In accordance with the Bohr's model, find the quantum number that characterizes the earth's revolution around the sun in an orbit of radius `1.5xx10^(11)m` with orbital speed `3xx10^4m//s`. (Mass of earth=`6.0xx10^(24)kg`)

Options

A. `5.98xx10^(86)`
B. `2.57xx10^(38)`
C. `8.57xx10^(64)`
D. `2.57xx10^(74)`

Answer

D

Explanation

Here, `r=1.5xx10^(11)m`,`v=3xx10^(4)ms^(-1)`
`m=6.0xx10^(24)kg`
Accroding to bohr's model
`mvr=(nh)/(2pi)rArrn=(2pi"mvr")/(h)`
`=(2xx22)/(7)xx(6xx10^(24)xx3xx10^(4)xx1.5xx10^(11))/(6.6xx10^(-34))=2.57xx10^(74)`