[Compton scattering] Solving for wavelength

[Rafiki]

Homework Statement

Question: Determine the energy of the incident photon when the recoiled electrons energy is measured to 3 MeV. The angle between the incident photon and the scattered photon is 60°.

Given and knowns:
θ=60° (Angle between incident photon $\gamma$ and scattered photon $\gamma'$)
Recoiled electron energy $E_e$= 3MeV
$\hbar$ = 6.6261$^{-34}$ Js
$m_e$=0.511 $\frac {MeV} {c^2}$ (rest mass for a electron)

Homework Equations

$E_i=E_\gamma+m_ec^2$
$E_f=E_{\gamma'}+E_e$
Compton scattering: $\lambda'-\lambda=\frac\hbar{m_ec}(1-cos\theta)$
$E_\gamma=\frac {\hbar c} {\lambda}$

The Attempt at a Solution

Starting out with $cos(60)=\frac 1 {2}$ and compton scattering i get
$\lambda'-\lambda= \frac \hbar {2m_ec}$ ⇔$\lambda'=\lambda+ \frac \hbar {2m_ec}$
Then $E_\gamma-E_{\gamma'}=(\gamma-1)m_ec^2$ and realize that $(\gamma-1)m_ec^2=E_e$.
Using $E_\gamma=\frac {\hbar c} {\lambda}$ gives us $E_e={\hbar c}( \frac 1 {\lambda} - \frac 1 {\lambda'})$.

Now rewrite to $\frac {E_e} {\hbar c}=( \frac 1 {\lambda} - \frac 1 {\lambda'})$ and substitute $\lambda'=\lambda+ \frac \hbar {2m_ec}$ ⇒ $\frac {E_e} {\hbar c}= \frac 1 {\lambda} - \frac1{\frac \hbar {2m_ec}+\lambda}$.

Solving for $\lambda$ ⇒...⇒ $\lambda=-\frac \hbar {4m_ec}\pm \sqrt{(\frac \hbar {4m_ec})^2+\frac {\hbar^2} {2E_em_e}}$

Is the final expression correct? Plugging in numbers gives me $\lambda= 326*10^{-6}$ nm which is a much lower/shorter wavelength than i expected it to be.
I asume it is reasonable within the gammaray territory but currently i do not have a feel for what values are reasonable or could be expected, so i am unsure if it is a matter of a error in the algebra, from a physics reasoning standpoint or plugging in numbers.
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Also sorry if my solution is a bit difficult to follow, this was my first time ever typing in latex (the guide on this site was extremely helpful and easy to follow though!).

As a bonus question, is it possible to get a () around the "$-$" in $\pm$ using latex?

 

[Rafiki]

I redid the algebra and it seems to check out. However i get the incident photons energy to 2.73 MeV, which seems strange to me since the recoiled electrons energy is 3.0 MeV. Meaning that the recoiled electrons energy is higher than the incident photons energy.
Can this occur without the electron having some initial speed? It wasn't specified in the question whether the electron was at rest or moving before the "collision".

 

[Rafiki]

I had converted poorly from electron volts to joules at one place, which gave me the contradicting numbers. Marking it as solved.