Question

In a neon discharge tube $2.8\times {10}^{18}N{e}^{+}$ Ne" ions move to the right per second while $1.2\times {10}^{18}$ electrons move to the left per second. Therefore, the current in the discharge tube is

• A. 0.64 A towards right
• B. 0.256 A towards right
• C. 0.64 Atowards left
• D. 0.256 A towards left

Answer 1

The correct option is A 0.64 A towards right

Answer is A.

Here, the total number of electrons that move in the discharge tube is $2.8\times {10}^{18}+1.2\times {10}^{18}=4\times {10}^{18}electrons.$
The charge of one electron is $1.6\times {10}^{-19}$.
$I=\frac{Q}{t}=\frac{4\times {10}^{18}\times 1.6\times {10}^{-19}}{1}=0.64Atowardstheright.$
Hence, the current in the discharge tube is 0.64 A towards right.