Question

Another use of distribution functions is determining the most probable value, which is done by realizing that at the distribution maximum the derivative of the distribution function with respect to the variable of interest is zero. Using this concept, determine the most probable value of \(x(0 \leq x \leq \infty)\) for the following function: $$P(x)=C x^{2} e^{-a x^{2}}$$ Compare your result to \(\langle x\rangle\) and \(\mathrm{x}_{\mathrm{rms}}\) when \(a=0.3\) (see Example Problem 29.13 ).

Short answer

To find the most probable value of x for the distribution function \(P(x) = Cx^{2}e^{-ax^{2}}\), we first find the derivative of P(x) with respect to x, then set it to zero, and solve for x. Doing so, we find that the most probable value of x is given by: \(x = \sqrt{\frac{1}{a}}\). Comparing this to the values of ⟨x⟩ and x_rms when a = 0.3, we find the most probable value of x ≈ 1.83, which is close to the average value ⟨x⟩ = 1.63 and very close to the root-mean-square value x_rms = 1.87.

Step-by-step solution

Differentiate P(x) with respect to x

First, let's differentiate the given function P(x) = Cx^2 e^(-ax^2) with respect to x, which is our variable of interest. Using the chain rule and product rule, the derivative, dP(x)/dx, can be expressed as follows: \( \displaystyle \frac{dP(x)}{dx} = C\left( 2 x e^{-ax^{2}} - 2 ax^{3} e^{-ax^{2}} \right) \) Now, we have the derivative of the function P(x).

Set the derivative equal to zero

In order to find the most probable value of x, we need to find the value of x that maximizes the distribution function, P(x). This occurs when the derivative is equal to zero, so let's set the derivative equal to zero: \( \displaystyle \frac{dP(x)}{dx} = C\left( 2 x e^{-ax^{2}} - 2 ax^{3} e^{-ax^{2}} \right) =0 \)

Solve for x

To solve for x, we can first factor out 2x e^(-ax^2) from the equation: \( 2 x e^{-ax^{2}} \left(1 - ax^{2} \right) = 0 \) Now, we have two possibilities to make the equation equal to zero: 1. \( 2 x e^{-ax^{2}} = 0 \) 2. \( 1 - ax^{2} = 0 \) The first equation has a trivial solution (x = 0), but this is one of the boundaries of x. The second equation gives us a non-trivial solution: \( 1 - ax^{2} = 0 \Rightarrow x^{2} = \frac{1}{a} \Rightarrow x = \sqrt{\frac{1}{a}} \) So, the most probable value of x is: \( x = \sqrt{\frac{1}{a}} \)

Compare to ⟨x⟩ and x_rms when a = 0.3

Now, let's find the most probable value of x when a = 0.3, and compare it with the given values of ⟨x⟩ and x_rms: \( x = \sqrt{\frac{1}{0.3}} \approx 1.83 \) Comparing to the given values, we have: - Most probable value of x: \( \approx 1.83 \) - ⟨x⟩: Example Problem 29.13 gives ⟨x⟩ = 1.63 - x_rms: Example Problem 29.13 gives \(x_{rms}\) = 1.87 From this comparison, we can see that the most probable value of x is close to but slightly higher than the average value ⟨x⟩ and very close to the root-mean-square value x_rms when a = 0.3.

Key concepts

Distribution Function

In statistics and probability theory, a distribution function describes how the values of a random variable are distributed. It is an equation or graph that shows the likelihood of each outcome. In physical chemistry, the distribution function often represents the distribution of particles across various states, such as energy levels or positions in space.

The function provided in the exercise, P(x) = C x^2 e^{-a x^2}, is an example of a distribution function. It shows the probability of finding a particle at position x. The constant C is a normalization factor ensuring that the total probability across all positions is equal to one. When a particle's position is more likely, the function yields higher values, and when less likely, lower values.

An important aspect of distribution functions is that they can express the Most Probable Value (MPV), which is the position where the probability of finding a particle is maximized. This value is particularly relevant in physical chemistry, where it can depict the most likely state of a particle in a system.

Derivative in Physical Chemistry

In physical chemistry, derivatives play a critical role in understanding how functions change with respect to variables. In the context of distribution functions, the derivative is used to find extrema—points where the function has either a maximum or minimum value.

The step by step solution of the exercise demonstrates the use of the derivative to find the most probable value in a distribution. By differentiating the probability distribution function P(x) with respect to x and setting the derivative equal to zero, we identify the point where the function reaches its maximum value. The zero derivative indicates that there is no increase or decrease at that specific value of x, signifying an extremum—ideally the most probable value.

This is an application of the first derivative test in calculus, which helps to identify local maxima and minima of functions. For the given function P(x), this process reveals that the most probable value of x is where the slope of the function is zero, indicating the peak of the probability distribution.

Expected Value and Root Mean Square

The expected value (often denoted as ⟨x⟩) and the root mean square value (x_rms) are statistical measures that help to describe the characteristics of distribution functions.

The expected value, or mean, represents the average outcome if an experiment were repeated many times. It is a central tendency measure that provides a single value summary of the probability distribution. Physically, it could represent the average position of a particle within a system.

The root mean square (x_rms) value, on the other hand, is a measure of the magnitude of a set of numbers. It gives insight into the variability of the distribution and incorporates both the mean and the dispersion of values. The x_rms value is particularly important in contexts where both the magnitude and the spread are significant, such as in quantum mechanics where it can describe the uncertainty in a particle's position.

In the exercise above, the comparison between the most probable value, the expected value, and the x_rms gives students a complete picture of the distribution's behavior. It demonstrates that while the most probable value indicates the peak of the distribution, both the expected value and x_rms take into account the general spread of values around it.

Probability Distribution Maximization

Probability distribution maximization involves finding the value that makes the distribution function reach its highest point. In other words, it's the process of identifying the most probable value that an outcome will occur, within a given probability distribution.

The exercise demonstrates this concept through a physical chemistry perspective. The function P(x) is differentiated and set to zero to find the most probable position x. This method, rooted in calculus, is often used in various fields, including economics and engineering, to optimize functions for maximum probability or efficiency.

Importantly, the maximization of the probability distribution function allows for the prediction of the most likely outcomes in a given system. Whether it's determining the position of a particle, the timing of an event, or the most efficient process, maximizing the probability distribution is a fundamental concept in theoretical and applied sciences.