The electron condactivety of a samiconductor increases `2480 nm` is incident on it . The hand gap in (eV) for the semicondactor is
A
`25 eV`
B
`1.1 eV`
C
`0.7 eV`
D
`0.5 eV`
Text Solution
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The correct Answer is:
D
Band gap = energy of photon of wavelength `2480 nm` so, `delta E = (hc)/(lambda) = ((6.63 xx 10^(-34) xx 3 xx 10^(8))/(2480 xx 10^(-9))) = (1)/(1.6 xx 10^(-19)) eV` `= 0.5 eV`
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