In a discharge tube when `200` volt potential difference is applied `6.25xx10^(18)` electrons move from cathode to anode and `3.125xx10^(18)` singly charged positive ions move from anode to cathode in one second. Then the power of tube is:

Options

A. `100 "watt"`
B. `200"watt"`
C. `300 "watt"`
D. `400"watt"`

Explanation

To find the power of the discharge tube, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Potential difference (V) = 200 volts
- Number of electrons moving from cathode to anode per second (N_e) = \(6.25 \times 10^{18}\)
- Number of singly charged positive ions moving from anode to cathode per second (N_i) = \(3.125 \times 10^{18}\)
- Charge of an electron (e) = \(1.6 \times 10^{-19}\) coulombs

2. **Calculate the Current (I):**
- The current due to the electrons moving from cathode to anode can be calculated using the formula:
\[ I_e = N_e \times e \]
- Substituting the values:
\[ I_e = 6.25 \times 10^{18} \times 1.6 \times 10^{-19} = 1.0 \, \text{A} \]

3. **Calculate the Current due to Positive Ions (I_i):**
- The current due to the positive ions moving from anode to cathode is:
\[ I_i = N_i \times e \]
- Substituting the values:
\[ I_i = 3.125 \times 10^{18} \times 1.6 \times 10^{-19} = 0.5 \, \text{A} \]

4. **Total Current (I):**
- The total current in the discharge tube is the sum of the currents due to electrons and positive ions:
\[ I = I_e + I_i = 1.0 \, \text{A} + 0.5 \, \text{A} = 1.5 \, \text{A} \]

5. **Calculate the Power (P):**
- The power of the tube can be calculated using the formula:
\[ P = V \times I \]
- Substituting the values:
\[ P = 200 \, \text{V} \times 1.5 \, \text{A} = 300 \, \text{W} \]

### Final Answer:
The power of the tube is **300 Watts**.

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