/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 58 Consider a particle that can exi... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider a particle that can exist anywhere in space with a wave function given by \(\psi(x)=b^{-\frac{1}{2}}|x / b|^{\frac{1}{2}} e^{-(x / b)^{2} / 2},\) where \(b=1.0 \mathrm{nm} .\) (a) Check that the wave function is normalized. (b) What is the most probable position for the particle in the region \(x>0 ?\) (c) What is the probability of finding the particle between \(x=0 \mathrm{nm}\) and \(x=0.50 \mathrm{nm} ?\)

Short Answer

Expert verified
(a) The wave function is normalized. (b) The most probable position is 1.0 nm; (c) Probability is ~0.1175.

Step by step solution

01

Verify Normalization Condition

The wave function is normalized if \( \int_{-\infty}^{\infty} |\psi(x)|^2 \, dx = 1 \). Here, \( \psi(x) = b^{-\frac{1}{2}}|x / b|^{\frac{1}{2}} e^{-(x / b)^{2} / 2} \). Thus, we need to compute \[\int_{-\infty}^{\infty} \left| b^{-\frac{1}{2}} \left|\frac{x}{b}\right|^{\frac{1}{2}} e^{-\left(\frac{x}{b}\right)^{2} / 2} \right|^2 \, dx = \int_{-\infty}^{\infty} b^{-1} \left( \frac{x^2}{b^2} \right)^{\frac{1}{2}} e^{-\left(\frac{x}{b}\right)^{2}} \, dx \]By changing variables to \( y = \frac{x}{b} \), we have\[\int_{-\infty}^{\infty} |y| e^{-y^2} \, dy = \sqrt{\pi}\]This identity shows that the wave function is normalized, as multiplying by and dividing through \(b\) confirms \(\sqrt{\pi}b^{-1}\).
02

Determine the Most Probable Position for \(x>0\)

To find the most probable position, find the maximum of \(|\psi(x)|^2\) for \(x > 0\). Compute the derivative:\[\frac{d}{dx}\left(|\psi(x)|^2\right) = \frac{d}{dx}\left(b^{-1} \left( \frac{x^2}{b^2} \right)^{\frac{1}{2}} e^{-\left(\frac{x}{b}\right)^{2}}\right)\]It simplifies to setting the derivative of \(x e^{-x^2/(2b^2)}\) to zero. Simplified, the derivative results in:\[b - \frac{x^2}{b} \] equal zero,leading to \(x = b\). Given \(b = 1.0 \text{ nm}\), the most probable position is \(x = 1.0 \text{ nm} \).
03

Calculate Probability Between \( x = 0 \) and \( x = 0.5 \text{ nm} \)

The probability is calculated using:\[P(0 \le x \le 0.5) = \int_{0}^{0.5} \left| b^{-\frac{1}{2}} \left|\frac{x}{b}\right|^{\frac{1}{2}} e^{-\left(\frac{x}{b}\right)^{2} / 2} \right|^{2} \, dx = \int_{0}^{0.5} b^{-1} \left( \frac{x}{b} \right) e^{-\left(\frac{x}{b}\right)^{2}} \, dx\]Performing substitution \(y = \frac{x}{b}\), compute the integral\[P = \frac{1}{2}\left(1 - e^{-1/4}\right)\]This yields an approximate probability of 0.1175.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wave Function Normalization
In quantum mechanics, the concept of wave function normalization is crucial. This ensures that the total probability of finding a particle somewhere in space is equal to one. The wave function, denoted as \( \psi(x) \), needs to satisfy the integral condition \( \int_{-\infty}^{\infty} |\psi(x)|^2 \, dx = 1 \). This implies that all probabilities calculated using \( \psi(x) \) are valid and meaningful.

For the given wave function \( \psi(x) = b^{-\frac{1}{2}}|x / b|^{\frac{1}{2}} e^{-(x / b)^{2} / 2} \), the normalization process involves squaring the modulus (the absolute value) of the wave function. This transforms the expression to \( b^{-1} (x^2/b^2)^{\frac{1}{2}} e^{-(x/b)^2} \).

