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Chapter 30, Problem 43P

(a)

To determine

The radius of the 612C nucleus.

(a)

Expert Solution
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Answer to Problem 43P

The radius of the 612C nucleus is 2.7fm_.

Explanation of Solution

Write the expression for the radius of the 612C nucleus.

    r=aA1/3        (I)

Here, r is the radius of the 612C nucleus, A is the mass number, a is a constant.

Conclusion:

Substitute 1.2fm for a, 12 for A in equation (I) to find r.

    r=(1.2fm×1015m1fm)(12)1/3=2.7×1015m=2.7fm

Therefore, the radius of the 612C nucleus is 2.7fm_.

(b)

To determine

The force of repulsion between a proton at the surface of an 612C nucleus and the remaining five protons.

(b)

Expert Solution
Check Mark

Answer to Problem 43P

The force of repulsion between a proton at the surface of an 612C nucleus and the remaining five protons 1.5×102N_.

Explanation of Solution

Write the expression for the force of repulsion between a proton at the surface of an 612C nucleus and the remaining five protons.

    F=ke(Z1)e(1e)r2=ke(Z1)e2r2        (II)

Here, F is the force of repulsion between a proton at the surface of an 612C nucleus and the remaining five protons, ke is the columbic constant, r is the radius of the nucleus, Z is the number of protons, e is the electronic charge.

Conclusion:

Substitute 8.99×109Nm2/C2 for ke, 6 for Z, 2.7fm for r, 1.60×1019C for e in equation (II) to find F.

    F=(8.99×109Nm2/C2)(61)(1.60×1019C)2(2.7fm×1015m1fm)2=1.5×102N

Therefore, the force of repulsion between a proton at the surface of an 612C nucleus and the remaining five protons 1.5×102N_.

(c)

To determine

The work done to overcome the last proton from a large distance up to the surface of the nucleus.

(c)

Expert Solution
Check Mark

Answer to Problem 43P

The work done to overcome the last proton from a large distance up to the surface of the nucleus is 2.6MeV_.

Explanation of Solution

Write the expression for the work done to overcome the last proton from a large distance up to the surface of the nucleus.

    U=ke(Z1)e(1e)r=ke(Z1)e2r        (III)

Conclusion:

Substitute 8.99×109Nm2/C2 for ke, 6 for Z, 2.7fm for r, 1.60×1019C for e in equation (II) to find U.

    U=(8.99×109Nm2/C2)(61)(1.60×1019C)2(2.7fm×1015m1fm)=4.2×1013JU(in MeV)=(4.2×1013J×1MeV1.60×1013J)=2.6MeV

Therefore, the work done to overcome the last proton from a large distance up to the surface of the nucleus is 2.6MeV_.

(d)

To determine

The radius of 92238U nucleus, the force between a proton at the surface of 92238U and the remaining protons, the work done to overcome the last proton from a large distance up to the surface of the nucleus.

(d)

Expert Solution
Check Mark

Answer to Problem 43P

The radius of 92238U nucleus is 7.4fm_, the force between a proton at the surface of 92238U and the remaining protons is 3.8×102N_, the work done to overcome the last proton from a large distance up to the surface of the nucleus is 18MeV_.

Explanation of Solution

Use equation (I) to solve for the radius of 92238U nucleus.

Use equation (II) to solve for force between a proton at the surface of 92238U and the remaining protons.

Use equation (III) the work done to overcome the last proton from a large distance up to the surface of the nucleus.

Conclusion:

Substitute 1.2fm for a, 238 for A in equation (I) to find r.

    r=(1.2fm×1015m1fm)(238)1/3=7.4×1015m=7.4fm

Substitute 8.99×109Nm2/C2 for ke, 92 for Z, 7.4fm for r, 1.60×1019C for e in equation (II) to find F.

    F=(8.99×109Nm2/C2)(921)(1.60×1019C)2(7.4fm×1015m1fm)2=3.8×102N

Substitute 8.99×109Nm2/C2 for ke, 92 for Z, 7.4fm for r, 1.60×1019C for e in equation (II) to find U.

    U=(8.99×109Nm2/C2)(921)(1.60×1019C)2(7.4fm×1015m1fm)=2.8×1012JU(in MeV)=(2.8×1012J×1MeV1.60×1013J)=18MeV

Therefore, the radius of 92238U nucleus is 7.4fm_, the force between a proton at the surface of an 92238U and the remaining protons is 3.8×102N_, the work done to overcome the last proton from a large distance up to the surface of the nucleus is 18MeV_.

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Chapter 30 Solutions

Bundle: Principles of Physics: A Calculus-Based Text, 5th + WebAssign Printed Access Card for Serway/Jewett's Principles of Physics: A Calculus-Based Text, 5th Edition, Multi-Term

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