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

We need to find the kinetic energy of the electron in the first excited state of hydrogen, expressed as $$x \times 10^{-1}$$ eV. For a hydrogen atom, the total energy, kinetic energy, and potential energy are related by $$E_n = -\frac{13.6}{n^2} \, \text{eV}$$, $$KE_n = -E_n = \frac{13.6}{n^2} \, \text{eV}$$, and $$PE_n = 2E_n = -\frac{27.2}{n^2} \, \text{eV}$$. The kinetic energy is always positive and equals the magnitude of the total energy, which follows from the virial theorem for a Coulomb potential: $$KE = -E$$ and $$PE = 2E$$.

Since the first excited state corresponds to $$n = 2$$, we have $$E_2 = -3.4 \, \text{eV}$$. Substituting into the expression for kinetic energy gives $$KE = -E_2 = -(-3.4) = 3.4 \, \text{eV}$$. Expressing this value in the form $$x \times 10^{-1}\, \text{eV}$$ yields $$3.4 \, \text{eV} = 34 \times 10^{-1} \, \text{eV}$$, so $$x = 34$$.

Option 34