The contribution in the total current flowing through a semiconductor due to electrons and holes are `3/4` and `1/4` respectively. If the drift velocity of electrons is `5/2` times that of holes at this temperature, then the ratio of concentration of electrons and holes is

Options

A. `6:5`
B. `5:6`
C. `3:2`
D. `2:3`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We know that the contributions to the total current from electrons and holes are given as:
- Contribution from electrons, \( I_e = \frac{3}{4} I \)
- Contribution from holes, \( I_h = \frac{1}{4} I \)

Additionally, we are informed that the drift velocity of electrons is \( \frac{5}{2} \) times that of holes:
- \( v_e = \frac{5}{2} v_h \)

### Step 2: Write the current equations
The current due to electrons and holes can be expressed using the formula:
\[ I = n \cdot A \cdot e \cdot v_d \]
where:
- \( n \) is the concentration of charge carriers (electrons or holes),
- \( A \) is the cross-sectional area,
- \( e \) is the charge of the electron,
- \( v_d \) is the drift velocity.

For electrons:
\[ I_e = n_e \cdot A \cdot e \cdot v_e \]

For holes:
\[ I_h = n_h \cdot A \cdot e \cdot v_h \]

### Step 3: Set up the ratio of currents
From the above equations, we can set up the ratio of the currents:
\[ \frac{I_e}{I_h} = \frac{n_e \cdot A \cdot e \cdot v_e}{n_h \cdot A \cdot e \cdot v_h} \]
The \( A \) and \( e \) cancel out:
\[ \frac{I_e}{I_h} = \frac{n_e \cdot v_e}{n_h \cdot v_h} \]

### Step 4: Substitute the known values
Substituting the known contributions to the total current:
\[ \frac{\frac{3}{4} I}{\frac{1}{4} I} = \frac{n_e \cdot v_e}{n_h \cdot v_h} \]
This simplifies to:
\[ \frac{3}{1} = \frac{n_e \cdot v_e}{n_h \cdot v_h} \]

### Step 5: Substitute the drift velocity relationship
Using the relationship \( v_e = \frac{5}{2} v_h \):
\[ \frac{3}{1} = \frac{n_e \cdot \frac{5}{2} v_h}{n_h \cdot v_h} \]
The \( v_h \) cancels out:
\[ \frac{3}{1} = \frac{5}{2} \cdot \frac{n_e}{n_h} \]

### Step 6: Rearranging to find the ratio of concentrations
Rearranging gives:
\[ \frac{n_e}{n_h} = \frac{3 \cdot 2}{5} = \frac{6}{5} \]

### Conclusion
Thus, the ratio of the concentration of electrons to the concentration of holes is:
\[ \frac{n_e}{n_h} = \frac{6}{5} \]