Answer:
P /K = 1,997 10⁻³⁶ s⁻¹
Explanation:
For this exercise let's start by finding the radiation emitted from the accelerator
[tex]\frac{dE}{dt}[/tex] = [tex]\frac{q^{2} a^{2} }{6\pi \epsilon_{o} c^{2} }[/tex]
the radius of the orbit is the radius of the accelerator a = r = 0.530 m
let's calculate
\frac{dE}{dt} = [(1.6 10⁻¹⁹)² 0.530²] / [6π 8.85 10⁻¹² (3 108)³]
P= \frac{dE}{dt}= 1.597 10⁻⁵⁴ W
Now let's reduce the kinetic energy to SI units
K = 5.0 10⁶ eV (1.6 10⁻¹⁹ J / 1 eV) = 8.0 10⁻¹⁹ J
the fraction of energy emitted is
P / K = 1.597 10⁻⁵⁴ / 8.0 10⁻¹⁹
P /K = 1,997 10⁻³⁶ s⁻¹
