For hydrogen atom, energy of an electron in first excited state is $-3.4 \mathrm{~eV}, \mathrm{K} . \mathrm{E}$. of the same electron of hydrogen atom is $x \mathrm{~eV}$. Value of $x$ is _________ $\times 10^{-1} \mathrm{~eV}$. (Nearest integer)

<p>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. <p>In a hydrogen atom, the total energy (E) of an electron in the nth state is given by:</p></p> <p> <p>$E_n = - \frac{13.6}{n^2} \mathrm{~eV}$</p> </p> <p>For the first excited state, $ n = 2 $. So, plugging in the value:</p> <p> <p>$E_2 = - \frac{13.6}{2^2} = - \frac{13.6}{4} = - 3.4 \mathrm{~eV}$</p> </p> <p>This value represents the total energy (E) of the electron in the first excited state. According to the virial theorem for an electron in a Coulomb potential (as in a hydrogen atom), the kinetic energy (K.E.) is equal to the negative of the total energy:</p> <p> <p>$\mathrm{K.E.} = - E$</p> </p> <p>Substituting the total energy we calculated:</p> <p> <p>$\mathrm{K.E.} = - (-3.4) = 3.4 \mathrm{~eV}$</p> </p> <p>Now, we need to find the value of $x$ in the form of $x \times 10^{-1} \mathrm{~eV}$.</p> <p> <p>$3.4 \mathrm{~eV} = 34 \times 10^{-1} \mathrm{~eV}$</p> </p> <p>Therefore, the value of $x$ is 34.</p>