1.5e14

Step 1: Calculate the time period of revolution in the ground state orbit of hydrogen. Given time period (T1) of ground state is T1 = 3.0 \times 10^{-16} s.
Step 2: The frequency (f1) is the reciprocal of the time period: f1 = \frac{1}{T1} = \frac{1}{3.0 \times 10^{-16}} s \approx 3.33 \times 10^{15} Hz.
Step 3: In the first excited state (n = 2), the frequency decreases by a factor of \frac{n^2}{n^2} = \frac{1^2}{2^2} = \frac{1}{4}. Thus, f2 = \frac{f1}{4} = \frac{3.33 \times 10^{15}}{4} = 0.833 \times 10^{15} Hz = 8.33 \times 10^{14} Hz.
Step 4: Therefore the frequency of revolution in the first excited state is approximately 1.5 \times 10^{14} Hz.