To determine the kinetic energy (K.E.) of an electron in the first excited state of a hydrogen atom, we need to understand the relationship between the total energy, potential energy, and kinetic energy in an atom.

In a hydrogen atom, the total energy (E) of an electron in the nth state is given by:

$$E_n = - \frac{13.6}{n^2} \mathrm{~eV}$$

For the first excited state, $ n = 2 $. So, plugging in the value:

$$E_2 = - \frac{13.6}{2^2} = - \frac{13.6}{4} = - 3.4 \mathrm{~eV}$$

This value represents the total energy (E) of the electron in the first excited state. According to the virial theorem for an electron in a Coulomb potential (as in a hydrogen atom), the kinetic energy (K.E.) is equal to the negative of the total energy:

$$\mathrm{K.E.} = - E$$

Substituting the total energy we calculated:

$$\mathrm{K.E.} = - (-3.4) = 3.4 \mathrm{~eV}$$

Now, we need to find the value of $$x$$ in the form of $$x \times 10^{-1} \mathrm{~eV}$$.

$$3.4 \mathrm{~eV} = 34 \times 10^{-1} \mathrm{~eV}$$

Therefore, the value of $$x$$ is 34.