1.

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 semi-conductor is

Answer»

0.9
0.7
0.5
1.1

Solution :The band GAP is GIVEN by `E_g=hv=(hc)/lamda`
Here, `lamda = 2480 nm = 2480 xx10^(-9) m = 248 xx10^(-8)m`
`:.E_g=(6.6xx10^(-34)xx3xx10^(8))/(248xx10^(-8))=7.984xx10^(-20)J`
`=(7.984xx10^(-20))/(1.6xx10^(-19))eV=0.499eV = 0.5 eV`


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