Match the List I with List II: LIST I | vedclass

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(B) $1$. Intrinsic Semiconductor: In an intrinsic semiconductor,the number of electrons in the conduction band equals the number of holes in the valen

Vedclass · Accepted Answer

$(A) \rightarrow II, (B) \rightarrow I, (C) \rightarrow III, (D) \rightarrow IV$

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(B) $1$. Intrinsic Semiconductor: In an intrinsic semiconductor,the number of electrons in the conduction band equals the number of holes in the valence band. Thus,the Fermi level lies exactly in the middle of the forbidden energy gap. So,$A \rightarrow II$.$2$. $n$-type semiconductor: In an $n$-type semiconductor,donor energy levels are introduced just below the conduction band. Consequently,the Fermi level shifts upward towards the conduction band. So,$B \rightarrow I$.$3$. $p$-type semiconductor: In a $p$-type semiconductor,acceptor energy levels are introduced just above the valence band. Consequently,the Fermi level shifts downward towards the valence band. So,$C \rightarrow III$.$4$. Metals: In metals,the valence band and conduction band overlap,and the Fermi level lies within the conduction band. So,$D \rightarrow IV$.Therefore,the correct matching is $(A \rightarrow II, B \rightarrow I, C \rightarrow III, D \rightarrow IV)$.

$LIST$ $I$	$LIST$ $II$
$A$. Intrinsic Semiconductor	$I$. Fermi-level near conduction band
$B$. $n$-type semiconductor	$II$. Fermi-level at middle
$C$. $p$-type semiconductor	$III$. Fermi-level near valence band
$D$. Metals	$IV$. Fermi-level inside conduction band

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