Falling Electron (on Nucleus)

The falling electron

Using the classic EM framework, a falling electron $e^-$ (a negative charge) emits energy/time as

$W= (2/3)*(a^2e^2/c^3)$ (erg/s) where $a$ is the acceleration of the falling electron in its circular path around the nucleus and $c$ is light’s speed. If initially the radius of the atom is $R= 1 A=10^{-10} m$ please find:

(a) The initial acceleration $a$.

(b) The frequency of the emitted radiation at $t=0$.

(c) What is the time interval necessary for the radius to go from $R=1 A$ to $R=0.5 A$, if the acceleration remains the same?

(d) When the radius is $R=0.5 A$ what is the radiation frequency?

falling electron on nucleus


a) The condition of equilibrium on the circular path is

Centripetal force = Electrical force

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

where $m =9.1*10^{-31} kg$ is the electron mass and $k=1/4\pi\epsilon0=9*10^9$ in SI units. Therefore the acceleration of the falling electron for the path radius $R=1 A$ is

$a = k*e^2/(mR^2) =9*10^9*(1.6*10^{-19})^2/9.1*10^{-31}*10^{-20}=2.53*10^{22} m/s^2$

b) From the equation $a= \omega^2*R$  or  $\omega= \sqrt{a/R}$ we find the frequency $F = \omega/(2*\pi) =1/(2*\pi)*\sqrt{a/R} = 2.53*10^{15} Hz$

c) The total energy for a certain radius $R$ is

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

and by writing the speed as $v=\omega*R$ and $\omega=\sqrt{a/R}$ we obtain finally the kinetic energy half of the potential energy in absolute value:

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

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

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

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

$=2.3*10^{-18} J$

and the power radiated is from the initial equation:

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

since $1 erg =10^{-7} J$ then $P =4*10^{-26} (J/s)$

Thus the time taken to fall from $R=1 A$ to $R=0.5 A$ is

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

d) When the radius is $R =0.5 A$ we have

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

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

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

Continue reading on accelerating charges.