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The ratio of the number of free electrons to holes n_(e)//n_(h) for two different materials A and B are 1 and lt 1 respectively. Name the type of semiconductors to which A and B belong. |
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Answer» SOLUTION :`n_(E)/n_(h) = 1 rArr n_(e) = n_(h) THEREFORE` Intrinsic SEMICONDUCTOR `n_(e)/n_(h) lt 1 rArr n_(e) lt n_(h) therefore` p type extrinsic semiconductor. |
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