Given data

• Time period in ground state orbit ($n=1$): $T_1 = 1.6 \times 10^{-16}\ \text{s}$

• We need the frequency of revolution in the first excited state, i.e. for $n=2$.

Use Bohr model relation

In Bohr's model:

• Radius: $r_n \propto n^2$
• Speed: $v_n \propto \frac{1}{n}$

Therefore, time period $T_n = \frac{2\pi r_n}{v_n} \propto \frac{n^2}{1/n} = n^3$

So, $T_n \propto n^3$

Find time period in first excited state

For $n=2$: $\frac{T_2}{T_1} = \left(\frac{2}{1}\right)^3 = 8$

Hence, $T_2 = 8T_1 = 8 \times 1.6 \times 10^{-16}$ $T_2 = 12.8 \times 10^{-16} = 1.28 \times 10^{-15}\ \text{s}$

Calculate frequency in first excited state

Frequency is reciprocal of time period: $f_2 = \frac{1}{T_2} = \frac{1}{1.28 \times 10^{-15}}$

$f_2 = \frac{1}{1.28} \times 10^{15} \approx 0.78125 \times 10^{15}$

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

Match with options

$7.8 \times 10^{14}\ \text{s}^{-1}$

This corresponds to Option C.

Comparison with stored answer

• Derived answer: C
• Stored correct answer: C

They match.