The energy of an electron in the first Bohr orbit of hydrogen atom is $$-2.18 \times 10^{-18}$$ J. Its energy in the third Bohr orbit is _____.

The energy of an electron in the $$n$$th Bohr orbit of hydrogen is given by:

$$E_n = \frac{E_1}{n^2}$$

Given $$E_1 = -2.18 \times 10^{-18}$$ J.

$$E_3 = \frac{E_1}{3^2} = \frac{E_1}{9} = \frac{-2.18 \times 10^{-18}}{9} = -0.2422 \times 10^{-18} \text{ J}$$

$$E_3 = \dfrac{1}{9}$$ of $$E_1$$. This means the energy in the third Bohr orbit is $$\dfrac{1}{9}$$th of the value of the energy in the first orbit.

The correct answer is Option A: $$\dfrac{1}{9}$$th of the value.