To solve this problem, we consider a particle in a one-dimensional infinite potential box. The particle's wavefunction for the lowest energy level ($n = 1$) is given by:

$\psi_1(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{\pi x}{L}\right)$

where $L = 1 \, \text{nm}$ is the length of the box.

Step 1: Probability of Finding the Particle

The probability of finding the particle within a specific range $[x_1, x_2]$ is given by the integral:

$P(x_1, x_2) = \int_{x_1}^{x_2} |\psi_1(x)|^2 dx$

Here, the range of interest is within $0.01 \, \text{nm}$ of the center of the box, meaning the range is $x_1 = 0.495 \, \text{nm}$ and $x_2 = 0.505 \, \text{nm}$.

Step 2: Substituting the Wavefunction

The probability becomes:

$P(0.495, 0.505) = \int_{0.495}^{0.505} \frac{2}{L} \sin^2\left(\frac{\pi x}{L}\right) dx$

Since $L = 1 \, \text{nm}$, this simplifies to:

$P(0.495, 0.505) = 2 \int_{0.495}^{0.505} \sin^2(\pi x) dx$

Using the trigonometric identity $\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$, the integral becomes:

$P(0.495, 0.505) = \int_{0.495}^{0.505} \left(1 - \cos(2\pi x)\right) dx$

Step 3: Compute the Integral

Splitting the integral:

$P(0.495, 0.505) = \int_{0.495}^{0.505} 1 \, dx - \int_{0.495}^{0.505} \cos(2\pi x) dx$

1. First term:

$\int_{0.495}^{0.505} 1 \, dx = 0.505 - 0.495 = 0.01$

2. Second term:

$\int_{0.495}^{0.505} \cos(2\pi x) dx = \frac{\sin(2\pi x)}{2\pi} \bigg|_{0.495}^{0.505}$

Evaluate the sine terms:

$\sin(2\pi \cdot 0.505) = \sin(1.01\pi) = -\sin(0.01\pi)$ $\sin(2\pi \cdot 0.495) = \sin(0.99\pi) = \sin(0.01\pi)$

Thus:

$\int_{0.495}^{0.505} \cos(2\pi x) dx = \frac{-\sin(0.01\pi) - \sin(0.01\pi)}{2\pi} = \frac{-2\sin(0.01\pi)}{2\pi} = \frac{-\sin(0.01\pi)}{\pi}$

For small angles, $\sin(\theta) \approx \theta$, so $\sin(0.01\pi) \approx 0.01\pi$. Therefore:

$\int_{0.495}^{0.505} \cos(2\pi x) dx \approx \frac{-0.01\pi}{\pi} = -0.01$

Step 4: Combine Results

Now, substitute back into the probability expression:

$P(0.495, 0.505) = 0.01 - (-0.01) = 0.02$

Final Answer:

The probability of finding the particle within $0.01 \, \text{nm}$ of the center of the box for the lowest-energy level is approximately:

$P \approx 0.02 \, \text{or} \, 2\%.$

Let me know if you'd like detailed clarifications or further assistance! Here are five related questions to deepen your understanding:

1. What is the probability density of the particle at the center of the box?
2. How does the probability change for higher energy levels ($n > 1$)?
3. Why is the wavefunction sinusoidal for a particle in a box?
4. What physical principles explain why the particle cannot be outside the box?
5. How does the size of the box affect the probability distribution?

Tip: Always verify small-angle approximations when simplifying trigonometric expressions in quantum mechanics problems.