$\lambda=\frac{h c}{E_{g}}$
where $\mathrm{E}_{\mathrm{g}}=$ energy gap of semiconductor
$=1.9\, \mathrm{eV}$
$=1.9 \times 1.6 \times 10^{-19} \,\mathrm{V}$
$\lambda=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{1.9 \times 1.6 \times 10^{-19}} \,\mathrm{m}$
$=6.5 \times 10^{-7}\, \mathrm{m}$
$=650 \times 10^{-9}\, \mathrm{m}$
$=650 \,\mathrm{nm}$
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Figure: $Image$
$(A)$ The particle enters Region $III$ only if its velocity $V>\frac{q / B}{m}$
$(B)$ The particle enters Region $III$ only if its velocity $\mathrm{V}<\frac{\mathrm{q} / \mathrm{B}}{\mathrm{m}}$
$(C)$ Path length of the particle in Region $II$ is maximum when velocity $V=\frac{q / B}{m}$
$(D)$ Time spent in Region $II$ is same for any velocity $V$ as long as the particle returns to Region $I$
[Given: In SI units $\frac{1}{4 \pi \in_0}=9 \times 10^9, \ln 2=0.7$. Ignore the area pierced by the wire.]
