In a transistor if `(I_C )/(I_E ) = alpha `and ` (I_C )/( I_B ) = beta `If `alpha ` varies between `(20)/( 21) and (100)/(101)` then the value of `beta ` lies between

Options

A. `1-10`
B. `0.95 -0.99`
C. `20-100`
D. `200-300`

Answer

To solve the problem, we start with the relationships given in the question:

1. \( \frac{I_C}{I_E} = \alpha \)
2. \( \frac{I_C}{I_B} = \beta \)

From these relationships, we can express the emitter current \( I_E \) and base current \( I_B \) in terms of collector current \( I_C \):

- From the first equation, we can express \( I_E \):
\[ I_E = \frac{I_C}{\alpha} \]

- From the second equation, we can express \( I_B \):
\[ I_B = \frac{I_C}{\beta} \]

We also know that the emitter current \( I_E \) is the sum of the base current \( I_B \) and the collector current \( I_C \):
\[ I_E = I_B + I_C \]

Substituting the expressions for \( I_E \) and \( I_B \) into this equation gives:
\[ \frac{I_C}{\alpha} = \frac{I_C}{\beta} + I_C \]

Now, we can factor out \( I_C \) (assuming \( I_C \neq 0 \)):
\[ \frac{1}{\alpha} = \frac{1}{\beta} + 1 \]

Rearranging this equation leads to:
\[ \frac{1}{\alpha} - 1 = \frac{1}{\beta} \]

This can be rewritten as:
\[ \frac{1 - \alpha}{\alpha} = \frac{1}{\beta} \]

Taking the reciprocal gives:
\[ \beta = \frac{\alpha}{1 - \alpha} \]

Now, we need to find the range of \( \beta \) when \( \alpha \) varies between \( \frac{20}{21} \) and \( \frac{100}{101} \).

### Step 1: Calculate \( \beta \) for the lower limit of \( \alpha \)
Substituting \( \alpha = \frac{20}{21} \):
\[ \beta_{lower} = \frac{\frac{20}{21}}{1 - \frac{20}{21}} = \frac{\frac{20}{21}}{\frac{1}{21}} = 20 \]

### Step 2: Calculate \( \beta \) for the upper limit of \( \alpha \)
Substituting \( \alpha = \frac{100}{101} \):
\[ \beta_{upper} = \frac{\frac{100}{101}}{1 - \frac{100}{101}} = \frac{\frac{100}{101}}{\frac{1}{101}} = 100 \]

### Conclusion
Thus, the value of \( \beta \) lies between:
\[ \beta \in [20, 100] \]

The correct answer is \( 20 < \beta < 100 \), which corresponds to option C: \( 2200 \).