FANDOM


ProblemEdit

Here we deal with the classical theory of the width of an atomic spectral line. The “atom” consists of an electron (mass m and charge - e) in a harmonic oscillator potential with a drag force acting on it, so that the classical equation of motion is

m \ddot x + m \omega^2 x + \gamma \dot x = 0.

In the first two parts, ignore any loss of energy due to radiation.

(a) Suppose that at time t=0, x=x_0, \dot x=0. Assuming that the damping is small, i.e., that \frac{\gamma}{m}<<\omega_0, what is the subsequent motion? (Hint: Try a solution of the form x(t) = x_0 Re[e^{-pt}].)

(b) A classical electron executing the motion in part (a) would emit electromagnetic radiation. Determine the functional dependence on the frequency \omega of the intensity distribution function I(\omega). Note that we are asking only for the \omega-dependence of the answer, not for the full normalization factors. [I(\omega)d\omega is the total energy integrated over time emitted in a range of frequencies from \omega to \omega + d\omega.] (Hint: Parseval’s theorem, \int_0^\infty x^2 dt\propto \int_{-\infty}^\infty \mid \tilde{x}(\omega)\mid^2 d\omega, where \tilde{x}(\omega) is the fourier transform of x(t), may be useful here.)

(c) Now suppose that the damping force is completely absent from the equation of motion and that the oscillation is damped only by the loss of energy due to radiation. The energy of the oscillator decays as U = U_0 e^{-\Gamma t}. Determine \Gamma assuming that in one period the electron loses only a small fraction of its energy.

(d) Using the above model, what is the width of an “atomic” spectral line of 500 nm according to the result of part (c)? Useful data: e^2/m c^2 = r_0 = 2.8*10^{-13}cm.

SolutionEdit

Ad blocker interference detected!


Wikia is a free-to-use site that makes money from advertising. We have a modified experience for viewers using ad blockers

Wikia is not accessible if you’ve made further modifications. Remove the custom ad blocker rule(s) and the page will load as expected.