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23 September, 17:23

Suppose that the resistance between the walls of a biological cell is 3.9 * 109 Ω. (a) What is the current when the potential difference between the walls is 84 mV? (b) If the current is composed of Na + ions (q = + e), how many such ions flow in 0.73 s?

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  1. 23 September, 17:38
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    (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.
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