In an n-p-n transistor `10^10` electron enter the emitter in `10^(-6)` s. If 2% of the electrons are lost in the base, the current amplification factor is

Explanation

We know that current `=` charge/time
The emitter current `(I_(E))` is given by `I_(E)=(Ne)/(t)=(10^(10)xx(1.6xx10^(-19)))/(10^(-6))=1.6mA`
the base `(I_(B))` is given by `I_(B)=(2)/(100)xx1.6=0.032mA`
In a transistor, `I_(E)=I_(B)+I_(C)`
`I_(C)=I_(E)-I_(B)=1.6-0.032=1.568mA`
Current transfer ratio `=(I_(C))/(I_(E))=(1.568)/(1.6)=0.98`
Current amplification factor `=(I_(C))/(I_(B))=(1.568)/(0.032)=49`.