* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download The potential energy outside the nucleus is
Internal energy wikipedia , lookup
Time in physics wikipedia , lookup
Yang–Mills theory wikipedia , lookup
Conservation of energy wikipedia , lookup
Aharonov–Bohm effect wikipedia , lookup
Renormalization wikipedia , lookup
Gibbs free energy wikipedia , lookup
Old quantum theory wikipedia , lookup
Density of states wikipedia , lookup
Path integral formulation wikipedia , lookup
Introduction to gauge theory wikipedia , lookup
Electrostatics wikipedia , lookup
Potential energy wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Nuclear physics wikipedia , lookup
Atomic nucleus wikipedia , lookup
Nuclear structure wikipedia , lookup
Hydrogen atom wikipedia , lookup
Perturbation theory wikipedia , lookup
The potential energy outside the nucleus is: kq2 U r r r r0 Where k is the Coloumb constant. The charge inside the sphere as a function of the radius f the sphere is (from volume ratios): qr 3 qi 3 r0 Thus the electric field is: E kqi kqr 3 r2 r0 And the potential is: kq r02 r 2 kq r02 r 2 kq 0 kq V r E r dr 3 rdr V r 0 3 3 r0 r0 r 2r0 2r0 r r0 r Thus the potential energy is: kq2 U r qV r 2r0 r 2 3 r0 The Hamiltonian of the unperturbed hydrogen atom is: p 2 kq2 H0 2 r And the new Hamiltonian is: p 2 kq2 H 2 2r0 r 2 3 r0 So we can rewrite the perturbed Hamiltonian as: H H0 H H kq2 r 2 3kq2 kq2 2r0 r 2r03 So all that is left is to calculate the perturbation correction for the ground wave function: 100 1,0,0 R10Y00 Because the ground-state wave-function has only radial dependence we get: E01 100 H 100 r0 E0 Y00 , d R10 r H r r 2 dr 1 2 2 0 But normalization requires: Y , d 1 2 00 And: R10 r R10 r0 2 e a03 r a0 2 a03 e r0 a0 But since r0 a0 we can conclude that the radial wave function is practically constant for r<r0: R10 r0 2 a 03 e r0 a0 2 a 03 So the integral simply becomes: r0 R10 r H r r 2 dr 2 0 4kq 2 a 03 r 4 0 kq 2 r 2 3kq 2 kq 2 2 r dr 2r0 r a 03 0 2r03 r r0 1 r0 4 4kq 2 1 r05 3 0 2 3 r03 r02 r dr r dr rdr 3 0 a03 2r03 5 2r0 3 2 2r0 0 2r0 0 4kq 2 3 a0 r02 r02 r02 2q 2 r02 2 5a03 10 2 Thus the correction is: r0 E0 R10 r 1 0 2 2kq2 r02 H r r dr 5a03 2 But the ground-state energy of te unperturbed atom is: E0 kq 2 13.6eV 2a 0 Hence: 2 1 E0 2 10 15 2q 2 r02 4 r0 4 3.9 10 9 eV E 13 . 6 0 3 10 5 a0 5 5a 0 0 . 53 10 For a hollow spherical shell, the calculation becomes much simpler. Retracing our steps we see that inside the shell the total charge is zero. Hence, from Gauss’ law, the electric field is zero and the potential is constant. The potential inside the shell is equal to the potential on the surface of the shell: kq V r0 So the potential energy inside is: U qV kq2 r0 Thus the Hamiltonian inside is: H p 2 kq 2 2 r0 Comparing this to the unperturbed Hamiltonian: p 2 kq2 H0 2 r We see that the perturbation is: H H0 H 1 1 H kq2 r r0 So the correction now is (again, only the radial integral counts and we simplify the ground-state wave function): r0 E 0 R10 r 1 0 2 H r r 2 dr 4kq 2 0 1 1 2 r dr a 03 0 r r0 r r 4kq 2 0 r 2 4kq 2 r 2 r 2 4kq 2 r 2 3 r dr 3 0 0 3 0 r0 3 a0 0 a0 2 a0 6 And: kq 2 E0 13.6eV 2a 0 Thus: 2 1 E0 2 10 15 8 r0 8 6.5 10 9 eV E 0 13.6 10 6 a0 6 0.53 10 The ratio of the muon’s mass to the electrons mass is: MeV m c 2 214 MeV me 0.51 2 c Bohr’s radius for the hydrogen is given by: 109 a0 2 kme q n q e Where qn and qe are the charge of the nucleus and the electron respectively (in the hydrogen atom they are obviously the same). Thus for a lead nucleus – muon system, the modified Bohr radius (that effects the ground-state wave function) is: a0 2 k 214me 82qe qe O a0 6 6 57 10 a 30 10 A 0 17.5 10 3 On the other hand, the radius of the nucleus is given by: r0 A 1 3 O 12 10 6 A O r0 71 10 6 A Therefore the spherical shell radius is twice as large as the orbital radius of the ground state. Thus our assumption that the wave function inside the nucleus is constant is no longer valid and we need to take into account the exponential term as well. R10 r 2 a 0 3 e r a0 So the integral for the perturbation is: r0 E 0 R10 r 1 0 2r 4kq 2 0 1 1 a0 2 H r r dr e r dr a 0 3 0 r r0 r 2 2 Which simplifies: 1 E0 4k 82q 2 a0 3 r0 re 0 2r a0 2r r 2 a0 dr e dr r 0 0 r0 Integrating by parts (or using integral tables) yields: x n e x dx 1 x n ex n x n 1 x e dx Thus: r0 re 2r a0 0 r0 2r 2r 2r 2r r a0 0 a0 a a0 a0 a0 a0 a dr e dr re e re 0 0 2r 0 2r 2r 4 0 2r 2r 0 0 a 2r e a0 ae a0 0 0 And by the same token: a0 a r 2 a0 0 r0 e dr 4r0 e 0 r0 2 r0 2r 2 2ar0 a e 0 2r 2 2a r a 2 0 0 0 0 0 So together: 1 E0 2r 2r 2r r r0 2 2 r 0 0 4k 82q 2 a 2 4k 82q 2 0 a0 r a0 a0 0 r0 r0 e a0 a0 e a0 re dr e dr 3 3 r r0 a0 a0 0 0 0 Using the fact that a0 0.36r0 the term in the exponential is a very negative number, thus the exponential terms disappear leaving us with: E01 4k 82q 2 0.16a0 2 a03 E01 1.28 k 82q 2 2a0 The ground state energy of this system (analogous to the hydrogen atom) is: E0 k 82q 2 2a0 Thus the correction due to the perturbation is larger than the unperturbed state. Thus the first order perturbation theory is totally inadequate to this case. In the first two cases the perturbation corrections were 10 orders of magnitude smaller that the non-perturbed energy, so un these cases the first order perturbation theory is applicable