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Chapter 42, Problem 23P
To determine

To show that the energy is E=2π22meL2(nx2+ny2+nz2).

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An electron is in a three-dimensional box. The x- and z-sides of the box have the same length, but the y-side has a different length. The two lowest energy levels are 2.24 eV and 3.47 eV, and the degeneracy of each of these levels (including the degeneracy due to the electron spin) is two. (a) What are the nX, nY, and nZ quantum numbers for each of these two levels? (b) What are the lengths LX, LY, and LZ for each side of the box? (c) What are the energy, the quantum numbers, and the degeneracy (including the spin degeneracy) for the next higher energy state?
An electron is in a state for which / 3. One allowed value of L, is h. L is described as a classical vector. At this value, what angle does the vector L make with the +z-axis? O 30.0° O 90.0° O 125° O 73.2°
For a "particle in a box" of length, L, the wavelength for the nth level is given by An 2L %3D 2п and the wave function is n(x) = A sin (x) = A sin (x). The energy levels are пп %3D n?h? given by En : %3D 8mL2 lPn(x)|2 is the probability of finding the particle at position x in the box. Since the particle must be somewhere in the box, the integral of this function over the length of the box must be equal to 1. This is the normalization condition and ensuring that this is the case is called “normalizing" the wave function. Find the value of A the amplitude of the wave function, that normalizes it. Write the normalized wave function for the nth state of the particle in a box.
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