The electric field across a cellular membrane is $1.0 \times 10^{7} \mathrm{N} / \mathrm{C}$ directed into the cell. (a) If a pore opens, which way do sodium ions (Na") flow- into the cell or out of the cell? (b) What is the magnitude of the electric force on the sodium ion? The charge on the sodium ion is \(+e\)

Given an electric field across a cellular membrane, we are asked to determine the direction of sodium ion flow when a pore opens. Since sodium ions have a positive charge (+e) and the electric field is directed into the cell, the positively charged sodium ions will be attracted to the negative direction of the electric field. As a result, the sodium ions will flow into the cell.

To find the magnitude of the electric force on the sodium ion, we can use Coulomb's law equation: \(F = qE\) where \(F\) is the electric force, \(q\) is the charge of the sodium ion, and \(E\) is the magnitude of the electric field. We are given the charge on a sodium ion (\(+e\)) and the electric field magnitude (\(1.0 \times 10^{7} \mathrm{N}/\mathrm{C}\)). Let's substitute the given values into the equation: \(F = (+e) (1.0 \times 10^{7} \mathrm{N}/\mathrm{C})\) Since the charge of an electron is \(e = 1.6 \times 10^{-19} \mathrm{C}\), the charge on a sodium ion is \(+e = +1.6 \times 10^{-19} \mathrm{C}\). Plug this value into our equation: \(F = (+1.6 \times 10^{-19} \mathrm{C}) (1.0 \times 10^{7} \mathrm{N}/\mathrm{C})\) Now, calculate the electric force: \(F = 1.6 \times 10^{-12} \mathrm{N}\) So, the magnitude of the electric force on the sodium ion is \(1.6 \times 10^{-12} \mathrm{N}\).