Question

Find the mean radiation power of an electron performing SHM of amplitude $0.1nm$ and frequency $\omega =6.5\times {10}^{14}{s}^{-1}$.

Answer 1

The radiation power of electron with momentum $p\left(t\right)$ is given by, $P=\frac{1}{4\pi {ϵ}_0}\frac{2(\frac{d^{2}p}{d{t}^2}{)}^2}{3c^3}$

where $(\frac{d^{2}p}{d{t}^2}{)}^2=e^2{a}^2{\omega }^4{\cos }^2\omega t$

The average value of it is $e^2{a}^2{\omega }^4\times 1/2$

Now, mean of radiation power,

$<P>=\frac{1}{4\pi {ϵ}_0}\times \frac{2}{3c^3}\times e^2{a}^2{\omega }^4\times 1/2$

$=(9\times {10}^9)\times \frac{1}{3(3\times {10}^8{)}^3}\times (1.6\times {10}^{-19}{)}^2(0.1\times {10}^{-9}{)}^2(6.5\times {10}^{14}{)}^4$

$=5.1\times {10}^{-15}W$