Explanation Given data: Length is 5 fm. n = 1 , 2 , 3 Formula used: The formula for energy is given by, E n = h 2 8 m L 2 n 2 (1) Here, h is the planks constant, which has a standard value as 6.63 × 10 − 34 J ⋅ s , m is mass of proton, which has a standard value as 1.67 × 10 − 27 Kg , L is the length of box, and n is the energy levels. Explanation: For n is 1, substitute 1.67 × 10 − 27 Kg for m , 6.63 × 10 − 34 J ⋅ s for h , 1 for n , and 5 fm for L in equation (1), E 1 = ( 6.63 × 10 − 34 J ⋅ s ) 2 8 × ( 1.67 × 10 − 27 Kg ) × ( 5 fm ) 2 × 1 2 = ( 6.63 × 10 − 34 J ⋅ s ) 2 8 × ( 1.67 × 10 − 27 Kg ) × ( 5 × 10 − 15 m ) 2 = 1.32 × 10 − 12 J For n is 2, substitute 1.67 × 10 − 27 Kg for m , 6.63 × 10 − 34 J ⋅ s for h , 2 for n , and 5 fm for L in equation (1), E 2 = ( 6.63 × 10 − 34 J ⋅ s ) 2 8 × ( 1.67 × 10 − 27 Kg ) × ( 5 fm ) 2 × 2 2 = ( 6

College Physics: A Strategic Approach (4th Edition)

4th Edition

ISBN: 9780134609034

Author: Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field

Publisher: PEARSON

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Chapter 28, Problem 45P

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The first three allowed energies of a photon in 5 fm long box.

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Chapter 28, Problem 44P

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Chapter 28 Solutions

College Physics: A Strategic Approach (4th Edition)

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The first three allowed energies of a photon in 5 fm long box.

Solution Summary: The author determines the first three allowed energies of a photon in 5 fm long box. The formula for energy is given by E_n=h28m

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The first three allowed energies of a photon in 5 fm long box.

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Chapter 28 Solutions