The radius of the C 6 12 nucleus.

Question

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Answer 1

Question

Chapter 13, Problem 10P

(a)

To determine

The radius of the C612 nucleus.

(a)

Expert Solution

Answer to Problem 10P

The radius of the C612 nucleus is 2.75×10−15 m .

Explanation of Solution

Write the equation for the nuclear radius.

R=R0A1/3 (I)

Here, R is the nuclear radius, R0 is constant with value 1.2×10−15 m and A is the mass number.

Conclusion:

The mass number of C612 is 12 .

Substitute 1.2×10−15 m for R0 and 12 for A in equation (I) to find R .

R=(1.2×10−15 m)(12)1/3=2.75×10−15 m

Therefore, the radius of the C612 nucleus is 2.75×10−15 m .

(b)

To determine

The force of repulsion between a proton at the surface of a C612 nucleus and the remaining five protons.

(b)

Expert Solution

Answer to Problem 10P

The force of repulsion between a proton at the surface of a C612 nucleus and the remaining five protons is 152 N .

Explanation of Solution

Write the equation for the Coulombic force.

F=kq1q2r2

Here, F is the force, k is the Coulomb constant, q1,q2 are the charges and r is the distance between the charges.

The charge of proton is +e and the charge o the remaining protons is (Z−1)e , where Z is the atomic number. The value of r will be equal to R .

Substitute +e for q1 , (Z−1)e for q2 and R for r in the above equation.

F=k(Z−1)e2R2 (II)

Here, e is the magnitude of the electronic charge.

Conclusion:

The value of k is 9×109 N⋅m2/C2 and the charge of the electron is 1.60×10−19 C . The atomic number of C612 is 6 .

Substitute 9×109 N⋅m2/C2 for k , 6 for Z , 1.60×10−19 C for e and 2.75×10−15 m for R in equation (II) to find F .

F=(9×109 N⋅m2/C2)(6−1)(1.60×10−19 C)2(2.75×10−15 m)2=152 N

Therefore, the force of repulsion between a proton at the surface of a C612 nucleus and the remaining five protons is 152 N .

(c)

To determine

The work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus.

(c)

Expert Solution

Answer to Problem 10P

The work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 2.62 MeV .

Explanation of Solution

The work that must be done will be equal to the electrostatic potential energy of the system.

Write the equation for the electrostatic potential energy.

U=kq1q2r

Here, U is the potential energy.

Substitute +e for q1 , (Z−1)e for q2 and R for r in the above equation.

U=k(Z−1)e2R (III)

Conclusion:

Substitute 9×109 N⋅m2/C2 for k , 6 for Z , 1.60×10−19 C for e and 2.75×10−15 m for R in equation (III) to find U .

U=(9×109 N⋅m2/C2)(6−1)(1.60×10−19 C)2(2.75×10−15 m)=4.19×10−13 J1 eV1.60×10−19 J1 MeV106 eV=2.62 MeV

Therefore, the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 2.62 MeV .

(d)

To determine

The radius of the U92238 nucleus, force of repulsion between a proton at the surface of a U92238 nucleus and the remaining protons and the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus.

(d)

Expert Solution

Answer to Problem 10P

The radius of the U92238 nucleus is 7.44×10−15 m , the force of repulsion between a proton at the surface of a nucleus U92238 and the remaining protons is 379 N and the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 17.6 MeV .

Explanation of Solution

The mass number of U92238 is 238 and the atomic number is 92 .

Conclusion:

Substitute 1.2×10−15 m for R0 and 238 for A in equation (I) to find R .

R=(1.2×10−15 m)(238)1/3=7.44×10−15 m

Substitute 9×109 N⋅m2/C2 for k , 92 for Z , 1.60×10−19 C for e and 7.44×10−15 m for R in equation (II) to find F .

F=(9×109 N⋅m2/C2)(92−1)(1.60×10−19 C)2(7.44×10−15 m)2=379 N

Substitute 9×109 N⋅m2/C2 for k , 92 for Z , 1.60×10−19 C for e and 7.44×10−15 m for R in equation (III) to find U .

