In a `n-p-n` transistor `10^(10)` electrons enter the emitter in `10^(-6)s`. `2%` of the elecrons are lost in the base. The current transfer ratio and the current amplification factor will be

Explanation

`I_(B)=(2)/(100)I_(E)`
`I_(E)=I_(B)+L_(C )=(2)/(100)L_(E)+I_(C )`
`I_(C )=(98)/(100)I_(E)`
Current transfer ratio `alpha=(I_(C ))/(I_(E))=(98)/(100)=0.98`
Current amplification factor `beta=(I_(c ))/(I_(B))=(98)/(2)=49`