In n – p – n transistor amplifier, the collector current is 9 mA. If 90% of the electrons emitted reach the collector, then:

the emitter current is 1 mA the base current is 10 mA $α = 0.9, β = 9.0$ $α = 9.0, β = 0.9$

$α = 0.9, β = 9.0$

Here, $I_C=9mA=\frac{90}{100}I_E$ $∴ I_E =10mA$ As $I_E = I_B + I_C$ $∴ I_B = I_E − I_C = 10 mA – 9 mA = 1 mA$ $α=\frac{I_C}{I_E}=\frac{9mA}{10mA}=0.9$ $β=\frac{I_C}{I_B}=\frac{9mA}{1mA}=9$