The ground state energy of hydrogen atom is $-13.6 \mathrm{eV}$. When its electron is in the first excited…
The ground state energy of hydrogen atom is $-13.6 \mathrm{eV}$. When its electron is in the first excited state, its excitation energy is
$3.4 \mathrm{eV}$
$6.8 \mathrm{eV}$
$10.2 \mathrm{eV}$
zero
Solution
Key Idea : Excitation energy is defined as the energy required to take the electron from ground level orbit to any higher order orbit (ie, $n=2,3,4 \ldots)$.
Given, ground state energy of hydrogen atom
$E_1=-13.6 \mathrm{eV}$
Energy of electron in first excited state (ie, $n=2$)$
E_2=-\frac{13.6}{(2)^2} \mathrm{eV}$
Therefore, excitation energy
$\begin{aligned}
\Delta E & =E_2-E_1 \\
& =-\frac{13.6}{4}-(-13.6) \\
& =-3.4+13.6=10.2 \mathrm{eV}
\end{aligned}$