The total energy of an electron in the first excited state of...

We have, Kinetic energy of electron,

$K = \dfrac{Ze^{2}}{8\pi{ε}_{0} \cdot r}$

also, Potential energy of electron,

$U = - \dfrac{1}{4\pi{ε}_{0}} \cdot \dfrac{Ze^{2}}{r}$

$∴$ Total energy

$E = K + U$

$= \dfrac{Ze^{2}}{8\pi{ε}_{0} \cdot r} - \dfrac{Ze^{2}}{4\pi{ε}_{0} \cdot r}$

$E = - \dfrac{Ze^{2}}{8\pi{ε}_{0}r}$ or $K = - E = - \left(-3. 4\right)$

$= 3. 4eV</p>$