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The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm, is incident on it. The band gap in (eV) for the semiconductor is

Option: 1

1.1 \mathrm{eV}


Option: 2

2.5 \mathrm{eV}


Option: 3

0.5 \mathrm{eV}


Option: 4

0.7 \mathrm{eV}


Answers (1)

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\mathrm{E_g=h \mathrm{v}=\frac{h c}{\lambda}=\left(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{2480 \times 10^{-9} \times 1.6 \times 10^{-19}}\right)=0.5 \mathrm{eV} }

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