The time period of revolution of electron in its ground state orbit in a hydrogen atom is $$1.6 \times 10^{-16}$$ s. The frequency of revolution of the electron in its first excited state (in s$$^{-1}$$) is:

We are told that for the ground state (principal quantum number $$n=1$$) the electron in hydrogen takes a time $$T_1 = 1.6 \times 10^{-16}\,\text{s}$$ to complete one revolution.

First we change this into frequency, because frequency is the reciprocal of time period.

By definition, $$\text{frequency} = \dfrac{1}{\text{time period}}.$$

So for the ground state we have

$$f_1 = \dfrac{1}{T_1} = \dfrac{1}{1.6 \times 10^{-16}}.$$

Carrying out the division,

$$f_1 = \dfrac{1}{1.6}\times 10^{16} = 0.625 \times 10^{16} = 6.25 \times 10^{15}\,\text{s}^{-1}.$$

Now we want the frequency for the first excited state, i.e. for $$n = 2.$$

In Bohr’s model the radius of the orbit varies as $$r_n \propto n^2$$ and the speed varies as $$v_n \propto \dfrac{1}{n}.$$

Using the formula for time period $$T_n = \dfrac{2\pi r_n}{v_n},$$ we substitute the proportionalities:

$$T_n \propto \dfrac{n^2}{1/n} = n^3.$$

Thus the time period increases as the cube of the quantum number, or equivalently, the frequency decreases as the cube of the quantum number:

$$f_n \propto \dfrac{1}{n^3}.$$

Therefore, when we move from $$n=1$$ to $$n=2,$$ the new frequency is obtained by dividing the ground-state frequency by $$2^3 = 8.$$

$$f_2 = \dfrac{f_1}{2^3} = \dfrac{6.25 \times 10^{15}}{8}.$$

Performing the division,

$$f_2 = 0.78125 \times 10^{15}\,\text{s}^{-1} = 7.8125 \times 10^{14}\,\text{s}^{-1}.$$

Rounded to the proper significant figures,

$$f_2 \approx 7.8 \times 10^{14}\,\text{s}^{-1}.$$

Hence, the correct answer is Option B.