The energy difference between n = 1 and n = 2 states if the length of the box is 3 fm .

Question

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Answer 1

Question

Chapter 28, Problem 37E

(a)

To determine

The energy difference between n=1 and n=2 states if the length of the box is 3 fm.

(a)

Expert Solution

Answer to Problem 37E

The energy difference is 68.4 MeV.

Explanation of Solution

Write the expression for the energy of the nth state of a particle confined in a box.

En = n2h28ma2 (I)

Here, En is the energy in the nth state, h is Planck’s constant, m is the mass of the particle and a is the dimension of the box.

Conclusion:

Substitute 2 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 3 fm for a in equation (I).

E2 = 22(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(3fm)2(1fm)2(10 −15 m)2= 1.46 × 10−11 J(1 eV1.6 × 10−19 J)= 9.12 × 107 eV= 91.2 MeV

Substitute 1 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 3 fm for a in expression (I).

E1 = 12(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(3fm)2(1fm)2(10 −15 m)2= 3.65 × 10−12 J(1 eV1.6 × 10−19 J)= 2.28 × 107 eV= 22.8 MeV

Evaluate the difference between E2 and E1.

E2 − E1 = (91.2 − 22.8) MeV= 68.4 MeV

Thus, the energy difference is 68.4 MeV.

(b)

To determine

The energy difference between n=1 and n=2 states if the length of the box is 8 fm.

(b)

Expert Solution

Answer to Problem 37E

The energy difference is 9.59 MeV.

Explanation of Solution

Write the expression for the energy of the nth state of a particle confined in a box.

En = n2h28ma2

Here, En is the energy in the nth state, h is Planck’s constant, m is the mass of the particle and a is the dimension of the box.

Conclusion:

Substitute 2 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 8 fm for a in expression (I).

E2 = 22(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(8fm)2(1fm)2(10 −15 m)2= 2.05 × 10−12 J(1 eV1.6 × 10−19 J)= 1.28 × 107 eV= 12.8 MeV

Substitute 1 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 8 fm for a in expression (I).

E1 = 12(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(8fm)2(1fm)2(10 −15 m)2= 5.13 × 10−13 J(1 eV1.6 × 10−19 J)= 3.21 × 106 eV= 3.21 MeV

Evaluate the difference between E2 and E1.

E2 − E1 = (12.8 − 3.21) MeV= 9.59 MeV

Thus, the energy difference is 9.59 MeV.

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Answer 2

Question

Chapter 28, Problem 37E

(a)

To determine

The energy difference between n=1 and n=2 states if the length of the box is 3 fm.

(a)

Expert Solution

Answer to Problem 37E

The energy difference is 68.4 MeV.

Explanation of Solution

Write the expression for the energy of the nth state of a particle confined in a box.

En = n2h28ma2 (I)

Here, En is the energy in the nth state, h is Planck’s constant, m is the mass of the particle and a is the dimension of the box.

Conclusion:

Substitute 2 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 3 fm for a in equation (I).

E2 = 22(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(3fm)2(1fm)2(10 −15 m)2= 1.46 × 10−11 J(1 eV1.6 × 10−19 J)= 9.12 × 107 eV= 91.2 MeV

Substitute 1 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 3 fm for a in expression (I).

E1 = 12(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(3fm)2(1fm)2(10 −15 m)2= 3.65 × 10−12 J(1 eV1.6 × 10−19 J)= 2.28 × 107 eV= 22.8 MeV

Evaluate the difference between E2 and E1.

E2 − E1 = (91.2 − 22.8) MeV= 68.4 MeV

Thus, the energy difference is 68.4 MeV.

(b)

To determine

The energy difference between n=1 and n=2 states if the length of the box is 8 fm.

(b)

Expert Solution

Answer to Problem 37E

The energy difference is 9.59 MeV.

Explanation of Solution

Write the expression for the energy of the nth state of a particle confined in a box.

En = n2h28ma2

Here, En is the energy in the nth state, h is Planck’s constant, m is the mass of the particle and a is the dimension of the box.

Conclusion:

Substitute 2 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 8 fm for a in expression (I).

E2 = 22(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(8fm)2(1fm)2(10 −15 m)2= 2.05 × 10−12 J(1 eV1.6 × 10−19 J)= 1.28 × 107 eV= 12.8 MeV

Substitute 1 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 8 fm for a in expression (I).

