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?
(b) Please find the initial frequency of radiation of the atom?
(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?
(d) For the electron radius of 0.5 Angstrom please find the frequency of radiation.
(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.)
Answer
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
$a =k*e^2/(mR^2) =10^{23} m/s^2$
$\omega =\sqrt{a/R} =4.5*10^{16} rad/s$
$F = \omega/(2*pi) =7.16*10^15 Hz$
