Question

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 \(1.5×10^{11} m\) with orbital speed \(3×10^4 m/s\). (Mass of earth\(=6.0×10^{24} kg\). )

Answer 1

The correct answer is: \(2.6×10^{74}.\)
Radius of the orbit of the Earth around the Sun, \(r=1.5×10^{11}m\)
Orbital speed of the Earth, \(v=3×10^4m/s\)
Mass of the Earth, \(m=6.0×10^{24}kg\)
According to Bohr's model,angular momentum is quantized and given as:
\(mvr=\frac{nh}{2π}\)
Where, h=Planck's constant\(=6.62×10^{-34}Js\)
n=Quantum number
\(∴n=\frac{mvr2π}{h}\)
\(=\frac{2π×6×10^{24}×3×10^4×1.5×10^{11}}{6.62×10^{-34}}\)
\(=25.61×10^{73}=2.6×10^{74}\)
Hence,the quantum number that characterizes the Earth' revolution is \(2.6×10^{74}.\)