# Electron in classical EM theory

### Classical electron

We know from theory that a charge (an electron) that is accelerating in an atom, gives out energy at a rate of $W=\frac{2}{3}\frac{a^{2}e^{2}}{c^{3}} (ergs/s)$  where $c$ is the light speed and $a$ is the particle acceleration.

Let us say that at the initial moment $t=0$, the radius of the H atom is 1 Angstrom. For simplicity we say that the electron moves in a circle in the atom.

(a) Please find the initial acceleration $a$ of the electron?

(c) Let us say that the acceleration $a$ is the same. Please find the time necessary for the radius to go from 1Angstrom to 0.5 Angstrom?

(I have found  the frequency and centripetal acceleration of the electron in the H atom.  I guess I have to use these equations to find the result.  A new perspective would help me.)

a) The Centripetal force = The Electrical force

$m*a = k*e^2/R^2$

($m =9.1*10^-31 kg$, $k=9*10^9$)

$a = k*e^2/(mR^2) =9*10^9*(1.6*10^{-19})^2/9.1*10^{-31}*10^{-20}=$

$=2.53*10^{22} m/s^2$

b) $a= \omega^2*R$

$\omega= sqrt{a/R}$

$F = \omega/(2*pi) =1/(2*\pi)*\sqrt{a/R} = 2.53*10^{15} Hz$

c) total E for R is

$E =mv^2/2 – k*e^2/R$ (kinetic +potential)

by writing

$v=\omega*R$ and $\omega=\sqrt{a/R}$ one obtains finally:

$mv^2/2 =(1/2)m*a/R*R^2 = (1/2)m*a*R =(1/2) *k*e^2/R$

Hence

$E =-(1/2)*k*e^2/R$

For $R1 =10^{-10} m$ and $R2 =0.5*10^{-10} m$ we have

$E1-E2 = 1/2*k*e^2*(1/R2-1/R1) =9*10^9*(1.6*10^{-19})^2*10^10 =2.3*10^{-18} J$

The power radiated by the electron is

$P = (2/3)a^2*e^2/c^3 =(2/3)*(2.53*10^{22})^2*(1.6*10^{-19})^2/27*10^{24} =4*10^{-19} erg/s$

since $1 erg =10^-7 J$

$P =4*10^{-26} J/s$

The time taken to fall from $R=1 Angstrom$ to $R=0.5 A$ is

time $t = (E1-E2)/P =2.3*10^{-18} /4*10^{-26} =5.68*10^7 sec$

d) at radius $R =0.5 A$ we have

$a =k*e^2/(mR^2) =10^{23} m/s^2$

$\omega =\sqrt{a/R} =4.5*10^{16} rad/s$

$F = \omega/(2*pi) =7.16*10^15 Hz$