1.

If one Al atom added to 10^(9) per atom of Si then charge on holes in one mole will be ……….

Answer»

`1.6xx10^(-10)C`
`1.6xx10^(-5)C`
`1.6xx10^(-8)C`
`0.6xx10^(-19)C`

Solution :`1.6xx10^(-5)C`
1 atom of Al PER `10^(9)` SI atom then at atom of `10^(23)` Si ?
`N=(1xx10^(23))/(10^(9))=10^(14)`
`therefore Q=Ne`
`=10^(14)xx1.6xx10^(-19)`
`therefore Q=1.6xx10^(-5)C`


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