U=(9×109 N⋅m2/C2)(92−1)(1.60×10−19 C)2(7.44×10−15 m)=2.82×10−12 J1 eV1.60×10−19 J1 MeV106 eV=17.6 MeV

Therefore, the radius of the U92238 nucleus is 7.44×10−15 m , the force of repulsion between a proton at the surface of a nucleus U92238 and the remaining protons is 379 N and the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 17.6 MeV .

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Answer 2

Question

Chapter 13, Problem 10P

(a)

To determine

The radius of the C612 nucleus.

(a)

Expert Solution

Answer to Problem 10P

The radius of the C612 nucleus is 2.75×10−15 m .

Explanation of Solution

Write the equation for the nuclear radius.

R=R0A1/3 (I)

Here, R is the nuclear radius, R0 is constant with value 1.2×10−15 m and A is the mass number.

Conclusion:

The mass number of C612 is 12 .

Substitute 1.2×10−15 m for R0 and 12 for A in equation (I) to find R .

R=(1.2×10−15 m)(12)1/3=2.75×10−15 m

Therefore, the radius of the C612 nucleus is 2.75×10−15 m .

(b)

To determine

The force of repulsion between a proton at the surface of a C612 nucleus and the remaining five protons.

(b)

Expert Solution

Answer to Problem 10P

The force of repulsion between a proton at the surface of a C612 nucleus and the remaining five protons is 152 N .

Explanation of Solution

Write the equation for the Coulombic force.

F=kq1q2r2

Here, F is the force, k is the Coulomb constant, q1,q2 are the charges and r is the distance between the charges.

The charge of proton is +e and the charge o the remaining protons is (Z−1)e , where Z is the atomic number. The value of r will be equal to R .

Substitute +e for q1 , (Z−1)e for q2 and R for r in the above equation.

F=k(Z−1)e2R2 (II)

Here, e is the magnitude of the electronic charge.

Conclusion:

The value of k is 9×109 N⋅m2/C2 and the charge of the electron is 1.60×10−19 C . The atomic number of C612 is 6 .

Substitute 9×109 N⋅m2/C2 for k , 6 for Z , 1.60×10−19 C for e and 2.75×10−15 m for R in equation (II) to find F .

F=(9×109 N⋅m2/C2)(6−1)(1.60×10−19 C)2(2.75×10−15 m)2=152 N

Therefore, the force of repulsion between a proton at the surface of a C612 nucleus and the remaining five protons is 152 N .

(c)

To determine

The work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus.

(c)

Expert Solution

Answer to Problem 10P

The work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 2.62 MeV .

Explanation of Solution

The work that must be done will be equal to the electrostatic potential energy of the system.

Write the equation for the electrostatic potential energy.

U=kq1q2r

Here, U is the potential energy.

Substitute +e for q1 , (Z−1)e for q2 and R for r in the above equation.

U=k(Z−1)e2R (III)

Conclusion:

Substitute 9×109 N⋅m2/C2 for k , 6 for Z , 1.60×10−19 C for e and 2.75×10−15 m for R in equation (III) to find U .

U=(9×109 N⋅m2/C2)(6−1)(1.60×10−19 C)2(2.75×10−15 m)=4.19×10−13 J1 eV1.60×10−19 J1 MeV106 eV=2.62 MeV

Therefore, the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 2.62 MeV .

(d)

To determine

The radius of the U92238 nucleus, force of repulsion between a proton at the surface of a U92238 nucleus and the remaining protons and the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus.

(d)

Expert Solution

Answer to Problem 10P

The radius of the U92238 nucleus is 7.44×10−15 m , the force of repulsion between a proton at the surface of a nucleus U92238 and the remaining protons is 379 N and the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 17.6 MeV .

Explanation of Solution

The mass number of U92238 is 238 and the atomic number is 92 .

Conclusion:

Substitute 1.2×10−15 m for R0 and 238 for A in equation (I) to find R .

R=(1.2×10−15 m)(238)1/3=7.44×10−15 m

Substitute 9×109 N⋅m2/C2 for k , 92 for Z , 1.60×10−19 C for e and 7.44×10−15 m for R in equation (II) to find F .

