An electron is in an infinite square well of width \(a\) : \((U(x)=\infty\) for \(x<0\) and \(x>a\) ). If the electron is in the first excited state, \(\Psi(x)=A \sin (2 \pi x / a),\) at what position is the probability function a maximum? a) 0 b) \(a / 4\) c) \(a / 2\) d) \(3 a / 4\) and \(3 a / 4\) e) at both \(a / 4\)

First, we find the probability function by squaring the given wave function. The probability function \(P(x)\) can be written as: \(P(x) = |\Psi(x)|^2 = (A \sin(2\pi x / a))^2 = A^2\sin^2(2\pi x / a)\)

Now, we calculate the derivative of the probability function with respect to x: \(\frac{dP(x)}{dx} = \frac{d}{dx}(A^2\sin^2(2\pi x / a))\) Using the chain rule for differentiation, we can break down this derivative as follows: \(\frac{dP(x)}{dx} = 2A^2 \sin(2\pi x / a) \cdot \cos(2\pi x / a) \cdot \frac{2\pi}{a}\)

To find the maximum of the probability function, we set the derivative equal to zero: \(0 = 2A^2 \sin(2\pi x / a) \cdot \cos(2\pi x / a) \cdot \frac{2\pi}{a}\) First, we note that \(A^2\) and \(\frac{2\pi}{a}\) are constants, which do not affect the condition for the function maximum. So we can remove them from the equation: \(0 = \sin(2\pi x / a) \cdot \cos(2\pi x / a)\) Now, we need to find the values of x that make the equation true. One way to do this is by using the double-angle trigonometric identity for sine: \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\) Comparing this to our equation, we see that: \(\theta = \pi x / a\) Thus, we can rewrite our equation as: \(0 = 2\sin(\pi x / a) \cos(\pi x / a)\) For this equation to be true, either \(\sin(\pi x / a)\) or \(\cos(\pi x / a)\) must be equal to zero. Let's solve for x in both cases: Case 1: \(\sin(\pi x / a) = 0\) \(\pi x / a = n\pi\) \(x = na\), where n is an integer. Case 2: \(\cos(\pi x / a) = 0\) \(\pi x / a = (2n+1)\pi/2\) \(x = (2n+1)a/2\), where n is an integer. Now, we need to choose the x values from these two cases that are within the range of the infinite square well (0 < x < a). According to case 1, x can be equal to 0 or a, but these are not within the limit of the well. In case 2, for n=0, we have x = a/2 which is within the limit. So, we choose the x value accordingly.

Having analyzed both cases, we find that the probability function, \(P(x)\), is at a maximum when \(x = a / 2\). The correct answer is: c) \(a / 2\)