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43 Pages

### Phys 2D soln6

Course: PHYS 2D 2D, Spring 2011
School: UCSD
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Word Count: 367

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/r2 1 The term originates from the angular momentum of the electron moving through the Coulomb potential. Uef f diverges as r 0 because 1/r2 goes to zero faster than the Coulomb term. The boundary conditions on this wavefunction are that it must go to zero at r = 0 and r = . This is the lowest energy state for = 1 because it only goes to zero at 0 and and nowhere in between. The principle quantum number for this...

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/r2 1 The term originates from the angular momentum of the electron moving through the Coulomb potential. Uef f diverges as r 0 because 1/r2 goes to zero faster than the Coulomb term. The boundary conditions on this wavefunction are that it must go to zero at r = 0 and r = . This is the lowest energy state for = 1 because it only goes to zero at 0 and and nowhere in between. The principle quantum number for this state is n = 2. The probability density is symmetric about the z-axis and is invariant under rotations about the z-axis. See the figure below. 2 The allowed energies will all be multiples of E0 2 2 /2mL2 . The easiest thing to do is try different combinations of n2 + n2 + n2 /4 and see which 5 are the lowest. 1 2 3 E n1 9E0 /4 1 3E0 1 17E0 /4 1 1 21E0 /4 2 1 6E0 2 1 n2 1 1 1 2 1 1 1 2 n3 1 2 3 1 1 4 2 2 deg. 1 1 1 2 3 3 Use the expectations values, L2 = 2 ( + 1), Lz = m , and the constraint m = - , - + 1, . . . , + to determine if a particular combination of angular momentum is allowed. L2 = 0 and Lz = 0 is allowed with = 0, m = 0 L2 = 2 Lz and = is not allowed because is not an integer in this case. L2 = 2 2 and Lz = 2 is not allowed because m = 2 > = 1 L2 = 3 2 and Lz = 0 is not allowed because is not an integer in this case. L2 = 6 2 and Lz = 2 is allowed with = 2, m = 2 4 The Schrdinger equation for this system is o 2 - 2m 2 R 2 R + r2 r r 2 + ( + 1) g - 2mr2 r R = ER with R = Ar exp(-r). To solve for E first calculate the derivatives of R R 1 = Ae-r (1 - r) = R - r r 2 2R = Ae-r (-2 + 2 r) = R - + 2 2 r r and plug them into the Schrdinger equation. o 2 - 2m 2 - 4 2 + 2 r r 2 R+ 1 ( + 1) g - 2mr2 r R = ER Notice that R now be eliminated. Match the terms is each side of the equation that have equal powers of r. For this equation to be true for a general r, these terms must be equivalent. - 2 + 2m r2 2 ( + 1) =0 2mr2 2 4 g - =0 2m r r 2 2 - From this we see 2m 2 = E =1 mg = 2 2 mg 2 E=- 2 8 2 Figure 1: The upper drawing a graph of the wavefunction as a function of the radius. The lower drawing is a sketch of the probability density for this state. 3
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