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 ]

Question

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 ] (A) 0.5 eV (B) 0.7 eV (C) 1.1 eV (D) 2.5 eV

Tardigrade Admin · Accepted Answer

λ textmax = 2 4 8 0 text nm = 24800 mathringA Energy (in eV ) =(12400 e V mathringA/λ ) textE=(1 2400/2 4 8 0 0) text eV textE=0 text.5 text eV

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 $(e V)$ for the semiconductor is [take $h c = 12400 e V \overset{˚}{A}$ ]

$0.5 e V$

$0.7 e V$

$1.1 e V$

$2.5 e V$

$λ_{max} = 2480 nm = 24800 \overset{˚}{A}$
Energy (in $e V$ ) $= \frac{12400 e V A ˚}{λ}$
$E = \frac{12400}{24800} eV$
$E= 0.5 eV$

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

0.5eV

0.7eV

1.1eV

2.5eV