Consider the solar system as a large atom. The quantum number (n) that characterises Earth's orbit (radius = 1.5 × $$10^{11}$$​ m) with Earth moving at an orbital speed of 3 × $$10^4$$​ m/s is (mass of Earth is 6 × $$10^{24}$$​ kg):

Correct option is B

​$$\text{To find the quantum number } n \text{ that characterizes Earth's orbit, we can use the following formula for angular momentum in the Bohr model of an atom:} \\mvr = \frac{nh}{2\pi} \\\text{Where:} \\m \text{ is the mass of the Earth,} \\v \text{ is the orbital speed of the Earth,} \\r \text{ is the radius of Earth's orbit,} \\n \text{ is the quantum number,} \\h \text{ is Planck's constant }$$$$\text{Step 1: Rearrange the equation to solve for } n: \\n = \frac{2\pi mvr}{h} \\\text{Step 2: Substitute the known values:} \\\text{Mass of Earth } m = 6 \times 10^{24} \, \text{kg}, \\\text{Orbital speed of Earth } v = 3 \times 10^4 \, \text{m/s}, \\\text{Radius of Earth's orbit } r = 1.5 \times 10^{11} \, \text{m}, \\\text{Planck's constant } h = 6.626 \times 10^{-34} \\\text{Substitute these values into the equation:} \\n = \frac{2\pi \times (6 \times 10^{24}) \times (3 \times 10^4) \times (1.5 \times 10^{11})}{6.626 \times 10^{-34}} \\\text{Step 3: Simplify the calculation:} \\n = \frac{2 \times 3.14 \times 6 \times 10^{24} \times 3 \times 10^4 \times 1.5 \times 10^{11}}{6.626 \times 10^{-34}} \\n = \frac{25.61 \times 10^{73}}{6.626 \times 10^{-34}} = 2.56 \times 10^{74} \\\text{Thus, the quantum number } n \text{ is } 2.56 \times 10^{74}.$$​​​