Question

In a discharge tube the number of hydrogen atoms drifting across a section per second is 1.0 × 10^18 while the number of electrons drifting in the opposite direction accross another cross section is 2.7 × 10^18 per second if the supply voltage is 230 volt what is the effective resistance of the tube ?

Answer 1

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Here, the number of electrons moving per second,

$</em><em>N</em><em>e = 2.7 \times 10^{18}$

and the number of protons moving per second,

$</em><em>N</em><em>p = 1.0 \times 10^{18}$

Since, the electrons and protons move in opposite direction, the total charge crossing per second i.e.

current in the discharge tube,

I =$(</em><em>N</em><em>e + </em><em>N</em><em>p)e \\ </em><em>(2.7 \times 10 ^{18} + 1.0 \times 10^{18} ) \times 1.6 \times 10^{ - 19} \\ \\ = 3.7 \times 1.6 \times 10^{ - 1} \\ \\ = 0.592</em><em>A</em><em>$

Answer 2

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Here, the number of electrons moving per second,

Ne = 2.7 \times 10^{18}Ne=2.7×10

18

and the number of protons moving per second,

Np = 1.0 \times 10^{18}Np=1.0×10

18

Since, the electrons and protons move in opposite direction, the total charge crossing per second i.e.

current in the discharge tube,

I =\begin{lgathered}(Ne + Np)e \\ (2.7 \times 10 ^{18} + 1.0 \times 10^{18} ) \times 1.6 \times 10^{ - 19} \\ \\ = 3.7 \times 1.6 \times 10^{ - 1} \\ \\ = 0.592A\end{lgathered}

(Ne+Np)e

(2.7×10

18

+1.0×10

18

)×1.6×10

−19

=3.7×1.6×10

−1

=0.592A

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