F=(9×109 N⋅m2/C2)(92−1)(1.60×10−19 C)2(7.44×10−15 m)2=379 N

Substitute 9×109 N⋅m2/C2 for k , 92 for Z , 1.60×10−19 C for e and 7.44×10−15 m for R in equation (III) to find U .

U=(9×109 N⋅m2/C2)(92−1)(1.60×10−19 C)2(7.44×10−15 m)=2.82×10−12 J1 eV1.60×10−19 J1 MeV106 eV=17.6 MeV

Therefore, the radius of the U92238 nucleus is 7.44×10−15 m , the force of repulsion between a proton at the surface of a nucleus U92238 and the remaining protons is 379 N and the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 17.6 MeV .

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Answer 3

Question

Chapter 13, Problem 10P

(a)

To determine

The radius of the C612 nucleus.

(a)

Expert Solution

Answer to Problem 10P

The radius of the C612 nucleus is 2.75×10−15 m .

Explanation of Solution

Write the equation for the nuclear radius.

R=R0A1/3 (I)

Here, R is the nuclear radius, R0 is constant with value 1.2×10−15 m and A is the mass number.

Conclusion:

The mass number of C612 is 12 .

Substitute 1.2×10−15 m for R0 and 12 for A in equation (I) to find R .

R=(1.2×10−15 m)(12)1/3=2.75×10−15 m

Therefore, the radius of the C612 nucleus is 2.75×10−15 m .

(b)

To determine

The force of repulsion between a proton at the surface of a C612 nucleus and the remaining five protons.

(b)

Expert Solution

Answer to Problem 10P

The force of repulsion between a proton at the surface of a C612 nucleus and the remaining five protons is 152 N .

Explanation of Solution

Write the equation for the Coulombic force.

F=kq1q2r2

Here, F is the force, k is the Coulomb constant, q1,q2 are the charges and r is the distance between the charges.

The charge of proton is +e and the charge o the remaining protons is (Z−1)e , where Z is the atomic number. The value of r will be equal to R .

Substitute +e for q1 , (Z−1)e for q2 and R for r in the above equation.

F=k(Z−1)e2R2 (II)

Here, e is the magnitude of the electronic charge.

Conclusion:

The value of k is 9×109 N⋅m2/C2 and the charge of the electron is 1.60×10−19 C . The atomic number of C612 is 6 .

Substitute 9×109 N⋅m2/C2 for k , 6 for Z , 1.60×10−19 C for e and 2.75×10−15 m for R in equation (II) to find F .

F=(9×109 N⋅m2/C2)(6−1)(1.60×10−19 C)2(2.75×10−15 m)2=152 N

Therefore, the force of repulsion between a proton at the surface of a C612 nucleus and the remaining five protons is 152 N .

(c)

To determine

The work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus.

(c)

Expert Solution

Answer to Problem 10P

The work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 2.62 MeV .

Explanation of Solution

The work that must be done will be equal to the electrostatic potential energy of the system.

Write the equation for the electrostatic potential energy.

U=kq1q2r

Here, U is the potential energy.

Substitute +e for q1 , (Z−1)e for q2 and R for r in the above equation.

U=k(Z−1)e2R (III)

Conclusion:

Substitute 9×109 N⋅m2/C2 for k , 6 for Z , 1.60×10−19 C for e and 2.75×10−15 m for R in equation (III) to find U .

U=(9×109 N⋅m2/C2)(6−1)(1.60×10−19 C)2(2.75×10−15 m)=4.19×10−13 J1 eV1.60×10−19 J1 MeV106 eV=2.62 MeV

Therefore, the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 2.62 MeV .

(d)

To determine

The radius of the U92238 nucleus, force of repulsion between a proton at the surface of a U92238 nucleus and the remaining protons and the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus.

(d)

Expert Solution

Answer to Problem 10P

The radius of the U92238 nucleus is 7.44×10−15 m , the force of repulsion between a proton at the surface of a nucleus U92238 and the remaining protons is 379 N and the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 17.6 MeV .

Explanation of Solution

The mass number of U92238 is 238 and the atomic number is 92 .

