Step 1: Given:
- Collector supply voltage \( V_C = 8 \, {V} \),
- Voltage drop across the resistor \( V_R = 0.6 \, {V} \),
- Resistor \( R_C = 500 \, \Omega \),
- Current gain factor \( \alpha = 0.96 \).
First, calculate the collector current using Ohm’s law:
\[
I_C = \frac{V_R}{R_C} = \frac{0.6}{500} = 1.2 \times 10^{-3} \, {A} = 1.2 \, {mA}
\]
Step 2: The relationship between the collector current and emitter current is given by:
\[
I_C = \alpha \cdot I_E
\]
where \( I_E \) is the emitter current. Rearranging the equation, we get:
\[
I_E = \frac{I_C}{\alpha} = \frac{1.2 \, {mA}}{0.96} = 1.25 \, {mA}
\]
Step 3: The base current \( I_B \) is related to the emitter current by:
\[
I_E = I_B + I_C
\]
Thus, the base current is:
\[
I_B = I_E - I_C = 1.25 \, {mA} - 1.2 \, {mA} = 0.05 \, {mA} = 50 \, \mu {A}
\]
Thus, the base current is \( \boxed{50 \, \mu A} \).