Q136 / 180 · KCET 2022 · Physics
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 $=6 \times 10^{24} \mathrm{~kg}$ ]
- A$2.57 \times 10^{38}$
- B$8.57 \times 10^{64}$
- Correct answerC$2.57 \times 10^{74}$
- D$5.98 \times 10^{86}$
Solution
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 $2.57 \times 10^{74}$.
Hence, the correct answer is option: 'C'.