Using Bohr's atomic model, derive an equation of the radius of the nth orbit of an electron.

Explanation

The atomic model of Bohr is shown in the figure.

Let mass of an electron m, charge e, linear speed in `n^(th)` orbit` v_(n)` and orbit with radius `r_(n)` and charge on nucleus Ze, where Ze = atomic number of an element.
The necessary centripetal force is provided Colombian attractive force between electron and nucleus. Thus.
`:. (mv_(n)^(2))/(r_(n))=(1)/(4pi epsi_(0))*((Ze)(e))/(r_(n)^(2))=(ze^(2))/(4pi epsi_(0)r_(n)^(2))`
where `epsi_(0)`= permittivity of medium
From Bohr hypothesis 2, if the anguli momentum of electron in `n^(th)` orbit is `L_(nm)`
`L_(n)=(nh)`
`:. mv_(n)r_(n)=(nh)/(pi) " "...(2)`
From equation (1),
`v_(n)(mv_(n)r_(n))=(Ze^(2))/(4pi epsi_(0))`
Putting the vlaue of equation (2),
`v_(n)-(nh)(2pi)-(Ze^(2))/(4pi epsi_(0))` [ From equation (2)]
`:.v_(n)=(Ze^(2))/(4pi epsi_(0))*(1)/((nh)/(2pi))`
`:.v_(n)=(1)/(n)*(Ze^(2))/(2h epsi_(0)) " "...(3)`
`:.v_(n) prop(Z)/(n)` where `e, 2h,epsi_(0)` constant
Hence orbital velocity in `n^(th)` orbit velocity decreases by a factor of n.
For hydrogen atom Z = 1,
`v_(n) prop (1)/(n)`
Now from equation (2).
`r_(n)=(nh)/(2pi mv_(n))`
Putting the value of eqution (3).
`r_(n)=(nhxxnxx2h epsi_(0))/(2pi mxxZe^(2))`
`:. r_(n)=(n^(2)h^(2) epsi_(0))/(pi mZe^(2))`
`:.r_(n) prop (n^(2))/(Z)[ :.` all other terms are contant]
For hydrogen Z=1
`:. r_(n)=(n^(2)h^(2) epsi_(0))/(pi me^(2))" "...(4)`
`:. r_(n) prop n^(2)` [ `:.` All other terms are constant] The size of the innermost orbit (n = 1) is called Bohr radius, represented by the symbol `a_(0)` thus Putting n = 1 in equation (4),
`r_(1)=a_(0)=n^(2)((h^(2) epsi_(0))/(pi me^(2)))`
putting the values of `h epsi_(0), pi` and.
`a_(0)=5.29xx10^(-11)m` or approximately `0.53Å`