Suppose that the resistance between the walls of a biological cell is 3.00 x 10^9 Ω.
(a) What is the current when the potential difference between the walls is 70 mV?
____________ A
(b) If the current is composed of Na+ ions (q = +e), how many such ions flow in 0.2 s?
___________ ions
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To find the current when the potential difference between the walls of the biological cell is 70 mV, we can use Ohm's Law, which states that the current (I) flowing through a conductor is equal to the potential difference (V) across the conductor divided by the resistance (R). The formula for Ohm's Law is:
I = V / R
(a) To find the current (I), we need to divide the potential difference (V) by the resistance (R). Let's plug in the values:
Potential difference (V) = 70 mV = 70 x 10^-3 V
Resistance (R) = 3.00 x 10^9 Ω
I = (70 x 10^-3 V) / (3.00 x 10^9 Ω)
Simplifying, we get:
I = 23.333 x 10^-12 A
Therefore, the current when the potential difference between the walls is 70 mV is 2.3333 nA (nanoAmperes).
(b) Now, to find the number of Na+ ions (q = +e) that flow in 0.2 seconds, we need to use the formula for current:
I = q / t
Where q is the charge, t is the time, and I is the current. Rearranging the formula, we have:
q = I x t
To find the number of ions, we need to divide the charge (q) by the charge of a single ion, which is +e.
Number of ions = q / (+e)
Plugging in the values:
I = 2.3333 x 10^-9 A (Converting from nanoAmperes to Amperes)
t = 0.2 s
+e = 1.6 x 10^-19 C (charge of a single ion)
q = (2.3333 x 10^-9 A) x (0.2 s)
Now, let's find the number of ions:
Number of ions = q / (+e)
Simplifying, we get:
Number of ions = [(2.3333 x 10^-9 A) x (0.2 s)] / (1.6 x 10^-19 C)
Calculating the result, we find the number of Na+ ions flowing in 0.2 s.