Question:
The ground state energy of hydrogen atom is $-13.6 \mathrm{eV}$. The energy of second excited state of $\mathrm{He}^{+}$ion in $\mathrm{eV}$ is:
Correct Option: , 3
Solution:
According to Bohr's model energy in $n^{\text {th }}$ state
$=-13.6 \times \frac{Z^{2}}{n^{2}} \mathrm{eV}$
For second excited state, of $\mathrm{He}^{+}, n=3$
$\therefore E_{3}\left(\mathrm{He}^{+}\right)=-13.6 \times \frac{2^{2}}{3^{2}} \mathrm{eV}=-6.04 \mathrm{eV}$