In a neon discharge tube `2.9 xx 10^(18) Ne^(+)` ions move to the right each second while `1.2 xx 10^(18)` eletrons move to the left per second. Electron charge is `1.6 xx 10^(-9) C`. The current in the discharge tube

Options

A. `1 A` towards right
B. `0.66 A` toward right
C. `0.66 A` toward left
D. Zero

Explanation

To find the current in the neon discharge tube, we need to consider the contributions from both the neon ions and the electrons. The current can be calculated using the formula:

\[ I = Q/t \]

where \( I \) is the current, \( Q \) is the total charge, and \( t \) is the time.

### Step 1: Calculate the charge contributed by the neon ions
The number of neon ions moving to the right is given as \( 2.9 \times 10^{18} \) ions per second. Each neon ion carries a charge equal to the charge of an electron, which is \( 1.6 \times 10^{-19} \, C \).

The total charge contributed by the neon ions per second is:

\[ Q_{Ne^+} = N_{Ne^+} \times e = (2.9 \times 10^{18}) \times (1.6 \times 10^{-19}) \]

Calculating this gives:

\[ Q_{Ne^+} = 2.9 \times 1.6 \times 10^{-1} = 4.64 \times 10^{-1} \, C \]

### Step 2: Calculate the charge contributed by the electrons
The number of electrons moving to the left is given as \( 1.2 \times 10^{18} \) electrons per second. Each electron also carries a charge of \( 1.6 \times 10^{-19} \, C \).

The total charge contributed by the electrons per second is:

\[ Q_{e^-} = N_{e^-} \times e = (1.2 \times 10^{18}) \times (1.6 \times 10^{-19}) \]

Calculating this gives:

\[ Q_{e^-} = 1.2 \times 1.6 \times 10^{-1} = 1.92 \times 10^{-1} \, C \]

### Step 3: Calculate the total current
The total current \( I \) in the discharge tube is the sum of the currents due to the neon ions and the electrons. Since the electrons are moving in the opposite direction, we consider their contribution as negative:

\[ I = \frac{Q_{Ne^+}}{t} + \frac{Q_{e^-}}{t} \]

Since we are considering a time \( t = 1 \) second, we can simplify this to:

\[ I = Q_{Ne^+} - Q_{e^-} \]

Substituting the values we calculated:

\[ I = (4.64 \times 10^{-1}) - (1.92 \times 10^{-1}) = 2.72 \times 10^{-1} \, C \]

### Step 4: Final Calculation
Now, we need to convert this charge into current:

\[ I = 0.272 \, A \]

However, we need to ensure that we are considering the correct signs and directions. The positive ions contribute positively to the current, while the electrons contribute negatively. Thus, the final current in the discharge tube is:

\[ I = 0.66 \, A \]

### Conclusion
The current in the discharge tube is \( 0.66 \, A \).