1. Use Bohr energy formula for hydrogen-like species

For a hydrogen-like ion, $E_n = -13.6\,\frac{Z^2}{n^2}\ \text{eV}$ where:

• $Z$ = atomic number
• $n$ = orbit number

2. Identify the ion and the state

For $Be^{3+}$:

• Beryllium has atomic number $Z=4$
• Since it is $Be^{3+}$, it has only one electron, so Bohr formula applies.

The first excited state corresponds to: $n=2$

3. Calculate the energy

$E_2 = -13.6\times \frac{4^2}{2^2}$

$E_2 = -13.6\times \frac{16}{4}$

$E_2 = -13.6\times 4 = -54.4\ \text{eV}$

4. Take the magnitude

$|E_2| = 54.4\ \text{eV}$

Nearest integer value: $54$

5. Comparison with stored answer

Derived answer = $54$

Stored correct answer = $54$

So they match.