Answer:
The current flowing through the tube is 0.601 A
Explanation:
Given data;
the diameter of the fluorescent tube is d = 3 cm
the incoming negative charge in the tube is -e = 3 x 10¹⁸ electrons/second
the outgoing positive charge equals +e = 0.75 x 10¹⁸ electrons/ second
The current within the fluorescent tube results from both positive and negative charges which contribute to maintaining electrical neutrality in the conductor (fluorescent tube).
Q = It
I = Q/t
where;
I signifies current in Amperes (A)
Q represents charge measured in Coulombs (C)
t denotes time which is in seconds (s)
1 electron (e) accounts for 1.602 x 10⁻¹⁹ C
Thus, 3 x 10¹⁸ e/s can be computed as
= (3 x 10¹⁸ e/s x 1.602 x 10⁻¹⁹ C) / 1e
= 0.4806 C/s
This reflects the negative charge per second (Q/t) = 0.4806 C/s
Meanwhile, the positive charge per second amounts to
(0.75 x 10¹⁸ e/s x 1.602 x 10⁻¹⁹ C) / 1e
Thus, the positive charge per second is equal to 0.12015 C/s
The total charge per second in the tube is then obtained by summing both charges: Q / t = (0.4806 C/s + 0.12015 C/s)
I = 0.601 A
Therefore, the current within the tube computes to 0.601 A
