For pure 'Ge' semiconductor quantity of 'e' and hole is `10^(19) e//m^(3)` if we doped donor impurity in it with density `10^(23) e//m^(3)` then quantity of hole `(e//m^(3))` in semiconductor is :-A. `10^(15)`B. `10^(19)`C. `10^(23)`D. `10^(27)`

For pure 'Ge' semiconductor quantity of 'e' and hole is `10^(19) e//m^

1.	For pure 'Ge' semiconductor quantity of 'e' and hole is `10^(19) e//m^(3)` if we doped donor impurity in it with density `10^(23) e//m^(3)` then quantity of hole `(e//m^(3))` in semiconductor is :-A. `10^(15)`B. `10^(19)`C. `10^(23)`D. `10^(27)`
Answer» Correct Answer - A `n_(h)=(n_(i)^(2))/(n_(e))=((10^(19))^(2))/(10^(23))=10^(15)`

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For pure 'Ge' semiconductor quantity of 'e' and hole is `10^(19) e//m^(3)` if we doped donor impurity in it with density `10^(23) e//m^(3)` then quantity of hole `(e//m^(3))` in semiconductor is :-A. `10^(15)`B. `10^(19)`C. `10^(23)`D. `10^(27)`

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