JEE Main 2025 — Structure Of Atom Question with Solution

To find the energy of an electron in the first excited state of a Be³⁺ ion, we can use the formula for the energy of an electron in a hydrogen-like atom:

${\mathrm{E}}_{\mathrm{T}}=-13.6\frac{Z^2}{n^2}\text{ eV}$

Given:

Energy of the first orbit (ground state) of the hydrogen atom: ${\mathrm{E}}_1=-13.6\text{ eV}$ (where $Z=1$ and $n=1$).

For Be³⁺:

Atomic number $Z=4$.

First excited state corresponds to $n=2$.

Calculation for Be³⁺:

The energy ratio between the hydrogen atom and the Be³⁺ ion can be written as:

$\frac{{\mathrm{E}}_{\mathrm{H}}}{{\mathrm{E}}_{{\mathrm{Be}}^{+3}}}=\frac{Z_1^2}{n_1^2}\times \frac{n_2^2}{Z_2^2}$

Substitute the known values:

$\frac{-13.6}{{\mathrm{E}}_{{\mathrm{Be}}^{+3}}}=\frac{1^2}{1^2}\times \frac{2^2}{4^2}$

$\frac{-13.6}{{\mathrm{E}}_{{\mathrm{Be}}^{+3}}}=\frac{1}{1}\times \frac{4}{16}$

Solving this gives:

${\mathrm{E}}_{{\mathrm{Be}}^{+3}}=-13.6\times 4=-54.4\text{ eV}$

The magnitude of the energy of the electron in the first excited state of Be³⁺ is $\left|-54.4\right|=54.4\text{ eV}$, which is approximately 54 eV when rounded to the nearest integer.