E1 = 12(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(8fm)2(1fm)2(10 −15 m)2= 5.13 × 10−13 J(1 eV1.6 × 10−19 J)= 3.21 × 106 eV= 3.21 MeV

Evaluate the difference between E2 and E1.

E2 − E1 = (12.8 − 3.21) MeV= 9.59 MeV

Thus, the energy difference is 9.59 MeV.

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Answer 3

Question

Chapter 28, Problem 37E

(a)

To determine

The energy difference between n=1 and n=2 states if the length of the box is 3 fm.

(a)

Expert Solution

Answer to Problem 37E

The energy difference is 68.4 MeV.

Explanation of Solution

Write the expression for the energy of the nth state of a particle confined in a box.

En = n2h28ma2 (I)

Here, En is the energy in the nth state, h is Planck’s constant, m is the mass of the particle and a is the dimension of the box.

Conclusion:

Substitute 2 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 3 fm for a in equation (I).

E2 = 22(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(3fm)2(1fm)2(10 −15 m)2= 1.46 × 10−11 J(1 eV1.6 × 10−19 J)= 9.12 × 107 eV= 91.2 MeV

Substitute 1 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 3 fm for a in expression (I).

E1 = 12(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(3fm)2(1fm)2(10 −15 m)2= 3.65 × 10−12 J(1 eV1.6 × 10−19 J)= 2.28 × 107 eV= 22.8 MeV

Evaluate the difference between E2 and E1.

E2 − E1 = (91.2 − 22.8) MeV= 68.4 MeV

Thus, the energy difference is 68.4 MeV.

(b)

To determine

The energy difference between n=1 and n=2 states if the length of the box is 8 fm.

(b)

Expert Solution

Answer to Problem 37E

The energy difference is 9.59 MeV.

Explanation of Solution

Write the expression for the energy of the nth state of a particle confined in a box.

En = n2h28ma2

Here, En is the energy in the nth state, h is Planck’s constant, m is the mass of the particle and a is the dimension of the box.

Conclusion:

Substitute 2 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 8 fm for a in expression (I).

E2 = 22(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(8fm)2(1fm)2(10 −15 m)2= 2.05 × 10−12 J(1 eV1.6 × 10−19 J)= 1.28 × 107 eV= 12.8 MeV

Substitute 1 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 8 fm for a in expression (I).

E1 = 12(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(8fm)2(1fm)2(10 −15 m)2= 5.13 × 10−13 J(1 eV1.6 × 10−19 J)= 3.21 × 106 eV= 3.21 MeV

Evaluate the difference between E2 and E1.

E2 − E1 = (12.8 − 3.21) MeV= 9.59 MeV

Thus, the energy difference is 9.59 MeV.

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Answer 4

Question

Chapter 28, Problem 37E

(a)

To determine

The energy difference between n=1 and n=2 states if the length of the box is 3 fm.

(a)

Expert Solution

Answer to Problem 37E

The energy difference is 68.4 MeV.

Explanation of Solution

Write the expression for the energy of the nth state of a particle confined in a box.

En = n2h28ma2 (I)

Here, En is the energy in the nth state, h is Planck’s constant, m is the mass of the particle and a is the dimension of the box.

Conclusion:

Substitute 2 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 3 fm for a in equation (I).

E2 = 22(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(3fm)2(1fm)2(10 −15 m)2= 1.46 × 10−11 J(1 eV1.6 × 10−19 J)= 9.12 × 107 eV= 91.2 MeV

Substitute 1 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 3 fm for a in expression (I).

E1 = 12(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(3fm)2(1fm)2(10 −15 m)2= 3.65 × 10−12 J(1 eV1.6 × 10−19 J)= 2.28 × 107 eV= 22.8 MeV

Evaluate the difference between E2 and E1.

E2 − E1 = (91.2 − 22.8) MeV= 68.4 MeV

Thus, the energy difference is 68.4 MeV.

(b)

To determine

The energy difference between n=1 and n=2 states if the length of the box is 8 fm.

(b)

Expert Solution

Answer to Problem 37E

The energy difference is 9.59 MeV.

Explanation of Solution

Write the expression for the energy of the nth state of a particle confined in a box.

En = n2h28ma2

Here, En is the energy in the nth state, h is Planck’s constant, m is the mass of the particle and a is the dimension of the box.

