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If an LED has to emit 662 nm wavelength of light thenwhat should be the band gap energy of its semiconductor? (h=6.62xx10^(-34)Js) |
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Answer» Solution :`LAMBDA=662xx10^(-9)m` `=6.62xx10^(-7)m` `h=6.62xx10^(-34)J` `lambda=(hc)/(E_(g))` `E_(g)=(hc)/(lambda)` `=(6.62xx10^(-34)xx 3xx10^(8))/(6.62xx10^(-7))` `THEREFORE E_(g)=3xx10^(-19)J` `therefore E_(g)=(3xx10^(-19))/(1.6xx10^(-19)) therefore E_(g)=1.875 eV` |
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