The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 $n m$ is incident on it. The band gap in ( $e V$ ) for the semiconductor is

Question

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- the semiconductor is0-90-70-51-1

Toppr Admin · Accepted Answer

Band gap- Eg-hc-x3BB-6-63-xD7-10-x2212-34-3-xD7-108-2480-xD7-10-x2212-9-xD7-1-6-xD7-10-x2212-19eV-0-5-xA0-eV

The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than $2480 n m$ is incident on it. The band gap (in $e V$ ) for the semiconductor is
$0.9$
$0.7$
$0.5$
$1.1$

Band gap, $E_{g} = \frac{h c}{λ} = \frac{(6.63 \times 10^{- 34}) (3 \times 10^{8})}{2480 \times 10^{- 9} \times 1.6 \times 10^{- 19}} e V = 0.5 e V$

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 is0.90.70.51.1

Band gap, Eg=hcλ=(6.63×10−34)(3×108)2480×10−9×1.6×10−19eV=0.5 eV

The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than $2480 n m$ is incident on it. The band gap (in $e V$ ) for the semiconductor is
$0.9$
$0.7$
$0.5$
$1.1$

Band gap, $E_{g} = \frac{h c}{λ} = \frac{(6.63 \times 10^{- 34}) (3 \times 10^{8})}{2480 \times 10^{- 9} \times 1.6 \times 10^{- 19}} e V = 0.5 e V$