In accordance with the Bohr's model, the quantum number that characterises the Earth's revolution around the Sun in an orbit of radius \(1.5 \times 10^{11} \mathrm{~m}\) with orbital speed \(3 \times 10^4 \mathrm{~ms}^{-1}\) is
[given, mass of Earth \(=6 \times 10^{24} \mathrm{~kg}\) ]

Given, \(v=3 \times 10^4 \mathrm{~m} / \mathrm{s}\)
\[
\begin{aligned}
r & =1.5 \times 10^{11} \mathrm{~m} \\
m_e & =6 \times 10^{24} \mathrm{~kg}
\end{aligned}
\]
According Bohr's atomic model,
Angular momentum \(=m v r=\frac{n h}{2 \pi}\)
where, \(h=\) Planck's constant \(=6.62 \times 10^{-34} \mathrm{~J}-\mathrm{s}\)
and \(n=\) quantum number
\[
\begin{aligned}
\therefore \quad n & =\frac{2 \pi\left(m_e v r\right)}{h} \\
& =\frac{2 \times 314 \times 6 \times 10^{24} \times 3 \times 10^4 \times 1.5 \times 10^{11}}{6.62 \times 10^{-34}} \\
& =2.57 \times 10^{74}
\end{aligned}
\]
Hence, the quantam number that characterises the Earth's revolution is \(2.57 \times 10^{74}\).