The energy of an electron in the ground state (n = 1) for He⁺ ion is –x J, then that for an electron in n = 2 state for Be³⁺ ion in J is

The correct option is -x.

According to Bohr's model of the hydrogen-like atom, the energy of an electron in the n^th orbit of a species with atomic number Z is given by the formula:

$E_n=-E_0\times \frac{Z^2}{n^2}$

where $E_0$ is a constant representing the ground-state energy of a hydrogen atom (Z = 1, n = 1).

Step 1: Calculate the energy for Helium ion (He⁺)
For He⁺, the atomic number Z = 2.
The electron is in the ground state, so n = 1.
Substituting these values into the energy equation:

$E_1\left({\mathrm{He}}^{+}\right)=-E_0\times \frac{2^2}{1^2}=-4E_0$

Given in the problem, this energy is equal to -x J:

$-4E_0=-x$

Therefore, we have:

$E_0=\frac{x}{4}$

Step 2: Calculate the energy for Beryllium ion (Be³⁺)
For Be³⁺, the atomic number Z = 4.
The electron is in the state n = 2.
Substituting these values into the energy equation:

$E_2\left({\mathrm{Be}}^{3+}\right)=-E_0\times \frac{4^2}{2^2}=-E_0\times \frac{16}{4}=-4E_0$

Step 3: Relate the two values
Substitute the value of $E_0$ from Step 1:

$E_2\left({\mathrm{Be}}^{3+}\right)=-4\times \left(\frac{x}{4}\right)=-x\text{J}$

Thus, the energy of the electron in the n = 2 state for the Be³⁺ ion is also -x J.