#### Question

Obtain an expression for the radius of Bohr orbit for H-atom.

#### Solution

Let us consider an electron revolving around the nucleus in a circular orbit of radius ‘r’.

According to Bohr’s first postulate, the centripetal force is equal to the electrostatic force of attraction. That is

`"mv"^2/"r"=1/(4piepsilon_o)xx"e"^2/"r"^2`

`"Or,""v"^2="e"^2/(4piepsilon_o"mr")` -------------------(1)

According to the Bohr's second postulate:

`"Angular momentum"= "n""h"/(2pi)`

`"mvr"="n""h"/(2pi)`

Or, `"v"="nh"/(2pi"mr")` -----------------(2)

Or, `"v"^2=("n"^2"h"^2)/(4pi^2"m"^2"r"^2)` ---------------------(3)

Comparing eqn (1) and eqn (3), we get

`"e"^2/(4piepsilon_o"mr")=("n"^2"h"^2)/(4pi^2"m"^2"r"^2)`

`"Or,""r"=(("h"^2epsilon_o)/(pi"me"^2))"n"^2` ----------------------(4)

This equation gives the radius of the n^{th} Bohr orbit.

`"For n"=1,"r"_1=(("h"^2epsilon_o)/(pi"me"^2))=0.537" ---------------(5)"`

`"In general,"" r"_n=(("h"^2epsilon_o)/(pi"me"^2))"n"^2`

The above equation gives the radius of Bohr orbit.