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A Ge specimen is dopped with Al. The con...

A `Ge` specimen is dopped with `Al`. The concentration of acceptor atoms is `~10^(21) at oms//m^(3)`. Given that the intrinsic concentration of electron hole pairs is `~10^(19)//m^(3)`, the concentration of electron in the speciman is

A

`10^(17)//ms^(3)`

B

`10^(15)//m^(3)`

C

`10^(4)//m^(3)`

D

`10^(2)//m^(3)`

Text Solution

Verified by Experts

The correct Answer is:
A
When Ge specimen is doped with Al, then concentration of acceptor atoms is alos called concentration of holes. Using formula `n_(i)^(2)=n_(0)p_(0)` where
`n_(1)`= concentratio of electron hole pair `=10^(19)m^(-3)`
`n_(0)`= concentration of electron
`p_(0)`= concentration of holes `=10^(21) "atom" m^(-3)`
`rArr (10^(-19))^(2)10^(21)xxn_(0) rArr n_(0)=10^(-17) m^(-3)`
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Knowledge Check

  • A Ge specimen is doped with Al. The concentration of acceptor atoms is ~ 10^(21) m^(-3) . Given that the intrinsic concentration of electron-hole pairs is ~ 10^(19) m^(-3) . The concentration of electrons in the specimen is

    A
    `10^(17) m^(-3)`
    B
    `10^(15) m^(-3)`
    C
    `10^(4) m^(-3)`
    D
    `10^(2) m^(-3)`

  • A Ge specimen is doped with Al. The concentration of electron - hole pair at this temperature is ~10^(19)m^(-3) , the concentration of electrons in the specimen is :

    A
    `10^(17)m^(-3)`
    B
    `10^(15)m^(-3)`
    C
    `10^(4)m^(-3)`
    D
    `10^(2)m^(-3)`

  • The concentration of H^(+) in 10^(-3) M NaOH is

    A
    `10^(-7)`
    B
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