In a neon discharge tube 2.9 × 1018 Ne +ions move to be the right per second while 1.2 × 108 electron move to the left per second electric charge is 1.6 × 10-19 C. The current in discharge tube is :

Question

In a neon discharge tube 2.9 × 1018 Ne +ions move to be the right per second while 1.2 × 108 electron move to the left per second electric charge is 1.6 × 10-19 C. The current in discharge tube is : (A) 0.66 A towards left (B) 0.66 A towards right (C) 1 A towards right (D) zero

Tardigrade Admin · Accepted Answer

Current due to both types of ions is in the same direction towards right so i=i1+i2 i=29 × 1018 × 1.6 × 10-19+1.2 × 1018 × 1.6 × 10-19 i=0.66 A

Q. In a neon discharge tube $2.9 \times 1 0^{18} N e^{+}$ ions move to be the right per second while $1.2 \times 1 0^{8}$ electron move to the left per second electric charge is $1.6 \times 1 0^{- 19} C$ . The current in discharge tube is :

0.66 A towards left

0.66 A towards right

1 A towards right

zero

Current due to both types of ions is in the same direction towards right so
$i = i_{1} + i_{2}$
$i = 29 \times 1 0^{18} \times 1.6 \times 1 0^{- 19} + 1.2 \times 1 0^{18} \times 1.6 \times 1 0^{- 19}$
$i = 0.66 A$

Q. In a neon discharge tube 2.9×1018Ne+ions move to be the right per second while 1.2×108 electron move to the left per second electric charge is 1.6×10−19C. The current in discharge tube is :