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Fund Quantum Mechanics Lect &amp; HW Solutions 42

Fund Quantum Mechanics Lect &amp; HW Solutions 42 - n...

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24 CHAPTER 3. BASIC IDEAS OF QUANTUM MECHANICS and see how big it is. Note that ¯ h = 1.054,57 10 34 J s. Express in units of eV, where one eV equals 1.602,18 10 19 J. Answer: E 1 = ¯ h 2 π 2 2 mℓ 2 x = (1 . 054 , 57 10 34 J s) 2 π 2 2 9 . 109 , 38 10 31 kg (2 10 10 m) 2 = 1 . 506 10 18 J or 9.4 eV. The true value is about 4.5 eV. That is in the ball park. 3.5.6.2 Solution pipee-b Question: Just for fun, plug macroscopic values, m = 1 kg and x = 1 m, into the expression for the ground state energy and see how big it is. Note that ¯ h = 1.054,57 10 34 J s. Answer: E 1 = ¯ h 2 π 2 2 mℓ 2 x = (1 . 054 , 57 10 34 J s) 2 π 2 2 1 kg 1 m 2 = 5 . 488 10 68 J or 3.4 10 49 eV. That energy is much less than you could ever hope to observe physically. A single photon of light would dwarf it by 50 orders of magnitude. 3.5.6.3 Solution pipee-c Question: What is the eigenfunction number, or quantum number,
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Unformatted text preview: n that produces a macro-scopic amount of energy, 1 J, for macroscopic values m = 1 kg and ℓ x = 1 m? With that many energy levels involved, would you see the diFerence between successive ones? Answer: Putting the generic expression for the eigenvalues, E n = n 2 ¯ h 2 π 2 2 mℓ 2 x equal to 1 J and plugging in the given numbers: n 2 (1 . 054 , 57 10 − 34 J s) 2 π 2 2 1 kg 1 m 2 = 1 J. Solving for n , you get n = 4.268,64 10 33 . Obviously, there is no way to distinguish that many energy levels. A calculator cannot even display all 34 digits of this number, even if you knew ¯ h to enough digits accuracy to compute 34 digits....
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