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Pure Si at 500 K has equal number of electron (n_(e )) and hole (n_(h)) concentrations of 1.5xx10^(16)m^(-3). Doping by indium increases n_(h) to 4.5xx10^(22)m^(-3). The doped semiconductor is of ………

Answer»

n-type semiconductor, electron CONCENTRATION`n_(e )=5XX10^(22)m^(-3)`
p-typesemiconductor, electron concentration `n_(e )=2.5xx10^(10)m^(-3)`
n-type semiconductor, electron concentration`n_(e )=2.5xx10^(23)m^(-3)`
p-type semiconductor, electron concentration `n_(e )=5xx10^(9)m^(-3)`

Solution :p-type semiconductor, electron concentration `n_(e )=5xx10^(9)m^(-3)`
`n_(i)^(2)=n_(e )n_(h)`
`THEREFORE (1.5xx10^(16))^(2)=n_(e )(4.5xx10^(22))`
`rArr n_(e )=0.5xx10^(10) or n_(e )=5xx10^(9)`
Now `n_(h)=4.5xx10^(22) rArr n_(n) GT gt n_(e )`
`therefore` Semiconductor is p-type and `n_( e)=5xx10^(9)m^(-3)`


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