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Pure Si at 500 K has equal number of electron (ne) and hole    \left ( n_{h} \right )concentrations of1.5\times 10^{16}m^{-3}.Dopping by indium increases n_{h} to 4.5\times 10^{22}m^{-3}.The doped semiconductor is of

  • Option 1)

    n-type with electron concentrations n_{e}=5\times 10^{22}m^{-3}

  • Option 2)

    p-type with electron concentrations n_{e}=2.5\times 10^{10}m^{-3}

  • Option 3)

    n-type with electron concentrations n_{e}=2.5\times 10^{23}m^{-3}

  • Option 4)

    P- type having electron concentrations n_{e}=5\times 10^{9}m^{-3} 

 

Answers (1)

best_answer

 

p - type semiconductor -

Here the base semiconductor (i.e.pure Si or Ge ) is mixed with tetravalent doping material (e.g. Al )

- wherein

1) nh >>ne

2) There is an acceptor level just above  the valence band .

 

 Since n_{h}> > n_{e} , hence it is a p type semiconductor.

 

n_{h}.n_{e}= n^{2} \Rightarrow n_{e}=\frac{n^{2}}{n_{h}}= \frac{(1.5X10^{16})}{4.5 \times 10^{22}}

n_{e}= \frac{2.25}{4.5} \times 10^{10}=5 \times 10^{9}/m^{3}

 


Option 1)

n-type with electron concentrations n_{e}=5\times 10^{22}m^{-3}

This option is incorrect

Option 2)

p-type with electron concentrations n_{e}=2.5\times 10^{10}m^{-3}

This option is incorrect

Option 3)

n-type with electron concentrations n_{e}=2.5\times 10^{23}m^{-3}

This option is incorrect

Option 4)

P- type having electron concentrations n_{e}=5\times 10^{9}m^{-3} 

This option is correct

Posted by

prateek

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