Conclusion:

Substitute 2 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 8 fm for a in expression (I).

E2 = 22(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(8fm)2(1fm)2(10 −15 m)2= 2.05 × 10−12 J(1 eV1.6 × 10−19 J)= 1.28 × 107 eV= 12.8 MeV

Substitute 1 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 8 fm for a in expression (I).

E1 = 12(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(8fm)2(1fm)2(10 −15 m)2= 5.13 × 10−13 J(1 eV1.6 × 10−19 J)= 3.21 × 106 eV= 3.21 MeV

Evaluate the difference between E2 and E1.

E2 − E1 = (12.8 − 3.21) MeV= 9.59 MeV

Thus, the energy difference is 9.59 MeV.

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Answer 5

Chapter 28, Problem 37E

(a)

To determine

The energy difference between n=1 and n=2 states if the length of the box is 3 fm.

(a)

Expert Solution

Answer to Problem 37E

The energy difference is 68.4 MeV.

Explanation of Solution

Write the expression for the energy of the nth state of a particle confined in a box.

En = n2h28ma2 (I)

Here, En is the energy in the nth state, h is Planck’s constant, m is the mass of the particle and a is the dimension of the box.

Conclusion:

Substitute 2 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 3 fm for a in equation (I).

E2 = 22(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(3fm)2(1fm)2(10 −15 m)2= 1.46 × 10−11 J(1 eV1.6 × 10−19 J)= 9.12 × 107 eV= 91.2 MeV

Substitute 1 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 3 fm for a in expression (I).

E1 = 12(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(3fm)2(1fm)2(10 −15 m)2= 3.65 × 10−12 J(1 eV1.6 × 10−19 J)= 2.28 × 107 eV= 22.8 MeV

Evaluate the difference between E2 and E1.

E2 − E1 = (91.2 − 22.8) MeV= 68.4 MeV

Thus, the energy difference is 68.4 MeV.

(b)

To determine

The energy difference between n=1 and n=2 states if the length of the box is 8 fm.

(b)

Expert Solution

Answer to Problem 37E

The energy difference is 9.59 MeV.

Explanation of Solution

Write the expression for the energy of the nth state of a particle confined in a box.

En = n2h28ma2

Here, En is the energy in the nth state, h is Planck’s constant, m is the mass of the particle and a is the dimension of the box.

Conclusion:

Substitute 2 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 8 fm for a in expression (I).

E2 = 22(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(8fm)2(1fm)2(10 −15 m)2= 2.05 × 10−12 J(1 eV1.6 × 10−19 J)= 1.28 × 107 eV= 12.8 MeV

Substitute 1 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 8 fm for a in expression (I).

E1 = 12(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(8fm)2(1fm)2(10 −15 m)2= 5.13 × 10−13 J(1 eV1.6 × 10−19 J)= 3.21 × 106 eV= 3.21 MeV

Evaluate the difference between E2 and E1.

E2 − E1 = (12.8 − 3.21) MeV= 9.59 MeV

Thus, the energy difference is 9.59 MeV.

Answer 6

Chapter 28, Problem 37E

(a)

To determine

The energy difference between n=1 and n=2 states if the length of the box is 3 fm.

(a)

Expert Solution

Answer to Problem 37E

The energy difference is 68.4 MeV.

Explanation of Solution

Write the expression for the energy of the nth state of a particle confined in a box.

En = n2h28ma2 (I)

Here, En is the energy in the nth state, h is Planck’s constant, m is the mass of the particle and a is the dimension of the box.

Conclusion:

Substitute 2 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 3 fm for a in equation (I).

E2 = 22(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(3fm)2(1fm)2(10 −15 m)2= 1.46 × 10−11 J(1 eV1.6 × 10−19 J)= 9.12 × 107 eV= 91.2 MeV

Substitute 1 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 3 fm for a in expression (I).

E1 = 12(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(3fm)2(1fm)2(10 −15 m)2= 3.65 × 10−12 J(1 eV1.6 × 10−19 J)= 2.28 × 107 eV= 22.8 MeV

Evaluate the difference between E2 and E1.

E2 − E1 = (91.2 − 22.8) MeV= 68.4 MeV

Thus, the energy difference is 68.4 MeV.

Answer 7

Chapter 28, Problem 37E

(a)

To determine

The energy difference between n=1 and n=2 states if the length of the box is 3 fm.

