Question

In an inert gas discharge tube $2.9\times {10}^{18}$ positive ions $\left(e^{+}\right)$ moves to the right through a cross-section of the tube each second, while $1.2\times {10}^{18}$ negative ions $\left(r^{-}\right)$ move to the left in the same time.The magnitude of current is

• A. 0.66A
• B. $0.66\times {10}^{-3}A$
• C. $0.66\times {10}^{-6}A$
• D. None of these

Answer 1

The correct option is A 0.66A

Given: In an inert gas discharge tube $2.9\times {10}^{18}$ positive ions $\left(e^{+}\right)$ moves to the right through a cross-section of the tube each second, while $1.2\times {10}^{18}$ negative ions $\left(r^{-}\right)$ move to the left in the same time.

To find the magnitude of the current

Solution:

We know,

Net Current, $I=I_{+}+i_{-}=\frac{\left(n_{+}\right)\left(q_{+}\right)}{t}+\frac{\left(n_{-}\right)\left(q_{-}\right)}{t}$

$⟹I=\frac{\left(n_{+}\right)}{t}\times e+\frac{\left(n_{-}\right)}{t}\times e⟹I=e(2.9\times {10}^{18}+1.2\times {10}^{18})⟹I=1.6\times {10}^{-19}(4.1\times {10}^{18})⟹I=6.56\times {10}^{-1}⟹I=0.66A$

is the required magnitude of the current