In a semiconductor material `(1//5)`th of the total current is carried by the holes and the remaining is carried by the electrons. The drift speed of electrons is twice that of holes at this temperature. Thr ratio between the number densities of electrons and holes is

Options

A. ` 21//6`
B. 5
C. `3//8`
D. 2

Explanation

To solve the problem, we need to find the ratio between the number densities of electrons (Ne) and holes (Nh) in a semiconductor material given the conditions of current distribution and drift speeds.

### Step-by-Step Solution:

1. **Define the Total Current (I)**:
Let the total current be denoted as \( I \).

2. **Determine the Current Carried by Holes and Electrons**:
According to the problem, \( \frac{1}{5} \) of the total current is carried by holes. Therefore, the current carried by holes \( I_h \) is:
\[ I_h = \frac{1}{5} I \]
The remaining current, which is carried by electrons \( I_e \), is:
\[ I_e = I - I_h = I - \frac{1}{5} I = \frac{4}{5} I \]

3. **Use the Formula for Current**:
The current carried by electrons can be expressed as:
\[ I_e = N_e \cdot A \cdot e \cdot V_e \]
where \( N_e \) is the number density of electrons, \( A \) is the cross-sectional area, \( e \) is the charge of an electron (approximately \( 1.6 \times 10^{-19} \) C), and \( V_e \) is the drift velocity of electrons.

Similarly, the current carried by holes can be expressed as:
\[ I_h = N_h \cdot A \cdot e \cdot V_h \]
where \( N_h \) is the number density of holes and \( V_h \) is the drift velocity of holes.

4. **Relate the Drift Velocities**:
The problem states that the drift speed of electrons is twice that of holes:
\[ V_e = 2V_h \]

5. **Set Up the Ratio of Currents**:
Now, 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 areas \( A \) and the charge \( e \) cancel out:
\[ \frac{I_e}{I_h} = \frac{N_e \cdot V_e}{N_h \cdot V_h} \]

6. **Substitute Known Values**:
Substitute \( I_e = \frac{4}{5} I \) and \( I_h = \frac{1}{5} I \):
\[ \frac{\frac{4}{5} I}{\frac{1}{5} I} = \frac{N_e \cdot V_e}{N_h \cdot V_h} \]
This simplifies to:
\[ 4 = \frac{N_e \cdot V_e}{N_h \cdot V_h} \]

7. **Substitute Drift Velocity Relation**:
Substitute \( V_e = 2V_h \):
\[ 4 = \frac{N_e \cdot (2V_h)}{N_h \cdot V_h} \]
The \( V_h \) cancels out:
\[ 4 = \frac{2N_e}{N_h} \]

8. **Rearranging to Find the Ratio**:
Rearranging gives:
\[ 2N_e = 4N_h \implies \frac{N_e}{N_h} = \frac{4}{2} = 2 \]

9. **Final Ratio**:
Thus, the ratio of the number densities of electrons to holes is:
\[ \frac{N_e}{N_h} = 2:1 \]

### Final Answer:
The ratio between the number densities of electrons and holes is \( 2:1 \).