Experimentally it is found that 12.8 eV energy is required to separate a hydrogen atom into a proton and an electron. So the orbital radius of the electron in a hydrogen atom is $$\dfrac{9}{x} \times 10^{-10}$$ m. The value of the $$x$$ is ______.
(1 eV = $$1.6 \times 10^{-19}$$ J, $$\dfrac{1}{4\pi\varepsilon_0} = 9 \times 10^9$$ $$\dfrac{Nm^2}{C^2}$$ and electronic charge = $$1.6 \times 10^{-19}$$ C)

Initially, we note that separating a hydrogen atom into a proton and an electron requires 12.8 eV of energy, and we aim to determine the value of $$x$$ such that the orbital radius is $$\frac{9}{x} \times 10^{-10}$$ m.

In electrostatic terms, the binding energy of the electron in a hydrogen atom is given by the relation:

$$|E| = \frac{ke^2}{2r}$$

Here, $$k = \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ N m}^2\text{/C}^2$$.

Rearranging this expression allows us to solve for the orbital radius $$r$$ as

$$r = \frac{ke^2}{2|E|}$$

We convert the given energy into joules by using $$1\text{ eV} = 1.6 \times 10^{-19}$$ J, yielding

$$|E| = 12.8\text{ eV} = 12.8 \times 1.6 \times 10^{-19}\text{ J} = 20.48 \times 10^{-19}\text{ J}.$$

Substituting the numerical values into the expression for $$r$$ gives:

$$r = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{2 \times 20.48 \times 10^{-19}}$$

At this point, we compute numerator and denominator separately:

Numerator: $$9 \times 10^9 \times 2.56 \times 10^{-38} = 23.04 \times 10^{-29}$$

Denominator: $$2 \times 20.48 \times 10^{-19} = 40.96 \times 10^{-19}$$

Therefore, the orbital radius becomes

$$r = \frac{23.04 \times 10^{-29}}{40.96 \times 10^{-19}} = \frac{23.04}{40.96} \times 10^{-10} = 0.5625 \times 10^{-10}\text{ m}$$

Expressing this result in the form $$\frac{9}{x} \times 10^{-10}\text{ m}$$ leads to

$$r = \frac{9}{16} \times 10^{-10}\text{ m}$$

By comparison, it follows that $$x = 16$$. Thus, the desired value is 16.