Answer
$ \approx 3.0 \times 10^{-14} J $
Work Step by Step
When it comes to Compton Scattering, the energy of Photon is $hc = 1240 eV . nm$
Given the initial particle wavelength $\lambda = 3.00 \times 10^{-12} m$
Substitute into initial energy equation (Remember to change unit to nm)
$ e = \frac{hc}{\lambda}$ = $\frac{1240 ev.nm}{0.003 nm}$ = $4.13 \times 10^{5} eV $
By getting the initial energy, we can calculate the Compton shift
$\Delta \lambda = \frac{h}{mc} (1-cos 90^{\circ})$
(Note that m is the mass of electron)
Since cos 90 is 0, modify the equation, times both h and mc with c
$\Delta \lambda = \frac{hc}{mc^{2}} $
since $hc = 1240 eV . nm$ and $mc^{2} = 511 \times 10^{3} eV $
so
$\Delta \lambda = \frac{1240 eV . nm}{511 \times 10^{3} eV}$ = 2.43 pm
Therefore, the new photon wavelength is
$\lambda$' = 3.00 pm + 2.43 pm = 5.43 pm
So the new photon energy is
$ e' = \frac{hc}{\lambda'}$ = $\frac{1240 ev.nm}{0.00543 nm}$ = $2.28 \times 10^{5} eV $
According to kinetic energy conservation principle,
$k_{e} = \Delta E = E - E' = 4.13 \times 10^{5} eV - 2.28 \times 10^{5} eV = 1.85 \times 10^{5} eV $
change unit to Joule
$ \approx 3.0 \times 10^{-14} J $
