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In a helium gas discharge tube $40 \times {10^{18}}H{e^ + }$ move towards the right side and n electron move towards the left side of the tube if the current in the tube be 8A towards the right then n=?
(A) $10 \times {10^{18}}$
(B) $3 \times {10^{19}}$
(C) $3 \times {10^{20}}$
(D) $3 \times {10^{21}}$

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Last updated date: 20th Jun 2024
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Answer
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Hint: Current flows when there is flow of charges. The amount of charge flowing will give you the current Also, the current is the total charge flowing with respect to time, so be careful while counting charges;

Complete solution:
Here, in this question the time is given to us. It is 1 second so we need to see how much charge flows in 1 second;
Here the 8 ampere current in the helium gas tube is due to the positive helium ions moving towards right and also because of the n electrons moving towards left.
We can mathematically write it as;
$8A = \dfrac{{40 \times {{10}^{18}} \times {q_{H{e^ + }}}C}}{{1\sec }}$$ + \dfrac{{n \times {q_{electron}}}}{{1\sec }}$ (equation:1)
Here ${q_{H{e^ + }}}$ is the magnitude of charge of a positive helium ion.
Hence, ${q_{H{e^ + }}} = 1.6 \times {10^{ - 19}}C$.
Also, ${q_{electron}}$ is the magnitude of charge of an electron.
So, ${q_{electron}} = 1.6 \times {10^{ - 19}}C$.
Thus, now we substitute these values in equation 1;
$8 = 40 \times {10^{18}} \times 1.6 \times {10^{ - 19}} + n \times 1.6 \times {10^{ - 19}}$
Solving this equation for n will give us the value of n;
$8 = \left( {40 \times {{10}^{18}} + n} \right)1.6 \times {10^{ - 19}}$
$40 \times {10^{18}} + n = \dfrac{8}{{1.6 \times {{10}^{ - 19}}}} = 50 \times {10^{18}}$
Hence,$n = \left( {50 - 40} \right){10^{18}} = 10 \times {10^{18}}$

Hence, option A is correct.

Note:(1) The current through a conductor is equal to charge flowing per second through its cross section.
(2) For conductors like the gas tube the positive and negative charge both were flowing and so the net current is the sum of the current due to flow of positive charge per second and the flow of negative charge per second.
(3) For conductors like the metal wire the positive charge is not mobile so we take into account the flow of negative charge only.