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

Emitter current `I_(e)=(Ne)/(y)=(10^(10)xx1.6xx10^(-19))/(10^(-6))=1.6mA`Base current `I_(b)=(2)/(100)xx1.6=0.032 mA`
but `I_(e)=I_(c)+I_(b):. I_(c)=I_(c)=I_(b)=1.6=1.6-0.032 = 1.568 mA`
`:. alpha=(I_(c))/(I_(e))=(1.568)/(1.6)=0.98 and beta=(I_(c))/(I_(b))=(1.568)/(0.032)=49`.