JEE Main 2024 — Structure Of Atom Question with Solution

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:

$$K.E.=-E$$

Substituting the total energy we calculated:

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

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.