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.

  • $\begingroup$ 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. $\endgroup$ – Poutnik Dec 15 '20 at 19:13
  • $\begingroup$ 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 $\endgroup$ – srijan Sri Dec 15 '20 at 20:01
  • $\begingroup$ If n is from 1 to 2 , it is 1st excited change.Why does n has to be = infinity ? $\endgroup$ – srijan Sri Dec 15 '20 at 20:07
  • 1
    $\begingroup$ n=infinity to make your error more obvious. :-) $\endgroup$ – Poutnik Dec 15 '20 at 20:07
  • 1
    $\begingroup$ Big one. I intentionally do not say which one, as you would learn more if you discover it yourself $\endgroup$ – Poutnik Dec 15 '20 at 20:09

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