An electron is confined between \(x=0\) and \(x=L\). The wave function of the electron is \(\psi(x)=A \sin (2 \pi x / L)\). The wave function is zero for the regions \(x<0\) and \(x>L\) a) Determine the normalization constant \(A\). b) What is the probability of finding the electron in the region \(0 \leq x \leq L / 3 ?\)

To determine the normalization constant A, we need to ensure the wave function is normalized. The wave function is normalized if: $$\int_{-\infty}^{+\infty}|\psi(x)|^2 dx = 1$$ For our given wave function, only the region 0 to L matters since the function is zero elsewhere. We have: $$\int_{0}^{L}|A \sin(\frac{2\pi x}{L})|^2 dx = 1$$ Squaring the absolute value, we get: $$\int_{0}^{L}A^2 \sin^2(\frac{2\pi x}{L}) dx = 1$$

We now need to evaluate the left side of the equation: $$\int_{0}^{L} A^2 \sin^2(\frac{2\pi x}{L}) dx$$ To do this, we can use the trigonometric identity \(\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}\): $$\int_{0}^{L} A^2 \frac{1 - \cos(\frac{4\pi x}{L})}{2} dx$$ Next, we can break this integral into two parts: $$A^2 \int_{0}^{L} \frac{1}{2} dx - A^2 \int_{0}^{L} \frac{\cos(\frac{4\pi x}{L})}{2} dx$$

Now we evaluate the two integrals: 1) For the first integral, we have: $$A^2 \int_{0}^{L} \frac{1}{2} dx = A^2 \cdot \frac{1}{2} \cdot L$$ 2) For the second integral, we have: $$A^2 \int_{0}^{L} \frac{\cos(\frac{4\pi x}{L})}{2} dx = A^2 \cdot \frac{1}{2} \cdot \frac{L}{4\pi} \cdot \left[\sin(\frac{4\pi x}{L})\right]_0^L = 0$$ So, the left side of the normalization equation is equal to: $$A^2 \cdot \frac{1}{2} \cdot L$$

We have: $$A^2 \cdot \frac{1}{2} \cdot L = 1$$ Solving for A, we get: $$A = \sqrt\frac{2}{L}$$

Now that we have the normalization constant, we can find the probability of finding the electron in the given region: $$P(0 \leq x \leq \frac{L}{3}) = \int_{0}^{\frac{L}{3}} |\psi(x)|^2 dx = \int_{0}^{\frac{L}{3}} (\sqrt\frac{2}{L} \sin(\frac{2\pi x}{L}))^2 dx$$ This simplifies to: $$P(0 \leq x \leq \frac{L}{3}) = \int_{0}^{\frac{L}{3}} \frac{2}{L} \sin^2(\frac{2\pi x}{L}) dx$$ Using the trigonometric identity again, we get: $$P(0 \leq x \leq \frac{L}{3}) = \int_{0}^{\frac{L}{3}} \frac{1}{L} - \frac{\cos(\frac{4\pi x}{L})}{L} dx$$ So, evaluating the integrals, we get: $$P(0 \leq x \leq \frac{L}{3}) = \frac{1}{3} - \frac{3}{4\pi}\frac{\sin(\frac{4\pi x}{L})}{4\pi}\Big|_0^{\frac{L}{3}} = \frac{1}{3}$$ Therefore, the probability of finding the electron in the region \(0 \leq x \leq \frac{L}{3}\) is \(\frac{1}{3}\).