Conclusion:

Substitute 1.2×10−15 m for R0 and 238 for A in equation (I) to find R .

R=(1.2×10−15 m)(238)1/3=7.44×10−15 m

Substitute 9×109 N⋅m2/C2 for k , 92 for Z , 1.60×10−19 C for e and 7.44×10−15 m for R in equation (II) to find F .

F=(9×109 N⋅m2/C2)(92−1)(1.60×10−19 C)2(7.44×10−15 m)2=379 N

Substitute 9×109 N⋅m2/C2 for k , 92 for Z , 1.60×10−19 C for e and 7.44×10−15 m for R in equation (III) to find U .

U=(9×109 N⋅m2/C2)(92−1)(1.60×10−19 C)2(7.44×10−15 m)=2.82×10−12 J1 eV1.60×10−19 J1 MeV106 eV=17.6 MeV

Therefore, the radius of the U92238 nucleus is 7.44×10−15 m , the force of repulsion between a proton at the surface of a nucleus U92238 and the remaining protons is 379 N and the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 17.6 MeV .

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Answer 4

Question

Chapter 13, Problem 10P

(a)

To determine

The radius of the C612 nucleus.

(a)

Expert Solution

Answer to Problem 10P

The radius of the C612 nucleus is 2.75×10−15 m .

Explanation of Solution

Write the equation for the nuclear radius.

R=R0A1/3 (I)

Here, R is the nuclear radius, R0 is constant with value 1.2×10−15 m and A is the mass number.

Conclusion:

The mass number of C612 is 12 .

Substitute 1.2×10−15 m for R0 and 12 for A in equation (I) to find R .

R=(1.2×10−15 m)(12)1/3=2.75×10−15 m

Therefore, the radius of the C612 nucleus is 2.75×10−15 m .

(b)

To determine

The force of repulsion between a proton at the surface of a C612 nucleus and the remaining five protons.

(b)

Expert Solution

Answer to Problem 10P

The force of repulsion between a proton at the surface of a C612 nucleus and the remaining five protons is 152 N .

Explanation of Solution

Write the equation for the Coulombic force.

F=kq1q2r2

Here, F is the force, k is the Coulomb constant, q1,q2 are the charges and r is the distance between the charges.

The charge of proton is +e and the charge o the remaining protons is (Z−1)e , where Z is the atomic number. The value of r will be equal to R .

Substitute +e for q1 , (Z−1)e for q2 and R for r in the above equation.

F=k(Z−1)e2R2 (II)

Here, e is the magnitude of the electronic charge.

Conclusion:

The value of k is 9×109 N⋅m2/C2 and the charge of the electron is 1.60×10−19 C . The atomic number of C612 is 6 .

Substitute 9×109 N⋅m2/C2 for k , 6 for Z , 1.60×10−19 C for e and 2.75×10−15 m for R in equation (II) to find F .

F=(9×109 N⋅m2/C2)(6−1)(1.60×10−19 C)2(2.75×10−15 m)2=152 N

Therefore, the force of repulsion between a proton at the surface of a C612 nucleus and the remaining five protons is 152 N .

(c)

To determine

The work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus.

(c)

Expert Solution

Answer to Problem 10P

The work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 2.62 MeV .

Explanation of Solution

The work that must be done will be equal to the electrostatic potential energy of the system.

Write the equation for the electrostatic potential energy.

U=kq1q2r

Here, U is the potential energy.

Substitute +e for q1 , (Z−1)e for q2 and R for r in the above equation.

U=k(Z−1)e2R (III)

Conclusion:

Substitute 9×109 N⋅m2/C2 for k , 6 for Z , 1.60×10−19 C for e and 2.75×10−15 m for R in equation (III) to find U .

U=(9×109 N⋅m2/C2)(6−1)(1.60×10−19 C)2(2.75×10−15 m)=4.19×10−13 J1 eV1.60×10−19 J1 MeV106 eV=2.62 MeV

Therefore, the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 2.62 MeV .

(d)

To determine

The radius of the U92238 nucleus, force of repulsion between a proton at the surface of a U92238 nucleus and the remaining protons and the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus.

