Fund Quantum Mechanics Lect & HW Solutions 62

Fund Quantum Mechanics Lect & HW Solutions 62 - 44...

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Unformatted text preview: 44 CHAPTER 4. SINGLE-PARTICLE SYSTEMS where a0 = 4πǫ0 h2 /me e2 , to find the ground state energy. Express in eV, where 1 eV equals ¯ 1.602,2 10−19 J. Values for the physical constants can be found at the start of this section and in the notations section. Answer: First verify the Bohr radius a0 = 4π 8.854 10−12 C2 /J m (1.054,6 10−34 J s)2 = 0.529,2 10−10 m 9.109 10−31 kg (1.602,2 10−19 C)2 Next, taking n = 1 for the ground state, E1 = − 1 1 eV (1.054,6 10−34 J s)2 = −2.179,9 10−18 J = −13.61eV −31 kg(0.529,2 10−10 m)2 12 2 9.109 10 1.602,2 10−19 J 4.2.3 Discussion of the eigenvalues 4.2.3.1 Solution hydc-a Question: If there are infinitely many energy levels E1 , E2 , E3 , E4 , E5 , E6 , . . ., where did they all go in the energy spectrum? Answer: The En for large values of n are all graphically indistinguishable from zero energy. The electron is almost free from the nucleus in those energy states. 4.2.3.2 Solution hydc-b Question: What is the value of energy level E2 ? And E3 ? Answer: No need to put all the numbers into h2 1 ¯ En = − 2me a2 n2 0 1 because the only difference between E2 and E1 is just a final factor 4 . So E2 = 1 E1 = −3.4 4 eV. Similarly, E3 = 1 E1 = −1.51 eV. 9 ...
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This note was uploaded on 01/06/2012 for the course PHY 3604 taught by Professor Dr.danielarenas during the Fall '11 term at UNF.

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