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

Question

The energy of electron in the ground state ($n=1$) for $He^+$ ion is $-x$ J, then that for an electron in $n=2$ state for $Be^{3+}$ ion in J is:

Accepted Answer

According to Bohr's theory for hydrogen-like species, the energy of an electron in the $n$-th orbit is given by the expression:
$$E_n = E_0 	imes \frac{Z^2}{n^2}$$
where $E_0$ is the ground state energy of hydrogen ($-2.18 	imes 10^{-18}$ J) and $Z$ is the atomic number.

1.  **For Helium ion ($He^+$):**
    *   Atomic number $Z = 2$.
    *   Ground state $n = 1$.
    *   Given Energy $E = -x$ J.
    $$E_{He^+} \propto \frac{2^2}{1^2} = 4$$
    So, $-x \propto 4$.

2.  **For Beryllium ion ($Be^{3+}$):**
    *   Atomic number $Z = 4$.
    *   Excited state $n = 2$.
    $$E_{Be^{3+}} \propto \frac{4^2}{2^2} = \frac{16}{4} = 4$$

3.  **Comparison:**
    Since the value proportional to the energy depends on $\frac{Z^2}{n^2}$, and for both cases this ratio is 4, the energies are equal.
    $$E_{Be^{3+}} = E_{He^+} = -x 	ext{ J}$$

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