In an n-p-n transistor `10^(10)` electrons enter the emitter in `10^(-6)`s. If 2% of the electrons are lost in the base, find the current transfer ratio and the current amplification factor.

Options

A. 0.02
B. 7
C. 33
D. 49

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the emitter current (IE)
The emitter current (IE) can be calculated using the formula:
\[ I_E = \frac{Q}{T} \]
where \( Q \) is the total charge of the electrons entering the emitter and \( T \) is the time interval.

Given:
- Number of electrons entering the emitter, \( n_E = 10^{10} \)
- Charge of an electron, \( e = 1.6 \times 10^{-19} \) coulombs
- Time, \( T = 10^{-6} \) seconds

First, we calculate the total charge \( Q \):
\[ Q = n_E \times e = 10^{10} \times 1.6 \times 10^{-19} \]
\[ Q = 1.6 \times 10^{-9} \text{ coulombs} \]

Now, substituting \( Q \) into the formula for \( I_E \):
\[ I_E = \frac{1.6 \times 10^{-9}}{10^{-6}} = 1.6 \times 10^{-3} \text{ A} = 1.6 \text{ mA} \]

### Step 2: Calculate the collector current (IC)
Since 2% of the electrons are lost in the base, 98% of the electrons will reach the collector. Thus, the collector current \( I_C \) can be calculated as:
\[ I_C = 0.98 \times I_E \]
Substituting the value of \( I_E \):
\[ I_C = 0.98 \times 1.6 \times 10^{-3} = 1.568 \times 10^{-3} \text{ A} = 1.568 \text{ mA} \]

### Step 3: Calculate the current transfer ratio (α)
The current transfer ratio \( \alpha \) is defined as:
\[ \alpha = \frac{I_C}{I_E} \]
Substituting the values we have:
\[ \alpha = \frac{1.568 \times 10^{-3}}{1.6 \times 10^{-3}} = 0.98 \]

### Step 4: Calculate the current amplification factor (β)
The current amplification factor \( \beta \) is given by:
\[ \beta = \frac{\alpha}{1 - \alpha} \]
Substituting the value of \( \alpha \):
\[ \beta = \frac{0.98}{1 - 0.98} = \frac{0.98}{0.02} = 49 \]

### Final Answers:
- Current transfer ratio \( \alpha = 0.98 \)
- Current amplification factor \( \beta = 49 \)