The ratio for the speed of the electron in the $$3^{rd}$$ orbit of $$He^+$$ to the speed of the electron in the $$3^{rd}$$ orbit of hydrogen atom will be :

The speed of an electron in the $$n^{th}$$ orbit of a hydrogen-like atom is given by:

$$v_n = \frac{Z e^2}{2 \varepsilon_0 n h}$$

This means the speed is proportional to $$\frac{Z}{n}$$, where $$Z$$ is the atomic number and $$n$$ is the orbit number.

For $$He^+$$ (Z = 2) in the 3rd orbit:

$$v_1 \propto \frac{Z_1}{n_1} = \frac{2}{3}$$

For Hydrogen (Z = 1) in the 3rd orbit:

$$v_2 \propto \frac{Z_2}{n_2} = \frac{1}{3}$$

The ratio of speeds is:

$$\frac{v_1}{v_2} = \frac{2/3}{1/3} = \frac{2}{1}$$

Therefore, the ratio of the speed of the electron in the 3rd orbit of $$He^+$$ to the speed in the 3rd orbit of hydrogen is $$2 : 1$$.

The correct answer is Option D.