The correction to the ground state energy of hydrogen due to the finite size of the nucleus.

Question

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Answer 1

Question

Chapter 7, Problem 7.33P

To determine

The correction to the ground state energy of hydrogen due to the finite size of the nucleus.

Expert Solution & Answer

Answer to Problem 7.33P

The correction to the ground state energy of hydrogen due to the finite size of the nucleus is $10−10E1$ .

Explanation of Solution

Using perturbation theory,

$H'=e28πε0(1b−1r) (0<r<b)$ (I)

And the energy correction is

$ΔE=〈ψ|H'|ψ〉$ (II)

The wave function of the ground state is

$ψ≡1πa3e−r/a$ (III)

Where $a=4πε0ℏ2me2$ .

Solving equation (II),

$ΔE=e28πε01πa3(4π)∫ab(1b−1r)e−r/ar2dr=e2πε0a3[1b∫abe−2r/ar2dr−∫abe−2r/ardr]=e2πε0a3[1b[−a2r2e−2r/a+a(a2)2e−2r/a(−2ra−1)]−[(a2)2e−2r/a(−2ra−1)]]0b=e2πε0a3[[−a2bb2e−2b/a+a34be−2b/a(−2ba−1)]−[(a2)2e−2b/a(−2ba−1)+a34b−a24]]$

Solving further,

$ΔE=e2πε0a3[e−2b/a(−ab2−a22−a34b+ab2+a24)+a24(ab−1)]=e2πε0a3[e−2b/a(−a24)(ab+1)+a24(ab−1)]=e24πε0a[e−2b/a(ab+1)+(1−ab)]$

Let $2ba=ε≪1$ , then $e−b/a≈1−ε+ε22−ε36+...$

Therefore, the above equation becomes,

$ΔE=e24πε0a[(1−2ε)+(a+2ε)(1−ε+ε22−ε36+..)]$

Solving the above equation,

$ΔE=e24πε0a2b23a2$

Since, the ground state energy of the hydrogen is

$E=E1=me42(4πε0)2ℏ2$

Where $a=4πε0ℏ2me2$ .

Therefore,

$Ea=−e22(4πε0)$

Keeping only the leading term,

$ΔEE=e24πε0(−2(4πε0)e2)2b23a2=−43(ba)2$

Substitute $a=5×10−11 m$ and $b=10−15 m$ in the above equation,

$ΔEE=−43(10−15 m5×10−11 m)2=−163×10−10≈−5×10−10$

The energy correction for the fine structure is $5×10−5$ and for the hyperfine structure is $3×10−8$ .

Therefore, the estimated correction to the ground state energy of hydrogen due to the finite size of the nucleus is very small compared to the fine structure and the hyperfine structure.

Conclusion:

Thus, the correction to the ground state energy of hydrogen due to the finite size of the nucleus is $10−10E1$ .

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Answer 2

Question

Chapter 7, Problem 7.33P

To determine

The correction to the ground state energy of hydrogen due to the finite size of the nucleus.

Expert Solution & Answer

Answer to Problem 7.33P

The correction to the ground state energy of hydrogen due to the finite size of the nucleus is $10−10E1$ .

Explanation of Solution

Using perturbation theory,

$H'=e28πε0(1b−1r) (0<r<b)$ (I)

And the energy correction is

$ΔE=〈ψ|H'|ψ〉$ (II)

The wave function of the ground state is

$ψ≡1πa3e−r/a$ (III)

Where $a=4πε0ℏ2me2$ .

