Questions: The solution to Schrodinger's wave equation for a particular situation is given by Ψ(x) = √(2/a) e^(-x/a0). Determine the probability of finding the particle between the following limits: (a) 0 ≤ x ≤ a0/4, (b) a0/4 ≤ x ≤ a0/2, (c) 0 ≤ x ≤ a0. Therefore (b) P = ∫ from (ao / 4) to (ao / 2) [(√(2/ao) exp (-x/ao))^2] dx = (2/ao) ∫ from (ao / 4) to (ao / 2) exp (-2x/ao) dx = (2/ao)((-ao)/2) exp (-2x/ao) from ao / 2 to ao / 4 or P = (-1)[exp (-1) - exp (-1/2)] which yields P = 0.239

The solution to Schrodinger's wave equation for a particular situation is given by Ψ(x) = √(2/a) e^(-x/a0). Determine the probability of finding the particle between the following limits: (a) 0 ≤ x ≤ a0/4, (b) a0/4 ≤ x ≤ a0/2, (c) 0 ≤ x ≤ a0. Therefore (b) P = ∫ from (ao / 4) to (ao / 2) [(√(2/ao) exp (-x/ao))^2] dx = (2/ao) ∫ from (ao / 4) to (ao / 2) exp (-2x/ao) dx = (2/ao)((-ao)/2) exp (-2x/ao) from ao / 2 to ao / 4 or P = (-1)[exp (-1) - exp (-1/2)] which yields P = 0.239

Transcript text: The solution to Schrodinger's wave equation for a particular situation is given by $\Psi(x)=\sqrt{\frac{2}{a}} e^{\frac{-x}{a_{0}}}$. Determine the probability of finding the particle between the following limits: (a) $0 \leq x \leq \frac{a_{0}}{4}$,(b) $\frac{a_{0}}{4} \leq x \leq \frac{a_{0}}{2}$,(c) $0 \leq x \leq a_{0}$ $\therefore$ (b) $P=\int_{a_{o} / 4}^{a_{o} / 2}\left[\sqrt{\frac{2}{a_{o}}} \exp \left(\frac{-x}{a_{o}}\right)\right]^{2} d x=\frac{2}{a_{o}} \int_{a_{o} / 4}^{a_{o} / 2} \exp \left(\frac{-2 x}{a_{o}}\right) d x=\left.\frac{2}{a_{o}}\left(\frac{-a_{o}}{2}\right) \exp \left(\frac{-2 x}{a_{o}}\right)\right|_{a_{o} / 2} ^{a_{o} / 4}$ or \[ P=(-1)\left[\exp (-1)-\exp \left(\frac{-1}{2}\right)\right] \] which yields \[ P=0.239 \]

Solution

Solution Steps

Step 1: Understand the Problem

We are given a wave function $\Psi(x) = \sqrt{\frac{2}{a_0}} e^{-\frac{x}{a_0}}$ and need to find the probability of a particle being within certain limits. The probability is calculated by integrating the square of the wave function over the given interval.

Step 2: Set Up the Integral for Probability

The probability $P$ of finding the particle between $x_1$ and $x_2$ is given by: \[ P = \int_{x_1}^{x_2} |\Psi(x)|^2 \, dx = \int_{x_1}^{x_2} \left(\sqrt{\frac{2}{a_0}} e^{-\frac{x}{a_0}}\right)^2 \, dx \] This simplifies to: \[ P = \frac{2}{a_0} \int_{x_1}^{x_2} e^{-\frac{2x}{a_0}} \, dx \]

Step 3: Solve for Each Interval

(a) Probability for $0 \leq x \leq \frac{a_0}{4}$

Set $x_1 = 0$ and $x_2 = \frac{a_0}{4}$: \[ P = \frac{2}{a_0} \int_{0}^{\frac{a_0}{4}} e^{-\frac{2x}{a_0}} \, dx \] The integral evaluates to: \[ P = \frac{2}{a_0} \left[ \frac{-a_0}{2} e^{-\frac{2x}{a_0}} \right]_{0}^{\frac{a_0}{4}} \] \[ = -\left[ e^{-\frac{1}{2}} - e^{0} \right] \] \[ = 1 - e^{-\frac{1}{2}} \] \[ \approx 0.3935 \]

(b) Probability for $\frac{a_0}{4} \leq x \leq \frac{a_0}{2}$

Set $x_1 = \frac{a_0}{4}$ and $x_2 = \frac{a_0}{2}$: \[ P = \frac{2}{a_0} \int_{\frac{a_0}{4}}^{\frac{a_0}{2}} e^{-\frac{2x}{a_0}} \, dx \] The integral evaluates to: \[ P = \frac{2}{a_0} \left[ \frac{-a_0}{2} e^{-\frac{2x}{a_0}} \right]_{\frac{a_0}{4}}^{\frac{a_0}{2}} \] \[ = -\left[ e^{-1} - e^{-\frac{1}{2}} \right] \] \[ = e^{-\frac{1}{2}} - e^{-1} \] \[ \approx 0.2394 \]

Set $x_1 = 0$ and $x_2 = a_0$: \[ P = \frac{2}{a_0} \int_{0}^{a_0} e^{-\frac{2x}{a_0}} \, dx \] The integral evaluates to: \[ P = \frac{2}{a_0} \left[ \frac{-a_0}{2} e^{-\frac{2x}{a_0}} \right]_{0}^{a_0} \] \[ = -\left[ e^{-2} - e^{0} \right] \] \[ = 1 - e^{-2} \] \[ \approx 0.8647 \]