Problem 4 - Physics

Question

Suppose that the resistance between the walls of a biological cell is \(5.0 \times 10^{9} \Omega\). (a) What is the current when the potential difference between the walls is \(75 \mathrm{mV} ?\) (b) If the current is composed of Na+ ions \((q=+e)\), how many such ions flow in \(0.50 \mathrm{~s}\) ?

Step-by-Step Solution

Answer

The current is \(1.5 \times 10^{-11} \mathrm{A}\). About \(4.7 \times 10^{7}\) Na+ ions flow in \(0.50 \mathrm{s}\).

Step 1: Understand the Problem

The problem asks us to calculate the current flowing through a biological cell's membrane when a potential difference is applied, and then to find out how many sodium ions (Na+) flow through the membrane in a given time.

Step 2: Identify Formula for Current

We use Ohm's Law, which states that the current \( I \) can be calculated using the formula \( I = \frac{V}{R} \), where \( V \) is the potential difference and \( R \) is the resistance.

Step 3: Calculate the Current

Given \( V = 75\,\mathrm{mV} = 75 \times 10^{-3}\,\mathrm{V} \) and \( R = 5.0 \times 10^{9} \Omega \), we can find \( I \):\[ I = \frac{75 \times 10^{-3}}{5.0 \times 10^{9}} = 1.5 \times 10^{-11}\,\mathrm{A} \]

Step 4: Calculate Charge Flow

From the formula \( Q = It \), where \( Q \) is the total charge, \( I \) is the current and \( t = 0.50 \mathrm{~s} \), calculate the total charge that flows in that time:\[ Q = 1.5 \times 10^{-11} \times 0.50 = 7.5 \times 10^{-12} \mathrm{C} \]

Step 5: Convert Charge to Number of Ions

Each Na+ ion carries a charge equal to the elementary charge \( e = 1.6 \times 10^{-19} \mathrm{C} \). The number of ions \( N \) is given by \( N = \frac{Q}{e} \):\[ N = \frac{7.5 \times 10^{-12}}{1.6 \times 10^{-19}} \approx 4.7 \times 10^{7} \text{ Na+ ions} \]

Key Concepts

Current Calculation

Calculating current is fundamental to understanding how electric circuits behave. In this exercise, we used Ohm's Law to calculate the current flowing through a biological cell membrane when a potential difference is applied.
Ohm's Law is given by the equation:

• \( I = \frac{V}{R} \)

where:

• \( I \) represents the current in amperes (A).
• \( V \) is the potential difference in volts (V).
• \( R \) is the resistance in ohms (\( \Omega \)).

Using these relationships, we calculated the current flowing through a cell's membrane when exposed to a potential difference of \( 75 \mathrm{mV} \) and a resistance of \( 5.0 \times 10^{9} \Omega \). The calculated current is extremely small: \( 1.5 \times 10^{-11} \mathrm{A} \). This highlights how sensitive biological systems are to electric currents.

Resistance in Biological Cells

Biological cells have walls made of a lipid bilayer that impedes the flow of ions, giving it a high resistance value. In this exercise, the resistance is a staggering \( 5.0 \times 10^{9} \Omega \).
This high resistance serves an important physiological purpose:

• It helps in maintaining the cell's internal environment by restricting the ion flow.
• It allows the cell to create a potential difference across its membrane to perform necessary cellular functions.

Understanding such resistance is crucial in cell biology and helps in designing medical treatments that involve electrical stimulation of cells.

Potential Difference

The potential difference, sometimes called voltage, is the driving force that pushes the charge through a circuit. In the context of a biological cell, it refers to the electrical potential energy difference across the cell membrane.
In our problem, the potential difference is \( 75 \mathrm{mV} \), equivalent to \( 75 \times 10^{-3} \mathrm{V} \).
A potential difference across cell membranes typically arises from:

• The uneven distribution of ions on either side of the membrane.
• Pumps and channels within the membrane that selectively allow ions to pass.

This potential difference plays a pivotal role in nerve impulse conduction and overall cellular function.

Charge Flow

Charge flow, or the movement of electric charge, is quantified by the total charge that passes through the circuit over time. The equation \( Q = It \) is used to calculate the flow of charge

• where \( Q \) is the total charge in coulombs (C), \( I \) is the current in amperes (A), and \( t \) is the time in seconds (s).

For this scenario, knowing the current \( 1.5 \times 10^{-11} \mathrm{A} \) and the time \( 0.50 \,\mathrm{s} \), the total charge that flows is \( 7.5 \times 10^{-12} \mathrm{C} \). Charge flow is essential in understanding cellular activities like nerve impulses, which rely on controlled charge transport.

Sodium Ions

Sodium ions (Na+) play critical roles in the physiological processes of cells, especially in the conduction of electrical signals along nerves. In this problem, the change in the number of Na+ ions flowing through the cell membrane was calculated.
Na+ ions are:

• Positively charged due to lacking one electron.
• Essential in establishing the electrochemical gradient across cellular membranes.

To calculate how many sodium ions cross the membrane, the total charge was divided by the charge of a single sodium ion, or elementary charge, which leads us to roughly \( 4.7 \times 10^{7} \) sodium ions flowing in \( 0.50 \,\mathrm{s} \). This showcases the vast number of ions involved even with relatively tiny amounts of current.

Elementary Charge

The elementary charge \( e \) is a fundamental constant and represents the smallest unit of electric charge that is considered indivisible. It is the charge of a single proton or the negative of that of a single electron.
In calculations involving ions, the elementary charge is a crucial factor. Each sodium ion, being \( Na^+ \), has a charge equal to \( +e \), where \( e \) is \( 1.6 \times 10^{-19} \mathrm{C} \).
Key points about the elementary charge:

• It's the building block of electric charge in particles.
• Crucial in determining the number of particles contributing to a given charge flow.

Through this fundamental knowledge, we determined how many sodium ions participated in creating the given charge flow across the cell membrane.