Question

The kinetic energy of the electron in first excited state of hydrogen atom is -

• A. $-2.4\text{eV}$
• B. $-3.4\text{eV}$
• C. $2.4\text{eV}$
• D. $3.4\text{eV}$

Answer 1

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

Kinetic energy of the electron is given by -

$KE=13.6\frac{Z^2}{n^2}\text{eV}$

For hydrogen atom, $Z=1$ and for first excited state, $n=2$.

$\therefore KE=13.6\times \frac{1^2}{2^2}=3.4\text{eV}$

Hence, option $\left(D\right)$ is the correct answer.