In a bipolar transistor, alpha is the ratio of?

Correct option is C

​$$\text{In a } \mathbf{Bipolar\ Junction\ Transistor\ (BJT)}, \alpha \text{ (alpha) represents the } \mathbf{current\ gain\ in\ the\ common\ base\ configuration} \text{ and is defined as the ratio of } \mathbf{collector\ current\ (I_C)} \text{ to } \mathbf{emitter\ current\ (I_E)}. \\\textbf{Key Concept:} \\\alpha = \frac{I_C}{I_E} \\\text{where:} \\\begin{aligned}&\bullet \quad I_C = \text{Collector current} \\&\bullet \quad I_E = \text{Emitter current} \\&\bullet \quad I_B = \text{Base current} \\&\bullet \quad I_E = I_C + I_B \quad \text{(Kirchhoff’s Current Law)}\end{aligned} \\\text{Since the } \mathbf{collector\ current\ (I_C)} \text{ is slightly less than the } \mathbf{emitter\ current\ (I_E)} \text{ due to base current loss, } \alpha \text{ is always less than 1.}$$​​