Solving equation (II),

$ΔE=e28πε01πa3(4π)∫ab(1b−1r)e−r/ar2dr=e2πε0a3[1b∫abe−2r/ar2dr−∫abe−2r/ardr]=e2πε0a3[1b[−a2r2e−2r/a+a(a2)2e−2r/a(−2ra−1)]−[(a2)2e−2r/a(−2ra−1)]]0b=e2πε0a3[[−a2bb2e−2b/a+a34be−2b/a(−2ba−1)]−[(a2)2e−2b/a(−2ba−1)+a34b−a24]]$

Solving further,

$ΔE=e2πε0a3[e−2b/a(−ab2−a22−a34b+ab2+a24)+a24(ab−1)]=e2πε0a3[e−2b/a(−a24)(ab+1)+a24(ab−1)]=e24πε0a[e−2b/a(ab+1)+(1−ab)]$

Let $2ba=ε≪1$ , then $e−b/a≈1−ε+ε22−ε36+...$

Therefore, the above equation becomes,

$ΔE=e24πε0a[(1−2ε)+(a+2ε)(1−ε+ε22−ε36+..)]$

Solving the above equation,

$ΔE=e24πε0a2b23a2$

Since, the ground state energy of the hydrogen is

$E=E1=me42(4πε0)2ℏ2$

Where $a=4πε0ℏ2me2$ .

Therefore,

$Ea=−e22(4πε0)$

Keeping only the leading term,

$ΔEE=e24πε0(−2(4πε0)e2)2b23a2=−43(ba)2$

Substitute $a=5×10−11 m$ and $b=10−15 m$ in the above equation,

$ΔEE=−43(10−15 m5×10−11 m)2=−163×10−10≈−5×10−10$

The energy correction for the fine structure is $5×10−5$ and for the hyperfine structure is $3×10−8$ .

Therefore, the estimated correction to the ground state energy of hydrogen due to the finite size of the nucleus is very small compared to the fine structure and the hyperfine structure.

Conclusion:

Thus, the correction to the ground state energy of hydrogen due to the finite size of the nucleus is $10−10E1$ .

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Answer 3

Question

Chapter 7, Problem 7.33P

To determine

The correction to the ground state energy of hydrogen due to the finite size of the nucleus.

Expert Solution & Answer

Answer to Problem 7.33P

The correction to the ground state energy of hydrogen due to the finite size of the nucleus is $10−10E1$ .

Explanation of Solution

Using perturbation theory,

$H'=e28πε0(1b−1r) (0<r<b)$ (I)

And the energy correction is

$ΔE=〈ψ|H'|ψ〉$ (II)

The wave function of the ground state is

$ψ≡1πa3e−r/a$ (III)

Where $a=4πε0ℏ2me2$ .

Solving equation (II),

$ΔE=e28πε01πa3(4π)∫ab(1b−1r)e−r/ar2dr=e2πε0a3[1b∫abe−2r/ar2dr−∫abe−2r/ardr]=e2πε0a3[1b[−a2r2e−2r/a+a(a2)2e−2r/a(−2ra−1)]−[(a2)2e−2r/a(−2ra−1)]]0b=e2πε0a3[[−a2bb2e−2b/a+a34be−2b/a(−2ba−1)]−[(a2)2e−2b/a(−2ba−1)+a34b−a24]]$

Solving further,

$ΔE=e2πε0a3[e−2b/a(−ab2−a22−a34b+ab2+a24)+a24(ab−1)]=e2πε0a3[e−2b/a(−a24)(ab+1)+a24(ab−1)]=e24πε0a[e−2b/a(ab+1)+(1−ab)]$

Let $2ba=ε≪1$ , then $e−b/a≈1−ε+ε22−ε36+...$

Therefore, the above equation becomes,

$ΔE=e24πε0a[(1−2ε)+(a+2ε)(1−ε+ε22−ε36+..)]$

Solving the above equation,

$ΔE=e24πε0a2b23a2$

Since, the ground state energy of the hydrogen is

$E=E1=me42(4πε0)2ℏ2$

Where $a=4πε0ℏ2me2$ .

Therefore,

$Ea=−e22(4πε0)$

Keeping only the leading term,

$ΔEE=e24πε0(−2(4πε0)e2)2b23a2=−43(ba)2$

Substitute $a=5×10−11 m$ and $b=10−15 m$ in the above equation,

$ΔEE=−43(10−15 m5×10−11 m)2=−163×10−10≈−5×10−10$

The energy correction for the fine structure is $5×10−5$ and for the hyperfine structure is $3×10−8$ .

