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

Question

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?

ytlu · 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 $n=infty$
$$
  E_infty = -frac{13.6}{infty^2} = 0 eV.
$$

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

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