Question

1. In a neon discharge tube 2.8 x10^18 Ne+ ions move
to the right per second while 1.2 x 10^18 electron
move to the left per second. Therefore, the current
in the discharge tube is
(1) 0.64 A towards night
(2) 0.256 A towards right
(3) 0.64 A towards left
(4) 0.256 A towards left​

Answer 1

Answer:

Given, rate of flow of change, of

Ne

+

ions, I

1

=2.9×10

18

×1.6×10

−19

[No. of ions × change of each ions]

=0.464A

of e

−

, I

2

=1.2×10

18

×1.6×10

−19

[conventional current is taken opposite to flow of e

−

]

=0.192A

Thus, net current

I=I

1

+I

2

=0.464+0.192

⇒I=0.656A towards right.

solution

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