Answer:
(a) 2.154×10⁻¹¹ A.
(b) 98300000.
Explanation:
(a)
Using Ohm's law,
V = IR ........................ Equation 1
Where V = Potential difference between the walls, I = current, R = Resistance between the walls.
Make I the subject of the equation
I = V/R...................... Equation 2
Given: V = 84 mV, = 0.084 V, R = 3.9×10⁹ Ω.
Substitute into equation 2
I = 0.084/(3.9×10⁹)
I = 2.154×10⁻¹¹ A.
(b)
I = q/t
q = It .................... Equation 1
Where q = quantity of electric charge, t = time.
Given: I = 2.154×10⁻¹¹ A, t = 0.73 s.
q = 2.154×10⁻¹¹×0.73
q = 1.572×10⁻¹¹ C.
The charge on an electron e = 1.6×10⁻¹⁹ C
n = q/e
where n = number of ions.
n = 1.572×10⁻¹¹/1.6×10⁻¹⁹
n = 9.83×10⁷
n = 98300000.
