In a fluorescent tube of diameter 3 cm, 3 1018 electrons and 0.75 1018 positive ions (with a charge of e) flow through a cross-sectional area each second. What is the current in the tube

The  current in the tube is 0.601 A

Explanation:

Given;

diameter of the fluorescent, d = 3 cm

negative charge flowing in the fluorescent tube, -e = 3 x 10¹⁸ electrons/second

positive charge flowing in the fluorescent tube, +e = 0.75 x 10¹⁸ electrons/ second

The current in the fluorescent tube is due to presence of positive and negative charges to create neutrality in the conductor (fluorescent tube).

Q = It

I = Q/t

where;

I is current in Ampere (A)

Q is charge in Coulombs (C)

t is time is seconds (s)

1 e = 1.602 x 10⁻¹⁹ C

3 x 10¹⁸ e/ s = ?

= (3 x 10¹⁸ e/s  x 1.602 x 10⁻¹⁹ C) / 1e

= 0.4806 C/s

negative charge per second (Q/t) = 0.4806 C/s

positive charge per second (Q/t) =  (0.75 x 10¹⁸ e/s  x 1.602 x 10⁻¹⁹ C) / 1e

positive charge per second (Q/t) = 0.12015 C/s

Total charge per second in the tube, Q / t = (0.4806 C/s + 0.12015 C/s)

                                                                I = 0.601 A

Therefore, the  current in the tube is 0.601 A

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