Question

Small electrostatic potentials. For a monovalent ion at \(T=300 \mathrm{~K}\), what is the value of \(\psi\) such that e \(\psi=k T\) ?

Short answer

\( \psi \ \approx 0.0259 \mathrm{V} \).

Step-by-step solution

Identify the Given Information

Extract the given values: Temperature, T = 300 K.

Recall the Relevant Equation

Use the relationship between the electrostatic potential \( \psi \ \) and other constants: \( e \psi = kT \ \).

Find the Value for Boltzmann's constant

Boltzmann's constant, \( k \), is approximately \( 1.38 \times 10^{-23} \ \mathrm{J/K} \).

Find the Electron Charge

The elementary charge, \( e \), is approximately \( 1.60 \times 10^{-19} \ \mathrm{C} \).

Rearrange the Formula

Solve for \( \psi \): \[ \psi = \frac{k T}{e} \. \]

Substitute the Values

Plug in the known values: \[ \psi = \frac{(1.38 \times 10^{-23} \ \mathrm{J/K}) \times 300 \ \mathrm{K}}{1.60 \times 10^{-19} \ \mathrm{C}} \. \]

Calculate the Result

Calculate \ \psi \ : \[ \psi = \frac {4.14 \times 10^{-21} \mathrm{J}}{1.60 \times 10^{-19} \mathrm{C}} = 0.0259 \ \mathrm{V} \. \]

Key concepts

Boltzmann's constant

Boltzmann's constant, often symbolized as \(k\), plays a crucial role in relating temperature to energy at the molecular level. Named after the physicist Ludwig Boltzmann, it is a fundamental constant in physics. The value of Boltzmann's constant is approximately \(1.38 \times 10^{-23} \ \text{J/K}\). This tiny number represents the energy corresponding to a temperature of one Kelvin (K) for each molecule.

Boltzmann's constant is essential in various fields of physics, including thermodynamics and statistical mechanics. In the context of our problem, it helps connect temperature to the electrostatic potential via energy terms. By using this constant, we can bridge macroscopic thermodynamic quantities, like temperature, with microscopic phenomena, such as the behavior of particles at the molecular level. This makes Boltzmann's constant a cornerstone in the study of how temperature influences molecular and atomic interactions.

Elementary charge

The elementary charge, denoted by \(e\), is the electric charge carried by a single proton and is one of the fundamental constants in nature. Its approximate value is \(1.60 \times 10^{-19} \ \text{C}\). This small value represents the charge of a single proton or, equivalently, the magnitude of the charge of a single electron (which is negative).

Understanding the elementary charge is vital in physics for characterizing electrical properties at the atomic and subatomic levels. For instance, in our exercise, the elementary charge helps to link the electrostatic potential \( \psi \) to the thermal energy via the formula \( e \psi = kT \). Given its fundamental nature, the elementary charge appears in numerous equations and applications, from calculating forces between charged particles to understanding the structure of atoms and molecules.

Temperature relation

The temperature relation in our context ties together thermal energy, electrostatic potential, and fundamental constants. Specifically, the relation \( e \psi = kT \) shows how the electrostatic potential \( \psi \) can be derived from known thermal energies.

At given temperature \( T \) of 300 K, the formula allows us to calculate \( \psi \) by dividing the product of Boltzmann's constant \( k \) and the temperature \( T \) by the elementary charge \( e \). This kind of relation is crucial in understanding how energy conversions operate at thermal equilibrium.

The calculation for our specific problem will thus be:
\ \[ \psi = \frac{kT}{e} \] where we have: \( k = 1.38 \times 10^{-23} \ \text{J/K} \), \( T = 300 \ \text{K} \), and \( e = 1.60 \times 10^{-19} \ \text{C} \). Plug these values into the relation to find \( \psi \) and we get:

\ \[ \psi = \frac {4.14 \times 10^{-21} \ \text{J}}{1.60 \times 10^{-19} \ \text{C}} = 0.0259 \text{V} \].
This simple relationship allows us to connect macroscopic quantities like temperature with microscopic potentials and holds significant importance in the fields of both physics and engineering.