An ionization chamber with parallel conducting plates as anode and cathode has `5 xx 10^(7)` electrons and the same number of singly-charged positive ions per `cm^(3)`. The electrons are moving at `0.4 m//s`. The current density from anode to cathodes `4 mu A//m^(2)`. The velocity of positive ions moving towards cathode is

Options

A. `0.4m//s`
B. `1.6m//s`
C. `zero`
D. `0.1m//s`

Answer

To solve the problem, we need to find the velocity of the singly charged positive ions moving towards the cathode in an ionization chamber. We are given the following information:

- Number density of electrons (n) = \(5 \times 10^7 \, \text{electrons/cm}^3\)
- Number density of positive ions = \(5 \times 10^7 \, \text{ions/cm}^3\)
- Velocity of electrons (v_e) = \(0.4 \, \text{m/s}\)
- Current density (J) = \(4 \, \mu A/m^2 = 4 \times 10^{-6} \, A/m^2\)

### Step-by-Step Solution:

1. **Convert number density from cm³ to m³:**
\[ n = 5 \times 10^7 \, \text{electrons/cm}^3 = 5 \times 10^7 \times 10^6 \, \text{electrons/m}^3 = 5 \times 10^{13} \, \text{electrons/m}^3 \]

2. **Use the formula for current density (J):**
The current density due to electrons and positive ions can be expressed as:
\[ J = n_e \cdot e \cdot v_e + n_i \cdot e \cdot v_i \]
where:
- \(n_e\) = number density of electrons
- \(n_i\) = number density of positive ions
- \(e\) = charge of an electron (\(1.6 \times 10^{-19} \, C\))
- \(v_e\) = velocity of electrons
- \(v_i\) = velocity of positive ions (which we need to find)

3. **Substituting the known values:**
Since \(n_e = n_i\):
\[ J = n \cdot e \cdot v_e + n \cdot e \cdot v_i \]
\[ J = n \cdot e (v_e + v_i) \]
Substituting the values:
\[ 4 \times 10^{-6} = (5 \times 10^{13}) \cdot (1.6 \times 10^{-19}) \cdot (0.4 + v_i) \]

4. **Calculate \(n \cdot e\):**
\[ n \cdot e = (5 \times 10^{13}) \cdot (1.6 \times 10^{-19}) = 8 \times 10^{-6} \, A/m^3 \]

5. **Substituting \(n \cdot e\) back into the equation:**
\[ 4 \times 10^{-6} = 8 \times 10^{-6} \cdot (0.4 + v_i) \]

6. **Solve for \(0.4 + v_i\):**
\[ 0.5 = 0.4 + v_i \]
\[ v_i = 0.5 - 0.4 = 0.1 \, m/s \]

### Final Result:
The velocity of the positive ions moving towards the cathode is \(0.1 \, m/s\).