= Solution
The emission mechanism is <Cherenkov radiation>. An <Electron> moving faster than the <phase velocity> $c/n$ of <light> in the material excites <electric polarization> disturbances whose radiation adds coherently on a cone. This does not require the <Electron> to exceed the <speed of light> in <vacuum>. Taking $n=1.4$ as constant over the detected optical band, the <Cherenkov cone angle> satisfies
$$
\cos\theta_C=\frac{c}{nv}=\frac1{n\beta},\qquad \beta=\frac vc.
$$
Emission requires $\beta>1/n$. The detector accepts $\theta_C<20^\circ$, so
$$
\frac1{n\beta}>\cos20^\circ,\qquad
\boxed{\frac1{1.4}<\beta<\frac1{1.4\cos20^\circ}.}
$$
Numerically,
$$
\boxed{0.71429c<v<0.76013c,\qquad
2.1414\times10^8<v<2.2788\times10^8\,\mathrm{m\,s^{-1}}.}
$$
The lower endpoint has vanishing <Cherenkov radiation> yield, and the upper endpoint is excluded by the stated strict angular acceptance. Faster <Electrons> emit at larger cone angles: at $v\simeq c$, $\theta_C\simeq\cos^{-1}(1/1.4)=44.4^\circ$, outside this detector's acceptance. The angular condition gives the accepted speed interval conditional on rays reaching the detector; no collection geometry or total signal yield is specified.
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