The total energy of an electron in the first excited state of hyrogen is about - 3.4 eV. Its kinetic energy in this state is :

The total energy $E$ of an electron in a hydrogen atom is given by $E=-\frac{13.6}{n^2}$ eV, where $n$ is the principal quantum number. For the first excited state, $n=2$, so $E=-\frac{13.6}{2^2}=-3.4$ eV. The kinetic energy $KE$ of the electron is related to the potential energy $PE$ by $E=KE+PE$. In a hydrogen atom, the potential energy is $-2$ times the total energy (since $E=\frac{1}{2}KE+PE$ and for a bound state, $PE=-2E$), so $PE=-2\times (-3.4)=6.8$ eV. Then, $KE=E-PE=-3.4-6.8=-10.2$ eV is incorrect because we should consider the relationship $E=KE+PE$ and the fact that $PE=-2E$ for a bound state implies $KE=E-PE=E+2E=3E$ for the relationship between $KE$ and $E$ in the context of atomic physics. However, the mistake here is in interpreting the relationship between total energy, kinetic energy, and potential energy for an electron in a bound state. The correct approach should consider that the total energy is negative, and the kinetic energy is positive but less than the absolute value of the total energy. For an electron in the first excited state with a total energy of $-3.4$ eV, to find the kinetic energy, we should recall that the potential energy is $-2$ times the total energy for a bound electron in a hydrogen-like atom, but this does not directly apply to calculating kinetic energy from total energy without considering the specific energy level. The kinetic energy of the electron in the $n^{th}$ state of hydrogen is given by $KE=\frac{13.6}{n^2}$ eV. For $n=2$, $KE=\frac{13.6}{4}=3.4$ eV. This positive value represents the kinetic energy of the electron in that state.