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Fund Quantum Mechanics Lect & HW Solutions 62

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

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44 CHAPTER 4. SINGLE-PARTICLE SYSTEMS where a 0 = 4 πǫ 0 ¯ h 2 /m e e 2 , 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 a 0 = 4 π 8 . 854 10 12 C 2 /J m (1 . 054 , 6 10 34 J s) 2 9 . 109 10 31 kg (1 . 602 , 2 10 19 C) 2 = 0 . 529 , 2 10 10 m Next, taking n = 1 for the ground state, E 1 = (1 . 054 , 6 10 34 J s) 2 2 9 . 109 10 31 kg(0 . 529 , 2 10 10 m) 2 1 1 2 = 2 . 179 , 9 10 18 J 1 eV 1 . 602 , 2 10 19 J = 13 . 61 eV 4.2.3 Discussion of the eigenvalues 4.2.3.1 Solution hydc-a Question: If there are infinitely many energy levels E 1 ,E 2 ,E 3 ,E 4 ,E 5 ,E 6 ,... , where did they
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