Speed of an electron in Bohr's $$7^{th}$$ orbit for Hydrogen atom is $$3.6 \times 10^6$$ m s$$^{-1}$$. The corresponding speed of the electron in $$3^{rd}$$ orbit, in m s$$^{-1}$$ is:

We need to find the speed of an electron in the 3rd orbit of hydrogen, given that the speed in the 7th orbit is $$3.6 \times 10^6$$ m/s.

Speed in Bohr's orbit:

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

$$v_n = \frac{v_0}{n}$$

where $$v_0$$ is a constant (speed in the first orbit).

This means speed is inversely proportional to the orbit number: $$v \propto \frac{1}{n}$$.

Using the ratio:

$$\frac{v_3}{v_7} = \frac{7}{3}$$

$$v_3 = v_7 \times \frac{7}{3} = 3.6 \times 10^6 \times \frac{7}{3} = 8.4 \times 10^6$$ m/s

The speed of the electron in the 3rd orbit is $$8.4 \times 10^6$$ m/s.

The correct answer is Option 4: $$8.4 \times 10^6$$ m/s.