Therefore, the estimated correction to the ground state energy of hydrogen due to the finite size of the nucleus is very small compared to the fine structure and the hyperfine structure.

Conclusion:

Thus, the correction to the ground state energy of hydrogen due to the finite size of the nucleus is $10−10E1$ .

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Answer 4

Question

Chapter 7, Problem 7.33P

To determine

The correction to the ground state energy of hydrogen due to the finite size of the nucleus.

Expert Solution & Answer

Answer to Problem 7.33P

The correction to the ground state energy of hydrogen due to the finite size of the nucleus is $10−10E1$ .

Explanation of Solution

Using perturbation theory,

$H'=e28πε0(1b−1r) (0<r<b)$ (I)

And the energy correction is

$ΔE=〈ψ|H'|ψ〉$ (II)

The wave function of the ground state is

$ψ≡1πa3e−r/a$ (III)

Where $a=4πε0ℏ2me2$ .

Solving equation (II),

$ΔE=e28πε01πa3(4π)∫ab(1b−1r)e−r/ar2dr=e2πε0a3[1b∫abe−2r/ar2dr−∫abe−2r/ardr]=e2πε0a3[1b[−a2r2e−2r/a+a(a2)2e−2r/a(−2ra−1)]−[(a2)2e−2r/a(−2ra−1)]]0b=e2πε0a3[[−a2bb2e−2b/a+a34be−2b/a(−2ba−1)]−[(a2)2e−2b/a(−2ba−1)+a34b−a24]]$

Solving further,

$ΔE=e2πε0a3[e−2b/a(−ab2−a22−a34b+ab2+a24)+a24(ab−1)]=e2πε0a3[e−2b/a(−a24)(ab+1)+a24(ab−1)]=e24πε0a[e−2b/a(ab+1)+(1−ab)]$

Let $2ba=ε≪1$ , then $e−b/a≈1−ε+ε22−ε36+...$

Therefore, the above equation becomes,

$ΔE=e24πε0a[(1−2ε)+(a+2ε)(1−ε+ε22−ε36+..)]$

Solving the above equation,

$ΔE=e24πε0a2b23a2$

Since, the ground state energy of the hydrogen is

$E=E1=me42(4πε0)2ℏ2$

Where $a=4πε0ℏ2me2$ .

Therefore,

$Ea=−e22(4πε0)$

Keeping only the leading term,

$ΔEE=e24πε0(−2(4πε0)e2)2b23a2=−43(ba)2$

Substitute $a=5×10−11 m$ and $b=10−15 m$ in the above equation,

$ΔEE=−43(10−15 m5×10−11 m)2=−163×10−10≈−5×10−10$

The energy correction for the fine structure is $5×10−5$ and for the hyperfine structure is $3×10−8$ .

Therefore, the estimated correction to the ground state energy of hydrogen due to the finite size of the nucleus is very small compared to the fine structure and the hyperfine structure.

Conclusion:

Thus, the correction to the ground state energy of hydrogen due to the finite size of the nucleus is $10−10E1$ .

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Answer 5

Chapter 7, Problem 7.33P

To determine

The correction to the ground state energy of hydrogen due to the finite size of the nucleus.

Expert Solution & Answer

Answer to Problem 7.33P

The correction to the ground state energy of hydrogen due to the finite size of the nucleus is $10−10E1$ .

Explanation of Solution

Using perturbation theory,

$H'=e28πε0(1b−1r) (0<r<b)$ (I)

And the energy correction is

$ΔE=〈ψ|H'|ψ〉$ (II)

The wave function of the ground state is

$ψ≡1πa3e−r/a$ (III)

Where $a=4πε0ℏ2me2$ .

