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In a NPN transistor, 108 electrons enter...

In a NPN transistor, 108 electrons enter the emitter in `10^(-8)` s . If `1%` electrons are lost in the base, the fraction of current that enters the collector and current amplification factor are respectively

A

`0.8` and 49

B

`0.9` and 90

C

`0.7` and 50

D

`0.99` and 99

Text Solution

Verified by Experts

The correct Answer is:
D
`I_(e)=(108xx1.6xx10^(-19))/(10^(-8))=172.8xx10^(-11)A`
`I_(B)=(1)/(100)xxI_(e)=0.01I_(e)`
and `I_(C)=0.99I_(e) :. beta=99`
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