A particle of mass \(m\) is confined to a box of width L. If the particle is in the first excited state, what are the probabilities of finding the particle in a region of width \(0.020 L\) around the given point \(x:\) (a) \(x=0.25 L ;\) (b) \(x=0.40 L ;(\mathrm{c}) x=0.75 L ;\) and (d) \(x=0.90 L\)

For a particle in an infinite potential well of width L, the wave function for the first excited state (n = 2) is given by: \[\psi_2(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{2 \pi x}{L}\right)\]

The probability density function is the square of the wave function: \[P_2(x) = \left|\psi_2(x)\right|^2 = \frac{2}{L} \sin^2 \left(\frac{2 \pi x}{L}\right)\]

To find the probability of the particle being in a region of width 0.020L around the given points, we need to integrate the probability density function from the lower bound (x - 0.010L) to the upper bound (x + 0.010L). This will be done for each case: (a) x = 0.25L \[P(x) = \int_{0.24L}^{0.26L} \frac{2}{L} \sin^2 \left(\frac{2\pi x}{L}\right)dx\] (b) x = 0.40L \[P(x) = \int_{0.39L}^{0.41L} \frac{2}{L} \sin^2 \left(\frac{2\pi x}{L}\right)dx\] (c) x = 0.75L \[P(x) = \int_{0.74L}^{0.76L} \frac{2}{L} \sin^2 \left(\frac{2\pi x}{L}\right)dx\] (d) x = 0.90L \[P(x) = \int_{0.89L}^{0.91L} \frac{2}{L} \sin^2 \left(\frac{2\pi x}{L}\right)dx\]

Now, we need to evaluate these integrals. We can substitute \(u = \frac{2\pi x}{L}\) and compute the integrals: (a) x = 0.25L \[P(x) = \int_{0.48\pi}^{0.52\pi} \frac{1}{\pi} \sin^2 (u)du\] (b) x = 0.40L \[P(x) = \int_{0.78\pi}^{0.82\pi} \frac{1}{\pi} \sin^2 (u)du\] (c) x = 0.75L \[P(x) = \int_{1.48\pi}^{1.52\pi} \frac{1}{\pi} \sin^2 (u)du\] (d) x = 0.90L \[P(x) = \int_{1.78\pi}^{1.82\pi} \frac{1}{\pi} \sin^2 (u)du\] After evaluating these integrals, we get: (a) P(0.25L) ≈ 0.0157 (b) P(0.40L) ≈ 0.0199 (c) P(0.75L) ≈ 0.0157 (d) P(0.90L) ≈ 0.0199 The probabilities of finding the particle in the regions specified around the points x = 0.25L, x = 0.40L, x = 0.75L, and x = 0.90L are approximately 1.57%, 1.99%, 1.57%, and 1.99% respectively.