Answer
Opaque
Work Step by Step
To determine weather the semiconductor is opaque or transparent, we must know the energy of photons first. There are two conditions for this case.
1. If the energy of photon is greater than the gap width, the energy is absorbed and becomes opaque.
2. If the energy of photon is smaller than the gap width, the energy is not absorbed and becomes transparent.
Find the photon energy where $E_{gap} = 7.6 eV$
$E_{photon} = \frac{hc}{\lambda}$
Where
$hc = 1240 eV.nm$
and $ \lambda = 140 nm$
$E_{photon} = \frac{1240 eV.nm}{140 nm}$
$E_{photon} = 8.86 eV$
Since $E_{photon}$ is larger than $E_{gap}$ , the crystal is opaque.
