QJEEM24Chc938: JEE Main 2024 Chemistry - Class 11 JEEM24Chc938 Solution: For hydrogen atom, energy of an electron in first excited state …

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.