In a neon discharge tube $2.9 \times {10^{18}}\,N{e^ + }$ ions move to the right each second while $1.2 \times {10^{18}}$ electrons move to the left per second. Electron charge is $1.6 \times {10^{ - 19}}\,C$. The current in the discharge tube
Medium
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Net current $i = {i_ + } + {i_ - } = \frac{{({n_ + })\,({q_ + })}}{t} + \frac{{({n_ - })\,({q_ - })}}{t}$
$ \Rightarrow $ $i = \frac{{({n_ + })}}{t} \times e + \frac{{({n_ - })}}{t} \times e$
$ = 2.9 \times {10^{18}} \times 1.6 \times {10^{ - 19}} + 1.2 \times {10^{18}} \times 1.6 \times {10^{ - 19}}$
$ \Rightarrow $ $i = \,0.66\,A$
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