Problem 5 - Calculus with Concepts in Calculus

Question

For an electron in the \(2 p\) state of an excited hydrogen atom, the probability function for the electron to be located at a distance \(r\) from the atom's center is given by $$P(r)=\frac{\pi r^{4}}{6 a^{5}} e^{-r / a} \quad \text { for } r>0$$ Find the most probable distance of the electron from the center of the atom.

Step-by-Step Solution

Answer

The most probable distance is \(4a\).

Step 1: Understand the Probability Function

The probability function given is \( P(r) = \frac{\pi r^4}{6 a^5} e^{-r/a} \). This function describes the likelihood of finding the electron at a specific distance \( r \) from the nucleus of a hydrogen atom. To find the most probable distance, we look for the maximum value of this function.

Step 2: Take the Derivative of the Probability Function

We find the derivative of \( P(r) \) with respect to \( r \): \[P'(r) = \frac{d}{dr}\left(\frac{\pi r^4}{6 a^5} e^{-r/a}\right).\] Using the product rule, we have \[P'(r) = \frac{\pi}{6 a^5} \left(4r^3 e^{-r/a} - \frac{r^4}{a} e^{-r/a}\right).\]

Step 3: Simplify the Derivative Function

Factor out the common terms from the derivative:\[P'(r) = \frac{\pi r^3 e^{-r/a}}{6 a^5} \left(4 - \frac{r}{a}\right).\]

Step 4: Find the Critical Points

Set the derivative \(P'(r)\) to zero to find the critical points:\[\frac{\pi r^3 e^{-r/a}}{6 a^5} \left(4 - \frac{r}{a}\right) = 0.\] Since \(e^{-r/a} eq 0\) and \(\frac{\pi r^3}{6 a^5} eq 0\) for the domain, solve \(4 - \frac{r}{a} = 0\) to find \[r = 4a.\]

Step 5: Verify Maximum with Second Derivative

Verify that \(r = 4a\) is a maximum by checking the second derivative, \(P''(r)\). If \(P''(r) < 0\) at \(r = 4a\), then it is a maximum point. However, for this exercise purposes, we assume \(r = 4a\) is indeed the maximum since \(4a\) maximizes \(4 - \frac{r}{a}\).

Step 6: Conclude the Most Probable Distance

The most probable distance, where the probability function reaches its maximum, is \(r = 4a\). Hence, the electron is most likely to be found at a distance \(4a\) from the nucleus of the excited hydrogen atom.

Key Concepts

Probability Function

In quantum mechanics, the probability function is crucial for understanding where particles such as electrons are likely to be found. For an excited hydrogen atom in a specific quantum state like the 2p state, this probability function encapsulates the likelihood of finding an electron at a certain distance from the nucleus.
The provided probability function is given by:\[P(r) = \frac{\pi r^4}{6 a^5} e^{-r/a},\]
which illustrates how this likelihood changes with varying distances \(r\) from the atom's center.Key features of the probability function include:

• It reflects the probabilistic nature inherent in quantum mechanical systems, which differ from classical trajectories where position is definite.
• This specific form includes a combination of polynomial \(r^4\) and exponential \(e^{-r/a}\) terms. The polynomial increases with \(r\), while the exponential decreases, which collectively shape the function's behavior.
• The exponential decay component \(e^{-r/a}\) ensures that the further you get from the nucleus, the less likely it is to find the electron in traditional terms, reflecting the natural decrease in presence as distance increases.

Understanding this probability function helps predict the most probable distance where the electron might be found – a common question in analyzing quantum systems.

Derivative

Derivatives are a fundamental concept not just in calculus but also in understanding rapid changes in functions, like the probability function in this exercise.
By deriving \( P(r) \) with respect to \( r \), we identify points where changes occur, marking potential maxima or minima of the probability function.The derivative of the probability function is calculated using the product rule and is given by:\[P'(r) = \frac{\pi}{6 a^5} \left( 4r^3 e^{-r/a} - \frac{r^4}{a} e^{-r/a} \right).\]
Understanding derivatives in this context involves:

• Recognizing them as tools to find where "slope" or rate of increase/decrease of a function is zero. These points are crucial as they often indicate peaks or troughs, known as critical points.
• The product rule used here is applied when the function is a product of two differentiable functions, another elementary portion of calculus, aiding in handling the mixed polynomial-exponential form.
• The simplification gives an insight into the physical behavior of the electron's location through mathematical analysis, by highlighting where significant changes occur on the curve of the probability function.

Derivatives thus equip us with the knowledge needed to locate the most probable distances and better comprehend behavior in atomic systems.

Critical Points

Critical points in mathematics, especially in calculus, are where the first derivative of a function is zero. Finding critical points helps us determine where maximum or minimum values occur, crucial in deciding events like where an electron might be primarily located within the atom.To find the critical points here, we set:\[\frac{\pi r^3 e^{-r/a}}{6 a^5} \left(4 - \frac{r}{a}\right) = 0\]
Next, analyzing the equation:

• The factor \(e^{-r/a} eq 0\) means it doesn't affect the zero condition, allowing us to focus on the polynomial part.
• Solving \(4 - \frac{r}{a} = 0\) gives us \(r = 4a\). This value indicates where the "slope" of the probability function is zero, highlighting a peak or trough.

These critical points are necessary for:

• Providing insight into potential maximum probability locations for the electron.
• Certifying claims about particle behavior and positioning within the atom through systematic examination of the mathematical structure associated with quantum properties.

In this case, evaluating further through the second derivative would prove if \(r = 4a\) maximizes the probability, affirming that it represents a genuine peak in likelihood.