Explanation Write the expression for energy of electron in two dimensional infinite square well. E n x n y = h 2 8 m L 2 ( n x 2 + n y 2 ) (I) Here, h is the plank constant, m is the mass of electron, L is the length of measured side in square well, n x is the energy level in x direction, n y is the energy level in y direction. Conclusion: Substitute 6.63 × 10 − 34 J ⋅ s for h , 9.11 × 10 − 31 kg for m , 1 for n x , 1 for n y , 1.25 nm for L in the Equation (I) to calculate E 11 . E 11 = ( 6.63 × 10 − 34 J ⋅ s ) 2 8 ( 9.11 × 10 − 31 kg ) ( 1.25 nm × 10 − 9 m 1 nm ) 2 ( 1 2 + 1 2 ) = 0.772 × 10 − 19 J = 0.772 × 10 − 19 J 1.6 × 10 − 19 C = 0.483 eV Substitute 6.63 × 10 − 34 J ⋅ s for h , 9.11 × 10 − 31 kg for m , 1 for n x , 2 for n y , 1.25 nm for L in the Equation (I) to calculate E 12 . E 12 = ( 6.63 × 10 − 34 J ⋅ s ) 2 8 ( 9.11 × 10 − 31 kg ) ( 1.25 nm × 10 − 9 m 1 nm ) 2 ( 1 2 + 2 2 ) = 1.93 × 10 − 19 J = 1.93 × 10 − 19 J 1.6 × 10 − 19 C = 1.21 eV Substitute 6.63 × 10 − 34 J ⋅ s for h , 9.11 × 10 − 31 kg for m , 2 for n x , 1 for n y , 1.25 nm for L in the Equation (I) to calculate E 21 . E 21 = ( 6.63 × 10 − 34 J ⋅ s ) 2 8 ( 9.11 × 10 − 31 kg ) ( 1.25 nm × 10 − 9 m 1 nm ) 2 ( 2 2 + 1 2 ) = 1.93 × 10 − 19 J = 1.93 × 10 − 19 J 1.6 × 10 − 19 C = 1.21 eV Substitute 6.63 × 10 − 34 J ⋅ s for h , 9.11 × 10 − 31 kg for m , 2 for n x , 2 for n y , 1.25 nm for L in the Equation (I) to calculate E 22 . E 22 = ( 6.63 × 10 − 34 J ⋅ s ) 2 8 ( 9.11 × 10 − 31 kg ) ( 1.25 nm × 10 − 9 m 1 nm ) 2 ( 2 2 + 2 2 ) = 3.09 × 10 − 19 J = 3.09 × 10 − 19 J 1.6 × 10 − 19 C = 1.93 eV Substitute 6.63 × 10 − 34 J ⋅ s for h , 9.11 × 10 − 31 kg for m , 2 for n x , 3 for n y , 1.25 nm for L in the Equation (I) to calculate E 23 . E 23 = ( 6.63 × 10 − 34 J ⋅ s ) 2 8 ( 9.11 × 10 − 31 kg ) ( 1.25 nm × 10 − 9 m 1 nm ) 2 ( 2 2 + 3 2 ) = 5.02 × 10 − 19 J = 5.02 × 10 − 19 J 1.6 × 10 − 19 C = 3

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ISBN: 9780134988566

Author: Wolfson, Richard

Publisher: Pearson Education,

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Chapter 35, Problem 46P

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The ten lowest energy levels for the electron.

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Chapter 35, Problem 45P

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Chapter 35 Solutions

Essential University Physics

Ch. 35.1 - Prob. 35.1GI Ch. 35.2 - Prob. 35.2GI Ch. 35.3 - Prob. 35.3GI Ch. 35.3 - Prob. 35.4GI Ch. 35.3 - Prob. 35.5GI Ch. 35.4 - Prob. 35.6GI Ch. 35 - Prob. 1FTD Ch. 35 - Prob. 2FTD Ch. 35 - Prob. 3FTD Ch. 35 - Prob. 4FTD

Ch. 35 - Prob. 5FTD Ch. 35 - Prob. 6FTD Ch. 35 - Prob. 7FTD Ch. 35 - What did Einstein mean by his re maxi, loosely...Ch. 35 - Prob. 9FTD Ch. 35 - Prob. 11E Ch. 35 - Prob. 12E Ch. 35 - Prob. 13E Ch. 35 - Prob. 14E Ch. 35 - Prob. 15E Ch. 35 - Prob. 16E Ch. 35 - Prob. 17E Ch. 35 - Prob. 18E Ch. 35 - Prob. 19E Ch. 35 - Prob. 20E Ch. 35 - Prob. 21E Ch. 35 - Prob. 22E Ch. 35 - Prob. 23E Ch. 35 - Prob. 24E Ch. 35 - Prob. 28E Ch. 35 - Prob. 29E Ch. 35 - Prob. 30E Ch. 35 - Prob. 31E Ch. 35 - Prob. 32E Ch. 35 - Prob. 33E Ch. 35 - Prob. 34E Ch. 35 - Prob. 35E Ch. 35 - Prob. 36P Ch. 35 - Prob. 37P Ch. 35 - Prob. 38P Ch. 35 - Prob. 39P Ch. 35 - Prob. 40P Ch. 35 - Prob. 41P Ch. 35 - Prob. 42P Ch. 35 - Prob. 43P Ch. 35 - Prob. 44P Ch. 35 - Prob. 45P Ch. 35 - Prob. 46P Ch. 35 - Prob. 47P Ch. 35 - Prob. 48P Ch. 35 - Prob. 49P Ch. 35 - Prob. 50P Ch. 35 - Prob. 51P Ch. 35 - Prob. 52P Ch. 35 - Prob. 53P Ch. 35 - Prob. 54P Ch. 35 - Prob. 55P Ch. 35 - Prob. 56P Ch. 35 - Prob. 57P Ch. 35 - Prob. 58P Ch. 35 - Prob. 59P Ch. 35 - Prob. 60P Ch. 35 - Prob. 61P Ch. 35 - Prob. 62PP Ch. 35 - Prob. 63PP Ch. 35 - Prob. 64PP Ch. 35 - Prob. 65PP

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