(d)

Expert Solution

Answer to Problem 10P

The radius of the U92238 nucleus is 7.44×10−15 m , the force of repulsion between a proton at the surface of a nucleus U92238 and the remaining protons is 379 N and the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 17.6 MeV .

Explanation of Solution

The mass number of U92238 is 238 and the atomic number is 92 .

Conclusion:

Substitute 1.2×10−15 m for R0 and 238 for A in equation (I) to find R .

R=(1.2×10−15 m)(238)1/3=7.44×10−15 m

Substitute 9×109 N⋅m2/C2 for k , 92 for Z , 1.60×10−19 C for e and 7.44×10−15 m for R in equation (II) to find F .

F=(9×109 N⋅m2/C2)(92−1)(1.60×10−19 C)2(7.44×10−15 m)2=379 N

Substitute 9×109 N⋅m2/C2 for k , 92 for Z , 1.60×10−19 C for e and 7.44×10−15 m for R in equation (III) to find U .

U=(9×109 N⋅m2/C2)(92−1)(1.60×10−19 C)2(7.44×10−15 m)=2.82×10−12 J1 eV1.60×10−19 J1 MeV106 eV=17.6 MeV

Therefore, the radius of the U92238 nucleus is 7.44×10−15 m , the force of repulsion between a proton at the surface of a nucleus U92238 and the remaining protons is 379 N and the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 17.6 MeV .

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Answer 5

Chapter 13, Problem 10P

(a)

To determine

The radius of the C612 nucleus.

(a)

Expert Solution

Answer to Problem 10P

The radius of the C612 nucleus is 2.75×10−15 m .

Explanation of Solution

Write the equation for the nuclear radius.

R=R0A1/3 (I)

Here, R is the nuclear radius, R0 is constant with value 1.2×10−15 m and A is the mass number.

Conclusion:

The mass number of C612 is 12 .

Substitute 1.2×10−15 m for R0 and 12 for A in equation (I) to find R .

R=(1.2×10−15 m)(12)1/3=2.75×10−15 m

Therefore, the radius of the C612 nucleus is 2.75×10−15 m .

(b)

To determine

The force of repulsion between a proton at the surface of a C612 nucleus and the remaining five protons.

(b)

Expert Solution

Answer to Problem 10P

The force of repulsion between a proton at the surface of a C612 nucleus and the remaining five protons is 152 N .

Explanation of Solution

Write the equation for the Coulombic force.

F=kq1q2r2

Here, F is the force, k is the Coulomb constant, q1,q2 are the charges and r is the distance between the charges.

The charge of proton is +e and the charge o the remaining protons is (Z−1)e , where Z is the atomic number. The value of r will be equal to R .

Substitute +e for q1 , (Z−1)e for q2 and R for r in the above equation.

F=k(Z−1)e2R2 (II)

Here, e is the magnitude of the electronic charge.

Conclusion:

The value of k is 9×109 N⋅m2/C2 and the charge of the electron is 1.60×10−19 C . The atomic number of C612 is 6 .

Substitute 9×109 N⋅m2/C2 for k , 6 for Z , 1.60×10−19 C for e and 2.75×10−15 m for R in equation (II) to find F .

F=(9×109 N⋅m2/C2)(6−1)(1.60×10−19 C)2(2.75×10−15 m)2=152 N

Therefore, the force of repulsion between a proton at the surface of a C612 nucleus and the remaining five protons is 152 N .

(c)

To determine

The work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus.

(c)

Expert Solution

Answer to Problem 10P

The work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 2.62 MeV .

Explanation of Solution

The work that must be done will be equal to the electrostatic potential energy of the system.

Write the equation for the electrostatic potential energy.

U=kq1q2r

Here, U is the potential energy.

Substitute +e for q1 , (Z−1)e for q2 and R for r in the above equation.

U=k(Z−1)e2R (III)

Conclusion:

Substitute 9×109 N⋅m2/C2 for k , 6 for Z , 1.60×10−19 C for e and 2.75×10−15 m for R in equation (III) to find U .

