question_answer

In accordance with the Bohr's model, find the quantum number that characterises the Earth's revolution around the sun in an orbit of radius \[1.5\times {{10}^{11}}m\]  with orbital speed \[3\times {{10}^{4}}\text{ }m/s.\] (Mass of earth \[=6\times {{10}^{24}}kg\])

Answer:

Given, radius of orbit,      \[r=1.5\times {{10}^{11}}m\] Orbital speed,     \[v=3\times {{10}^{4}}\text{ }m/s\] Mass of Earth,    \[M=6\times {{10}^{24\,}}kg\] Angular momentum, \[Mvr=\frac{nh}{2\pi }\] or \[n=\frac{2\pi vrM}{h}\] [where, n is the quantum number of the orbit] \[=\frac{2\times 3.14\times 3\times {{10}^{4}}\times 1.5\times {{10}^{11}}\times 6\times {{10}^{24}}}{6.63\times {{10}^{-34}}}\] \[=2.57\times {{10}^{74}}\] or \[n=2.6\times {{10}^{74}}\] Thus, the quantum number is \[2.6\times {{10}^{74}}\] which is too large.