Solving equation (II),

$ΔE=e28πε01πa3(4π)∫ab(1b−1r)e−r/ar2dr=e2πε0a3[1b∫abe−2r/ar2dr−∫abe−2r/ardr]=e2πε0a3[1b[−a2r2e−2r/a+a(a2)2e−2r/a(−2ra−1)]−[(a2)2e−2r/a(−2ra−1)]]0b=e2πε0a3[[−a2bb2e−2b/a+a34be−2b/a(−2ba−1)]−[(a2)2e−2b/a(−2ba−1)+a34b−a24]]$

Solving further,

$ΔE=e2πε0a3[e−2b/a(−ab2−a22−a34b+ab2+a24)+a24(ab−1)]=e2πε0a3[e−2b/a(−a24)(ab+1)+a24(ab−1)]=e24πε0a[e−2b/a(ab+1)+(1−ab)]$

Let $2ba=ε≪1$ , then $e−b/a≈1−ε+ε22−ε36+...$

Therefore, the above equation becomes,

$ΔE=e24πε0a[(1−2ε)+(a+2ε)(1−ε+ε22−ε36+..)]$

Solving the above equation,

$ΔE=e24πε0a2b23a2$

Since, the ground state energy of the hydrogen is

$E=E1=me42(4πε0)2ℏ2$

Where $a=4πε0ℏ2me2$ .

Therefore,

$Ea=−e22(4πε0)$

Keeping only the leading term,

$ΔEE=e24πε0(−2(4πε0)e2)2b23a2=−43(ba)2$

Substitute $a=5×10−11 m$ and $b=10−15 m$ in the above equation,

$ΔEE=−43(10−15 m5×10−11 m)2=−163×10−10≈−5×10−10$

The energy correction for the fine structure is $5×10−5$ and for the hyperfine structure is $3×10−8$ .

Therefore, the estimated correction to the ground state energy of hydrogen due to the finite size of the nucleus is very small compared to the fine structure and the hyperfine structure.

Conclusion:

Thus, the correction to the ground state energy of hydrogen due to the finite size of the nucleus is $10−10E1$ .

Answer 6

Chapter 7, Problem 7.33P

To determine

The correction to the ground state energy of hydrogen due to the finite size of the nucleus.

Expert Solution & Answer

Answer to Problem 7.33P

The correction to the ground state energy of hydrogen due to the finite size of the nucleus is $10−10E1$ .

Explanation of Solution

Using perturbation theory,

$H'=e28πε0(1b−1r) (0<r<b)$ (I)

And the energy correction is

$ΔE=〈ψ|H'|ψ〉$ (II)

The wave function of the ground state is

$ψ≡1πa3e−r/a$ (III)

Where $a=4πε0ℏ2me2$ .

Solving equation (II),

$ΔE=e28πε01πa3(4π)∫ab(1b−1r)e−r/ar2dr=e2πε0a3[1b∫abe−2r/ar2dr−∫abe−2r/ardr]=e2πε0a3[1b[−a2r2e−2r/a+a(a2)2e−2r/a(−2ra−1)]−[(a2)2e−2r/a(−2ra−1)]]0b=e2πε0a3[[−a2bb2e−2b/a+a34be−2b/a(−2ba−1)]−[(a2)2e−2b/a(−2ba−1)+a34b−a24]]$

Solving further,

$ΔE=e2πε0a3[e−2b/a(−ab2−a22−a34b+ab2+a24)+a24(ab−1)]=e2πε0a3[e−2b/a(−a24)(ab+1)+a24(ab−1)]=e24πε0a[e−2b/a(ab+1)+(1−ab)]$

Let $2ba=ε≪1$ , then $e−b/a≈1−ε+ε22−ε36+...$

Therefore, the above equation becomes,

$ΔE=e24πε0a[(1−2ε)+(a+2ε)(1−ε+ε22−ε36+..)]$

Solving the above equation,

$ΔE=e24πε0a2b23a2$

Since, the ground state energy of the hydrogen is

$E=E1=me42(4πε0)2ℏ2$

Where $a=4πε0ℏ2me2$ .

