The energy of an electron in first Bohr orbit of H-atom is -13.6 eV. The magnitude of energy value of electron in the first excited state of $$Be^{3+}$$ is _____ eV. (nearest integer value)

For any one-electron (hydrogen-like) species, the Bohr energy expression is
$$E_n = -13.6\,\text{eV}\,\frac{Z^{2}}{n^{2}}$$ where
$$Z$$ = atomic number  and  $$n$$ = principal quantum number.

Given: energy of the first Bohr orbit of hydrogen $$\bigl(Z = 1,\; n = 1\bigr)$$ is $$-13.6$$ eV, which matches the above formula and confirms its use.

For $$Be^{3+}$$ we have $$Z = 4$$ (beryllium’s atomic number).
The first excited state corresponds to the second orbit, so $$n = 2$$.

Substitute these values:
$$E_{2} = -13.6\,\text{eV}\,\frac{4^{2}}{2^{2}}$$

Compute numerator and denominator:
$$4^{2} = 16,\quad 2^{2} = 4$$

Therefore,
$$E_{2} = -13.6\,\text{eV}\,\frac{16}{4} = -13.6\,\text{eV}\,\times 4 = -54.4\,\text{eV}$$

The question asks for the magnitude (absolute value):
$$|E_{2}| = 54.4\,\text{eV}$$

To the nearest integer, the required energy magnitude is 54 eV.