The energy of an electron in first Bohr's orbit of H atom is $$-13.6$$ eV. The energy value of electron in the first excited state of Li$$^{2+}$$ is:

The energy of an electron in a hydrogen-like atom is given by the formula:

$$ E_n = -\frac{13.6 Z^2}{n^2} \text{ eV} $$

where $$ E_n $$ is the energy in electron volts (eV), $$ Z $$ is the atomic number, and $$ n $$ is the principal quantum number.

For the hydrogen atom (H), the given energy in the first Bohr's orbit (where $$ n = 1 $$) is -13.6 eV. Since hydrogen has $$ Z = 1 $$, this matches the formula: $$ E_1 = -\frac{13.6 \times 1^2}{1^2} = -13.6 \text{ eV} $$.

Now, we need to find the energy for the Li²⁺ ion. Lithium (Li) has an atomic number $$ Z = 3 $$. The Li²⁺ ion has lost two electrons, leaving only one electron, so it is a hydrogen-like system. Therefore, we can use the same energy formula.

The question asks for the energy in the first excited state. The first excited state corresponds to the second orbit, so $$ n = 2 $$.

Substituting $$ Z = 3 $$ and $$ n = 2 $$ into the formula:

$$ E_2 = -\frac{13.6 \times (3)^2}{(2)^2} $$

First, calculate $$ (3)^2 $$:

$$ (3)^2 = 3 \times 3 = 9 $$

Next, calculate $$ (2)^2 $$:

$$ (2)^2 = 2 \times 2 = 4 $$

Now substitute these values:

$$ E_2 = -\frac{13.6 \times 9}{4} $$

Multiply 13.6 by 9:

$$ 13.6 \times 9 = 122.4 $$

Now divide by 4:

$$ \frac{122.4}{4} = 30.6 $$

So the expression becomes:

$$ E_2 = -30.6 \text{ eV} $$

Therefore, the energy value of the electron in the first excited state of Li²⁺ is -30.6 eV.

Comparing with the options:

A. 27.2 eV

B. -30.6 eV

C. 30.6 eV

D. -27.2 eV

The value -30.6 eV corresponds to option B.

Hence, the correct answer is Option B.