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  4. The electrical conductivity of a semiconductor increases when electromagnetic radiation of a wavelength shorter than 2480 nm is incident on it. The bandgap in (eV) for the semiconductor is [take hc=12400 eV mathringA ]

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Q. The electrical conductivity of a semiconductor increases when electromagnetic radiation of a wavelength shorter than $2480\,nm$ is incident on it. The bandgap in $(eV)$ for the semiconductor is [take $hc=12400\,eV\,\mathring{A}$ ]

NTA AbhyasNTA Abhyas 2022

Solution:

$\lambda _{\text{max}} = 2 4 8 0 \text{ nm} = 24800\, \mathring{A}$
Energy (in $eV$ ) $=\frac{12400\, e V \mathring{A}}{\lambda }$
$\text{E}=\frac{1 2400}{2 4 8 0 0}\text{ eV}$
$\text{E=}0\text{.}5\,\text{ eV}$