U=(9×109 N⋅m2/C2)(6−1)(1.60×10−19 C)2(2.75×10−15 m)=4.19×10−13 J1 eV1.60×10−19 J1 MeV106 eV=2.62 MeV

Therefore, the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 2.62 MeV .

(d)

To determine

The radius of the U92238 nucleus, force of repulsion between a proton at the surface of a U92238 nucleus and the remaining protons and the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus.

(d)

Expert Solution

Answer to Problem 10P

The radius of the U92238 nucleus is 7.44×10−15 m , the force of repulsion between a proton at the surface of a nucleus U92238 and the remaining protons is 379 N and the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 17.6 MeV .

Explanation of Solution

The mass number of U92238 is 238 and the atomic number is 92 .

Conclusion:

Substitute 1.2×10−15 m for R0 and 238 for A in equation (I) to find R .

R=(1.2×10−15 m)(238)1/3=7.44×10−15 m

Substitute 9×109 N⋅m2/C2 for k , 92 for Z , 1.60×10−19 C for e and 7.44×10−15 m for R in equation (II) to find F .

F=(9×109 N⋅m2/C2)(92−1)(1.60×10−19 C)2(7.44×10−15 m)2=379 N

Substitute 9×109 N⋅m2/C2 for k , 92 for Z , 1.60×10−19 C for e and 7.44×10−15 m for R in equation (III) to find U .

U=(9×109 N⋅m2/C2)(92−1)(1.60×10−19 C)2(7.44×10−15 m)=2.82×10−12 J1 eV1.60×10−19 J1 MeV106 eV=17.6 MeV

Therefore, the radius of the U92238 nucleus is 7.44×10−15 m , the force of repulsion between a proton at the surface of a nucleus U92238 and the remaining protons is 379 N and the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 17.6 MeV .

Answer 6

Chapter 13, Problem 10P

(a)

To determine

The radius of the C612 nucleus.

(a)

Expert Solution

Answer to Problem 10P

The radius of the C612 nucleus is 2.75×10−15 m .

Explanation of Solution

Write the equation for the nuclear radius.

R=R0A1/3 (I)

Here, R is the nuclear radius, R0 is constant with value 1.2×10−15 m and A is the mass number.

Conclusion:

The mass number of C612 is 12 .

Substitute 1.2×10−15 m for R0 and 12 for A in equation (I) to find R .

R=(1.2×10−15 m)(12)1/3=2.75×10−15 m

Therefore, the radius of the C612 nucleus is 2.75×10−15 m .

Answer 7

Chapter 13, Problem 10P

(a)

To determine

The radius of the C612 nucleus.

Answer 8

(a)

Expert Solution

Answer to Problem 10P

The radius of the C612 nucleus is 2.75×10−15 m .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the equation for the nuclear radius.

R=R0A1/3 (I)

Here, R is the nuclear radius, R0 is constant with value 1.2×10−15 m and A is the mass number.

Conclusion:

The mass number of C612 is 12 .

Substitute 1.2×10−15 m for R0 and 12 for A in equation (I) to find R .

R=(1.2×10−15 m)(12)1/3=2.75×10−15 m

Therefore, the radius of the C612 nucleus is 2.75×10−15 m .

Answer 13

(b)

To determine

The force of repulsion between a proton at the surface of a C612 nucleus and the remaining five protons.

(b)

Expert Solution

Answer to Problem 10P

The force of repulsion between a proton at the surface of a C612 nucleus and the remaining five protons is 152 N .

Explanation of Solution

Write the equation for the Coulombic force.

F=kq1q2r2

Here, F is the force, k is the Coulomb constant, q1,q2 are the charges and r is the distance between the charges.

The charge of proton is +e and the charge o the remaining protons is (Z−1)e , where Z is the atomic number. The value of r will be equal to R .

Substitute +e for q1 , (Z−1)e for q2 and R for r in the above equation.

F=k(Z−1)e2R2 (II)

Here, e is the magnitude of the electronic charge.

Conclusion:

The value of k is 9×109 N⋅m2/C2 and the charge of the electron is 1.60×10−19 C . The atomic number of C612 is 6 .

Substitute 9×109 N⋅m2/C2 for k , 6 for Z , 1.60×10−19 C for e and 2.75×10−15 m for R in equation (II) to find F .

