Bohr Model from 2nd Law

Please derive the Bohr model of atom starting from the second Law of Physics ($F=m*a$)

Assume hydrogen atom. $Z=+1$ (a single charge $e$ is the nucleus)

From Newton law and Coulomb law

$m*a =\frac{K e^2}{R^2}$   or  $a =\frac{K e^2}{m R^2}$

For circular motion

$a =\frac{v^2}{R}$ that is   $\frac{v^2}{R} = \frac{K e^2}{m R^2}$  or   $v^2 = \frac{K e^2}{m R}$       (1)

Angular moment is $L =R*P =m v R$
$L = \sqrt{K m R*e^2}$

$L$ is quantified means   $L= n * \hbar$ ($n$ is the quantum number)

$n^2*\hbar^2 = K m R*e^2$   or     $R = \frac{n^2*\hbar^2}{K m e^2}$    (2)

Back into (1) one has:
$v^2 =\frac{Ke^2}{mR} = \frac{Ke^2)^2}{n^2*\hbar^2}$

$En =Ek -U$   or     $En =\frac{mv^2}{2} – \frac{Ke^2}{R} = \frac{mv^2}{2} – mv^2 = -\frac{mV^2}{2}$
$En =- \frac{m(Ke^2)}{2*(n^2*\hbar^2)}$   (3)

[ Taken from Wiki one has (for hydrogen Z=1)]:

$E=\frac{Zke^2}{2r_n}=-\frac{Z^2(ke^2)^2m_e}{2\hbar^2n^2}\approx \frac{-13.6Z^2}{n^2} eV$

From equation (2):

$R = n^2*\hbar^2 /(K m e^2)$    (2)

$R1 = \hbar^2/(K m e^2) =5.3*10^{-11} m =0.53 A$

From equation (3) we have

$E1 =- m(K e^2)/ [2*\hbar^2] =-13.6 eV$