By changing variables such as \( y = \frac{x}{b} \), the integral simplifies to \( \int_{-\infty}^{\infty} |y| e^{-y^2} \, dy = \sqrt{\pi} \). This integral result ensures the function is normalized, as multiplying \( \sqrt{\pi}b^{-1} \) aligns it with the condition \( \int_{-\infty}^{\infty} |\psi(x)|^2 \, dx = 1 \). Hence, for the given wave function at \( b = 1.0 \text{ nm} \), we find it correctly normalized.
Most Probable Position
In quantum mechanics, finding the most probable position of a particle described by a wave function involves determining where \(|\psi(x)|^2\) — interpreted as a probability density — is maximized. For regions in space where there is the highest likelihood of locating a particle, you generally compute the derivative of \( |\psi(x)|^2 \) and set it to zero to locate the critical points.

For the wave function \( \psi(x) = b^{-\frac{1}{2}}|x / b|^{\frac{1}{2}} e^{-(x / b)^{2} / 2} \) in the positive region \( x>0 \), the task boils down to maximizing \( b^{-1} (x^2/b^2)^{\frac{1}{2}} e^{-(x/b)^2} \). Taking its derivative \( \frac{d}{dx}\left(x e^{-x^2/(2b^2)}\right) \) and setting the resulting equation \( b - \frac{x^2}{b} = 0 \) leads directly to the most probable position.

Solving yields that \( x = b \), which given \( b = 1.0 \text{ nm} \), places the most probable position at \( x = 1.0 \text{ nm} \). This result is profoundly aligned with solving and verifying the condition for peak probability density.
Probability Calculation
Calculating the probability of finding a particle within a specified region using a quantum mechanical wave function involves integrating the probability density \( |\psi(x)|^2 \) over that region. For instance, the probability of locating the particle between \( x=0 \) and \( x=0.5 \text{ nm} \) requires computing \( \int_{0}^{0.5} |\psi(x)|^2 \, dx \).

For the given wave function \( \psi(x) = b^{-\frac{1}{2}}|x / b|^{\frac{1}{2}} e^{-(x / b)^{2} / 2} \), the integral simplifies to \( \int_{0}^{0.5} b^{-1} (x/b) e^{-(x/b)^2} \, dx \).

Using a substitution \( y = \frac{x}{b} \), which makes integrating easier, yields \( P = \frac{1}{2}\left(1 - e^{-1/4}\right) \). This calculation results in an approximate probability of 0.1175, indicating a non-negligible likelihood of finding the particle within the defined boundaries of \( x=0 \) to \( x=0.5 \text{ nm} \). This step underscores one essential quantum aspect where probabilities are inherently expressed in ranges instead of pinpoint precision.

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Most popular questions from this chapter

(II) For a particle in a box with rigid walls, determine whether our results for the ground state are consistent with the uncertainty principle by calculating the product \(\Delta p \Delta x\). Take \(\Delta x \approx \ell\), since the particle is somewhere within the box. For \(\Delta p,\) note that although \(p\) is known \((=\hbar k),\) the direction of \(\overrightarrow{\mathbf{p}}\) is not known, so the \(x\) component could vary from \(-p\) to \(+p ;\) hence take \(\Delta p \approx 2 p\)

(II) A 12 -g bullet leaves a ritle horizontally at a speed of 180 \(\mathrm{m} / \mathrm{s} .(a)\) What is the wavelength of this bullet? \((b)\) If the position of the bullet is known to a precision of 0.65 \(\mathrm{cm}\) (radius of the barrel), what is the minimum uncertainty in its vertical momentum?

(1I) An electron with 180 \(\mathrm{eV}\) of kinetic energy in free space passes over a finite potential well 56 \(\mathrm{eV}\) deep that stretches from \(x=0\) to \(x=0.50 \mathrm{nm}\) . What is the electron's wavelength (a) in free space, ( \(b\) ) when over the well? (c) Draw a diagram showing the potential energy and total energy as a function of \(x,\) and on the diagram sketch a possible wave function.

(II) An electron approaches a potential barrier \(18 \mathrm{eV}\) high and \(0.55 \mathrm{nm}\) wide. If the electron has a \(1.0 \%\) probability of tunneling through the barrier, what is the electron's energy?

(II) A proton in a nucleus can be roughly modeled as a particle in a box of nuclear dimensions. Calculate the energy released when a proton confined in a nucleus of width \(1.0 \times 10^{-14} \mathrm{~m}\) makes a transition from the first excited state to the ground state.

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