\(E_{p\,e}=h.\left\{1-\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\right\}.f_{cir}\)
\(E_{p\,e}=-E_{p\,a}\)
The energy of the absorbed photon that set off the oscillation about the center has less energy than the particle in circular motion. As the energy of the absorbed photon decrease, the term
\(cos(\theta)\to\cfrac{1}{\sqrt{2}}\)
subsequent emitted photons has progressively less energy. At,
\(cos(\theta)=\cfrac{1}{\sqrt{2}}\)
no photons are emitted nor absorbed as the particle oscillates. Beyond this point the particle first absorb energy then emit energy during its oscillation.
What happens when the initially absorbed photon has higher energy than the particle in circular motion?
\(x_v\gt x_1\)
It is impossible to displace the circular motion in the direction perpendicular to its plane. It is possible that in the first instant, the particle switches to circular motion with radius \(x=x_1\) and is displaced along a radial line by \(x=x_v\). The particle oscillate along this radial line. So,
\(cos(\theta)=\cfrac{x_1}{x_v}\)
and the particle emit/absorb energy as previously.
