I intentionally do not say which one, as you would learn more if you discover it yourself Commented Dec 15, 2020 at 20:09

The potential energy ($$U$$) of an electron in nth orbit of hydrogen atom is given below : $$U = \frac{-27.2}{n^2} \space (in \space eV)$$ $$So, \space -6.8 = \frac{-27.2}{n^2}$$ $$\implies n=2$$ As the electron is in second orbit it is said to be in its first excited state.