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.
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