In a NPN transistor, 108 electrons enter the emitter in `10^(-8)` s . If `1%` electrons are lost in the base, the fraction of current that enters the collector and current amplification factor are respectively

Options

A. `0.8` and 49
B. `0.9` and 90
C. `0.7` and 50
D. `0.99` and 99

Explanation

To solve the problem step-by-step, we will follow these calculations:

### Step 1: Calculate the emitter current (I_E)
The number of electrons entering the emitter is given as \( N = 10^8 \) electrons in \( t = 10^{-8} \) seconds.

The charge of one electron \( e \) is \( 1.6 \times 10^{-19} \) coulombs.

The total charge \( Q \) entering the emitter can be calculated as:
\[ Q = N \times e = 10^8 \times 1.6 \times 10^{-19} \text{ C} \]
Now, substituting the values:
\[ Q = 1.6 \times 10^{-11} \text{ C} \]

The emitter current \( I_E \) can be calculated using the formula:
\[ I_E = \frac{Q}{t} = \frac{1.6 \times 10^{-11}}{10^{-8}} = 1.6 \times 10^{-3} \text{ A} = 1.6 \text{ mA} \]

### Step 2: Calculate the base current (I_B)
Since 1% of the electrons are lost in the base, the base current \( I_B \) can be calculated as:
\[ I_B = 1\% \text{ of } I_E = \frac{1}{100} I_E = \frac{1}{100} \times 1.6 \times 10^{-3} = 1.6 \times 10^{-5} \text{ A} = 16 \mu A \]

### Step 3: Calculate the collector current (I_C)
Since 99% of the electrons enter the collector, the collector current \( I_C \) can be calculated as:
\[ I_C = 99\% \text{ of } I_E = \frac{99}{100} I_E = \frac{99}{100} \times 1.6 \times 10^{-3} = 1.584 \times 10^{-3} \text{ A} = 1.584 \text{ mA} \]

### Step 4: Calculate the current amplification factor (β)
The current amplification factor \( \beta \) is given by:
\[ \beta = \frac{I_C}{I_B} \]
Substituting the values we have:
\[ \beta = \frac{1.584 \times 10^{-3}}{1.6 \times 10^{-5}} = 99 \]

### Final Results
1. The fraction of current that enters the collector:
\[ \text{Fraction entering collector} = \frac{I_C}{I_E} = \frac{1.584 \times 10^{-3}}{1.6 \times 10^{-3}} = 0.99 \]
2. The current amplification factor \( \beta \):
\[ \beta = 99 \]

### Summary
The fraction of current that enters the collector is \( 0.99 \) and the current amplification factor is \( 99 \).

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