= Solution
For a particle of <rest mass> $m_0$ and <velocity> $\mathbf u$, the <relativistic energy> and <momentum> are
$$
\boxed{E=\gamma_um_0c^2,\qquad\mathbf p=\gamma_um_0\mathbf u,\qquad
\gamma_u=(1-u^2/c^2)^{-1/2}.}
$$
For a <photon> of wavelength $\lambda$ travelling along $\widehat{\mathbf n}$,
$$
\boxed{E_\gamma=\frac{hc}{\lambda},\qquad\mathbf p_\gamma=\frac h\lambda\widehat{\mathbf n},\qquad E_\gamma=c|\mathbf p_\gamma|.}
$$
Here $h$ is <Planck constant> and $c$ is the <speed of light>.
Let $\varepsilon$ and $\varepsilon'$ be the incident and scattered <photon> <energies>. The stationary <electron> initially has <energy> $m_ec^2$. By <four-momentum conservation>, its final <energy> and <momentum> are
$$
E_e'=m_ec^2+\varepsilon-\varepsilon',\qquad
\mathbf p_e'=\frac{\varepsilon}{c}\widehat{\mathbf n}
-\frac{\varepsilon'}c\widehat{\mathbf n}',
\quad\widehat{\mathbf n}\cdot\widehat{\mathbf n}'=\cos\theta.
$$
Use the electron's <relativistic energy-momentum relation>, rather than assuming its recoil is nonrelativistic:
$$
(m_ec^2+\varepsilon-\varepsilon')^2
=m_e^2c^4+\varepsilon^2+\varepsilon'^2-2\varepsilon\varepsilon'\cos\theta.
$$
Cancelling common terms gives
$$
m_ec^2(\varepsilon-\varepsilon')
=\varepsilon\varepsilon'(1-\cos\theta),\qquad
\frac1{\varepsilon'}-\frac1\varepsilon=\frac{1-\cos\theta}{m_ec^2}.
$$
Since $\lambda=hc/\varepsilon$, the <Compton wavelength shift> is
$$
\boxed{\lambda'-\lambda=\frac{h}{m_ec}(1-\cos\theta).}
$$
The word elastic here means that the final <electron> has the same <rest mass> as initially and the complete photon-electron system conserves total <energy>; the <photon> alone loses <energy> to recoil.
For backscattering, $1-\cos\theta=2$, and hence
$$
\boxed{\varepsilon'=\frac{\varepsilon}{1+2\varepsilon/(m_ec^2)}
=\frac{0.511}{2.511}\,\mathrm{MeV}=0.2035\,\mathrm{MeV}.}
$$
Choose the incident direction as $+x$, so that the scattered <photon> travels in the $-x$ direction. An observer moving with <velocity> $+\beta c\widehat{\mathbf x}$ meets that <photon> head-on. The <Lorentz transformation> of <four-momentum> gives the observed <energy>
$$
\varepsilon_{\rm obs}=\gamma_\beta(\varepsilon'-\beta c p_x')
=\gamma_\beta\varepsilon'(1+\beta)
=\varepsilon'\sqrt{\frac{1+\beta}{1-\beta}}.
$$
This is the approaching-observer <relativistic Doppler effect>. Set $D=\varepsilon/\varepsilon'=2.511/0.511$ and solve:
$$
\boxed{\beta=\frac{D^2-1}{D^2+1}=0.9205\ \text{approximately}.}
$$
The observer must move \b[opposite to the scattered photon], along the original incident direction, at approximately $0.9205c$. Moving with the scattered <photon> would redshift it instead.
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