Accurate answer is (a) 0.8mA
Paint a vivid picture of understanding with: Collector current, IC2=αF×IQ/(1+e^Vd⁄VT),
VT = Volts equivalent of temperature = 25mv,
⇒ Vd = V1-V2 =2.078v-2.06v=0.018v (equ1)
Substituting equation 1,
⇒ Vd/VT = 0.018v/25mv = 0.72v (equ2)
Substituting equation 2,
⇒ IC2= 1×2.4mA/(1+e^0.72) = 2.4mA/(1+2.05) = 0.8mA.