Answer 8

(a)

Expert Solution

Answer to Problem 37E

The energy difference is 68.4 MeV.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the expression for the energy of the nth state of a particle confined in a box.

En = n2h28ma2 (I)

Here, En is the energy in the nth state, h is Planck’s constant, m is the mass of the particle and a is the dimension of the box.

Conclusion:

Substitute 2 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 3 fm for a in equation (I).

E2 = 22(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(3fm)2(1fm)2(10 −15 m)2= 1.46 × 10−11 J(1 eV1.6 × 10−19 J)= 9.12 × 107 eV= 91.2 MeV

Substitute 1 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 3 fm for a in expression (I).

E1 = 12(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(3fm)2(1fm)2(10 −15 m)2= 3.65 × 10−12 J(1 eV1.6 × 10−19 J)= 2.28 × 107 eV= 22.8 MeV

Evaluate the difference between E2 and E1.

E2 − E1 = (91.2 − 22.8) MeV= 68.4 MeV

Thus, the energy difference is 68.4 MeV.

Answer 13

(b)

To determine

The energy difference between n=1 and n=2 states if the length of the box is 8 fm.

(b)

Expert Solution

Answer to Problem 37E

The energy difference is 9.59 MeV.

Explanation of Solution

Write the expression for the energy of the nth state of a particle confined in a box.

En = n2h28ma2

Here, En is the energy in the nth state, h is Planck’s constant, m is the mass of the particle and a is the dimension of the box.

Conclusion:

Substitute 2 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 8 fm for a in expression (I).

E2 = 22(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(8fm)2(1fm)2(10 −15 m)2= 2.05 × 10−12 J(1 eV1.6 × 10−19 J)= 1.28 × 107 eV= 12.8 MeV

Substitute 1 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 8 fm for a in expression (I).

E1 = 12(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(8fm)2(1fm)2(10 −15 m)2= 5.13 × 10−13 J(1 eV1.6 × 10−19 J)= 3.21 × 106 eV= 3.21 MeV

Evaluate the difference between E2 and E1.

E2 − E1 = (12.8 − 3.21) MeV= 9.59 MeV

Thus, the energy difference is 9.59 MeV.

Answer 14

(b)

To determine

The energy difference between n=1 and n=2 states if the length of the box is 8 fm.

Answer 15

(b)

Expert Solution

Answer to Problem 37E

The energy difference is 9.59 MeV.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write the expression for the energy of the nth state of a particle confined in a box.

En = n2h28ma2

Here, En is the energy in the nth state, h is Planck’s constant, m is the mass of the particle and a is the dimension of the box.

Conclusion:

Substitute 2 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 8 fm for a in expression (I).

E2 = 22(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(8fm)2(1fm)2(10 −15 m)2= 2.05 × 10−12 J(1 eV1.6 × 10−19 J)= 1.28 × 107 eV= 12.8 MeV

Substitute 1 for n, 6.626 × 10−34 J⋅s for h, 1.67 × 10−27 kg for m and 8 fm for a in expression (I).

E1 = 12(6.626 × 10−34 J⋅s)28(1.67 × 10−27 kg)(8fm)2(1fm)2(10 −15 m)2= 5.13 × 10−13 J(1 eV1.6 × 10−19 J)= 3.21 × 106 eV= 3.21 MeV

Evaluate the difference between E2 and E1.

E2 − E1 = (12.8 − 3.21) MeV= 9.59 MeV

Thus, the energy difference is 9.59 MeV.

Answer 20

Ch. 28 - Prob. 1RQ Ch. 28 - Prob. 2RQ Ch. 28 - Prob. 3RQ Ch. 28 - Prob. 4RQ Ch. 28 - Prob. 5RQ Ch. 28 - Prob. 6RQ Ch. 28 - Prob. 7RQ Ch. 28 - Prob. 8RQ Ch. 28 - Prob. 9RQ Ch. 28 - Prob. 10RQ

The energy difference between n = 1 and n = 2 states if the length of the box is 3 fm .

The energy difference between n=1 and n=2 states if the length of the box is 3 fm.

Answer to Problem 37E

Explanation of Solution

The energy difference between n=1 and n=2 states if the length of the box is 8 fm.

Answer to Problem 37E

Explanation of Solution

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Chapter 28 Solutions