Question

The total energy of an electron in the ground state of the hydrogen atom is $-13.6\text{eV}.$ The kinetic energy of an electron in the first excited state is

• A. $3.4\text{eV}$
• B. $6.8\text{eV}$
• C. $13.6\text{eV}$
• D. $1.7\text{eV}$

Answer 1

The correct option is A $3.4\text{eV}$

The energy of hydrogen atom when the electron revolves in the $n^{th}$ orbit is,

$E_n=-13.6\frac{Z^2}{n^2}\text{eV}=-\frac{13.6}{n^2}\text{eV}[\because Z=1]$

For ground state, $(n=1)$

$E_1=\frac{-13.6}{1^2}\text{eV}=-13.6\text{eV}$

For the first excited state, $(n=2)$

$E_2=\frac{-13.6}{2^2}\text{eV}=-3.4\text{eV}$

The total energy of an electron in the orbit is equal to the negative of its kinetic energy.i.e.,

$\therefore {\text{(K.E)}}_2=-{\text{E}}_2=3.4\text{eV}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.

Why this question ? Key idea: The relation between the kinetic, potential, and total energy of a state in the hydrogen atom is, $\left|\begin{array}{c}{\text{(K.E)}}_n\end{array}\right|=\left|\begin{array}{c}{\text{(E)}}_n\end{array}\right|=\frac{1}{2}\left|\begin{array}{c}{\text{(P.E)}}_n\end{array}\right|$