Question

What will be the energy corresponding to the first excited state of a hydrogen atom if the potential energy of the atom is taken to be 10 eV when the electron is widely separated from the proton? Can we still write En = E1/n2, or rn = a0 n2?

Answer 1

- 13.4 eV is the energy corresponding to the first excited state of a hydrogen atom when the potential energy is 10 eV.

Yes, we can still write $\mathrm{E}_{n}=\mathrm{E}_{1} / n^{2}, \text { or } r_{n}=a_{0} n^{2}$.

Explanation:

Energy of $n^{t h}$ state of hydrogen is given by

$E_{n}=-\frac{13.6}{n^{2}} e V$

Energy of first excited state $(n=2)$ $of hydrogen,$ $E_{1}=-\frac{13.6}{4} \mathrm{eV}=-3.4 \mathrm{eV}$

This relationship remains true when energy at the reference point is zero. The energy reference point is typically the energy of the atom when the electron is completely isolated from the proton. In this question the atom's potential energy is taken as 10 eV. So here our reference point energy is 10 eV when the electron is widely isolated from the proton. Earlier the energy of the first excited state was -3.4 eV, When that point of reference had zero resources but now as the reference point has changed so the first excited state's energy will also change by the corresponding amount.

Thus, $E_{I}=-3.4 \mathrm{eV}-10 \mathrm{eV}=-13.4 \mathrm{eV}$.

We are still writing $\mathrm{E}_{n}=\mathrm{E}_{1} / n^{2}, \text { or } r_{n}=a_{0} n^{2}$ because these formulas are independent of the refrence point enegy.