In an $n-p-n$ transistor,$200$ electrons enter the emitter in $10^{-8} \ s$. If $1 \%$ of electrons are lost in the base,then the current that enters the emitter and the current amplification factor are respectively $\left[e=1.6 \times 10^{-19} \ C\right]$

• A
  $2 \times 10^{-10} \ A$ and $49$
• B
  $3.2 \times 10^{-9} \ A$ and $99$
• C
  $1.6 \times 10^{-19} \ A$ and $90$
• D
  $1.7 \times 10^{-11} \ A$ and $70$