*Question E6A08:
What is the term for the frequency at which the grounded-base current gain of a bipolar junction transistor has decreased to 0.7 of the gain obtainable at 1 kHz?*  

*Correct Answer is  D: Alpha cutoff frequency*  

### AI Response: (may contain errors)  

The correct answer is **D: Alpha cutoff frequency**.

### Explanation:

1. **Correct Answer: Alpha cutoff frequency**
   - The **alpha cutoff frequency** is the frequency at which the grounded-base current gain ($\alpha$) of a bipolar junction transistor (BJT) falls to 0.7 of its low-frequency value. This is a key parameter in understanding the frequency response of a BJT. At this frequency, the transistor's ability to amplify current is reduced to 70% of its maximum at low frequencies (e.g., 1 kHz).

2. **Why the Other Choices are Incorrect:**
   - **A: Corner frequency**
     - The corner frequency typically refers to the point in a filter's frequency response where the output power drops to half its maximum value, or -3 dB point. It is not specifically related to the current gain of a BJT.
   - **B: Alpha rejection frequency**
     - This term is not standard in the context of BJTs or frequency response. It does not describe the reduction in current gain.
   - **C: Beta cutoff frequency**
     - The beta cutoff frequency refers to the frequency at which the current gain ($\beta$) of a BJT falls to 0.7 of its low-frequency value. This is different from the alpha cutoff frequency, which deals with the grounded-base configuration.

### Useful Formula:

- The relationship between $\alpha$ and $\beta$ in a BJT is given by:
  $$ \alpha = \frac{\beta}{\beta + 1} $$
  
  This formula shows how the current gain in different configurations (common base vs. common emitter) is related. Understanding these relationships helps in analyzing transistor behavior at various frequencies.