The potential energy of an electron in the hy | vedclass

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(A) The relationship between total energy $(E_n)$ and potential energy $(PE)$ is given by $E_n = \frac{1}{2} PE$. Substituting the given value: $E_n =

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(A) The relationship between total energy $(E_n)$ and potential energy $(PE)$ is given by $E_n = \frac{1}{2} PE$. Substituting the given value: $E_n = \frac{-6.8 \ eV}{2} = -3.4 \ eV$. The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is $E_n = \frac{-13.6 \ eV}{n^2}$. Equating the two: $\frac{-13.6}{n^2} = -3.4$,which gives $n^2 = 4$,so $n = 2$. The state $n = 1$ is the ground state,and $n = 2$ corresponds to the first excited state.

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

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