Question

The Maxwell-Boltzmann distribution $f(v_x)$ of one-dimensional velocities $v_x$ at temperature T is [Given: A is a normalization constant such that $ \int_{-\infty}^{\infty} f(v_x) dv_x = 1$, and $k_B$ is the Boltzmann constant]

• A. $Aexp(-mv_x^2/2k_BT)$
• B. $Aexp(-mv_x^2/k_BT)$
• C. $Av_x^2exp(-mv_x^2/2k_BT)$
• D. $Av_x^2exp(-mv_x^2/k_BT)$

Answer 1

The Correct Option is A

The problem involves identifying the one-dimensional Maxwell-Boltzmann distribution of velocities, specifically for the component of velocity \( v_x \), at a given temperature \( T \). We are given that \( A \) is the normalization constant and \( k_B \) is the Boltzmann constant. Let's break down the reasoning to find the correct expression among the given options.

Firstly, the Maxwell-Boltzmann distribution for one-dimensional velocities is generally given by:

• \(f(v_x) = A \exp \left( -\frac{mv_x^2}{2k_BT} \right)\),

where:

• \(m\) is the mass of the particle,
• \(k_B\) is the Boltzmann constant,
• \(T\) is the absolute temperature,
• \(v_x\) is the velocity component in one dimension,
• \(\exp\) denotes the exponential function.

Now, let's examine the options:

1. \(A \exp \left(-\frac{mv_x^2}{2k_BT}\right)\)
2. \(A \exp \left(-\frac{mv_x^2}{k_BT}\right)\)
3. \(A v_x^2 \exp \left(-\frac{mv_x^2}{2k_BT}\right)\)
4. \(A v_x^2 \exp \left(-\frac{mv_x^2}{k_BT}\right)\)

The first option matches the one-dimensional Maxwell-Boltzmann velocity distribution form. The exponent in the distribution contains a factor of \( \frac{1}{2} \) in the denominator, which is typical in the Gaussian like distribution for velocity, representing \( \frac{mv_x^2}{2k_BT} \).

Options 2, 3, and 4 introduce variations that do not match the classical form of the one-dimensional velocity distribution.

Thus, the correct option is indeed:

• \(A \exp \left(-\frac{mv_x^2}{2k_BT}\right)\)

This justifies that among the given options, the correct expression for the Maxwell-Boltzmann distribution of one-dimensional velocities at temperature \( T \) is:

\(A \exp \left( -\frac{mv_x^2}{2k_BT} \right)\)