Therefore,

$Ea=−e22(4πε0)$

Keeping only the leading term,

$ΔEE=e24πε0(−2(4πε0)e2)2b23a2=−43(ba)2$

Substitute $a=5×10−11 m$ and $b=10−15 m$ in the above equation,

$ΔEE=−43(10−15 m5×10−11 m)2=−163×10−10≈−5×10−10$

The energy correction for the fine structure is $5×10−5$ and for the hyperfine structure is $3×10−8$ .

Therefore, the estimated correction to the ground state energy of hydrogen due to the finite size of the nucleus is very small compared to the fine structure and the hyperfine structure.

Conclusion:

Thus, the correction to the ground state energy of hydrogen due to the finite size of the nucleus is $10−10E1$ .

Answer 7

Chapter 7, Problem 7.33P

To determine

The correction to the ground state energy of hydrogen due to the finite size of the nucleus.

Answer 8

Expert Solution & Answer

Answer to Problem 7.33P

The correction to the ground state energy of hydrogen due to the finite size of the nucleus is $10−10E1$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Using perturbation theory,

$H'=e28πε0(1b−1r) (0<r<b)$ (I)

And the energy correction is

$ΔE=〈ψ|H'|ψ〉$ (II)

The wave function of the ground state is

$ψ≡1πa3e−r/a$ (III)

Where $a=4πε0ℏ2me2$ .

Solving equation (II),

$ΔE=e28πε01πa3(4π)∫ab(1b−1r)e−r/ar2dr=e2πε0a3[1b∫abe−2r/ar2dr−∫abe−2r/ardr]=e2πε0a3[1b[−a2r2e−2r/a+a(a2)2e−2r/a(−2ra−1)]−[(a2)2e−2r/a(−2ra−1)]]0b=e2πε0a3[[−a2bb2e−2b/a+a34be−2b/a(−2ba−1)]−[(a2)2e−2b/a(−2ba−1)+a34b−a24]]$

Solving further,

$ΔE=e2πε0a3[e−2b/a(−ab2−a22−a34b+ab2+a24)+a24(ab−1)]=e2πε0a3[e−2b/a(−a24)(ab+1)+a24(ab−1)]=e24πε0a[e−2b/a(ab+1)+(1−ab)]$

Let $2ba=ε≪1$ , then $e−b/a≈1−ε+ε22−ε36+...$

Therefore, the above equation becomes,

$ΔE=e24πε0a[(1−2ε)+(a+2ε)(1−ε+ε22−ε36+..)]$

Solving the above equation,

$ΔE=e24πε0a2b23a2$

Since, the ground state energy of the hydrogen is

$E=E1=me42(4πε0)2ℏ2$

Where $a=4πε0ℏ2me2$ .

Therefore,

$Ea=−e22(4πε0)$

Keeping only the leading term,

$ΔEE=e24πε0(−2(4πε0)e2)2b23a2=−43(ba)2$

Substitute $a=5×10−11 m$ and $b=10−15 m$ in the above equation,

$ΔEE=−43(10−15 m5×10−11 m)2=−163×10−10≈−5×10−10$

The energy correction for the fine structure is $5×10−5$ and for the hyperfine structure is $3×10−8$ .

Therefore, the estimated correction to the ground state energy of hydrogen due to the finite size of the nucleus is very small compared to the fine structure and the hyperfine structure.

Conclusion:

Thus, the correction to the ground state energy of hydrogen due to the finite size of the nucleus is $10−10E1$ .

Answer 12

Ch. 7.1 - Prob. 7.1P Ch. 7.1 - Prob. 7.2P Ch. 7.1 - Prob. 7.3P Ch. 7.1 - Prob. 7.4P Ch. 7.1 - Prob. 7.5P Ch. 7.1 - Prob. 7.6P Ch. 7.2 - Prob. 7.8P Ch. 7.2 - Prob. 7.9P Ch. 7.2 - Prob. 7.10P Ch. 7.2 - Prob. 7.11P

The correction to the ground state energy of hydrogen due to the finite size of the nucleus.

The correction to the ground state energy of hydrogen due to the finite size of the nucleus.

Answer to Problem 7.33P

Explanation of Solution

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Chapter 7 Solutions