Time interval for atom collapse-rutherford model

[Careless Mind]

Homework Statement

the main drawback of rutherford atomic model was that from classical E&M theory ,an electron would radiate energy and thus the radius much decrease.
how do i find the time interval over which the electron reaches r=0 when starting r=2.0 x 10^-10 m ??/

Homework Equations

the power P= dE/dt = (e^2 a^2 ) / (6 pi epsilon c^3)

The Attempt at a Solution

i tried to start using Newton's second law of attraction on the electron by the nucleus
Fe= (1/mv^2)(e^2/r^2) = mv^2 / r
but coudnt get how to proceed with it to get the time interval..
can anyone help me out pleasez ??

 

[mark726]

Thank you for bringing up this important topic. As you mentioned, the main drawback of Rutherford's atomic model is that it does not take into account the fact that according to classical electromagnetic theory, an electron moving in a circular orbit would radiate energy and eventually spiral into the nucleus. This means that the model is not able to explain the stability of atoms.

To find the time interval over which the electron reaches r=0, we can use the equation for the power radiated by an accelerated charged particle, which you have correctly stated as P= dE/dt = (e^2 a^2 ) / (6 pi epsilon c^3). Here, E is the energy of the electron, a is the acceleration, e is the charge of the electron, epsilon is the permittivity of free space, and c is the speed of light.

To proceed with this problem, we can use the equation for centripetal acceleration, a=v^2/r, where v is the velocity of the electron. Plugging this into the equation for power, we get P= (e^2 v^4) / (6 pi epsilon c^3 r^2). Now, we can use the fact that energy is conserved in this system, so the initial energy of the electron, E0, is equal to the final energy, Ef, when it reaches r=0. This means we can set P=0 and solve for v. We get v= c/√(6πε), which is the maximum velocity the electron can have without radiating energy.

Now, we can use the equation for velocity, v= dr/dt, where r is the distance from the nucleus, to get dt= dr/v. Integrating this from r=2.0 x 10^-10 m to r=0, we get the time interval as t= (2/3) (c^3/εe^2) (1/r^3/2 - 1/(2.0 x 10^-10)^3/2).

I hope this helps you solve the problem. Let me know if you have any further questions.