In a npn transistor $10^{10}$ electrons enter the emitter in $10^{–6} s$. 4% of the electrons are lost in the base. The current transfer ratio will be

0.96

No. of electrons reaching the collector, $n_C=\frac{96}{100}×10^{10}=0.96×10^{10}$

Emitter current, $I_E=\frac{n_E×e}{t}$

Collector current, $I_C=\frac{n_C×e}{t}$

∴ Current transfer ratio, $α=\frac{I_C}{I_E}=\frac{n_C}{n_E}=\frac{0.96×10^{10}}{10^{10}}=0.96$