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 $2.6×10^{74}.$