F=(9×109 N⋅m2/C2)(6−1)(1.60×10−19 C)2(2.75×10−15 m)2=152 N

Therefore, the force of repulsion between a proton at the surface of a C612 nucleus and the remaining five protons is 152 N .

Answer 14

(b)

To determine

The force of repulsion between a proton at the surface of a C612 nucleus and the remaining five protons.

Answer 15

(b)

Expert Solution

Answer to Problem 10P

The force of repulsion between a proton at the surface of a C612 nucleus and the remaining five protons is 152 N .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write the equation for the Coulombic force.

F=kq1q2r2

Here, F is the force, k is the Coulomb constant, q1,q2 are the charges and r is the distance between the charges.

The charge of proton is +e and the charge o the remaining protons is (Z−1)e , where Z is the atomic number. The value of r will be equal to R .

Substitute +e for q1 , (Z−1)e for q2 and R for r in the above equation.

F=k(Z−1)e2R2 (II)

Here, e is the magnitude of the electronic charge.

Conclusion:

The value of k is 9×109 N⋅m2/C2 and the charge of the electron is 1.60×10−19 C . The atomic number of C612 is 6 .

Substitute 9×109 N⋅m2/C2 for k , 6 for Z , 1.60×10−19 C for e and 2.75×10−15 m for R in equation (II) to find F .

F=(9×109 N⋅m2/C2)(6−1)(1.60×10−19 C)2(2.75×10−15 m)2=152 N

Therefore, the force of repulsion between a proton at the surface of a C612 nucleus and the remaining five protons is 152 N .

Answer 20

(c)

To determine

The work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus.

(c)

Expert Solution

Answer to Problem 10P

The work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 2.62 MeV .

Explanation of Solution

The work that must be done will be equal to the electrostatic potential energy of the system.

Write the equation for the electrostatic potential energy.

U=kq1q2r

Here, U is the potential energy.

Substitute +e for q1 , (Z−1)e for q2 and R for r in the above equation.

U=k(Z−1)e2R (III)

Conclusion:

Substitute 9×109 N⋅m2/C2 for k , 6 for Z , 1.60×10−19 C for e and 2.75×10−15 m for R in equation (III) to find U .

U=(9×109 N⋅m2/C2)(6−1)(1.60×10−19 C)2(2.75×10−15 m)=4.19×10−13 J1 eV1.60×10−19 J1 MeV106 eV=2.62 MeV

Therefore, the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 2.62 MeV .

Answer 21

(c)

To determine

The work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus.

Answer 22

(c)

Expert Solution

Answer to Problem 10P

The work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 2.62 MeV .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

The work that must be done will be equal to the electrostatic potential energy of the system.

Write the equation for the electrostatic potential energy.

U=kq1q2r

Here, U is the potential energy.

Substitute +e for q1 , (Z−1)e for q2 and R for r in the above equation.

U=k(Z−1)e2R (III)

Conclusion:

Substitute 9×109 N⋅m2/C2 for k , 6 for Z , 1.60×10−19 C for e and 2.75×10−15 m for R in equation (III) to find U .

U=(9×109 N⋅m2/C2)(6−1)(1.60×10−19 C)2(2.75×10−15 m)=4.19×10−13 J1 eV1.60×10−19 J1 MeV106 eV=2.62 MeV

Therefore, the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 2.62 MeV .

Answer 27

(d)

To determine

The radius of the U92238 nucleus, force of repulsion between a proton at the surface of a U92238 nucleus and the remaining protons and the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus.

(d)

Expert Solution

Answer to Problem 10P

The radius of the U92238 nucleus is 7.44×10−15 m , the force of repulsion between a proton at the surface of a nucleus U92238 and the remaining protons is 379 N and the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 17.6 MeV .

Explanation of Solution

The mass number of U92238 is 238 and the atomic number is 92 .

Conclusion:

Substitute 1.2×10−15 m for R0 and 238 for A in equation (I) to find R .

R=(1.2×10−15 m)(238)1/3=7.44×10−15 m

Substitute 9×109 N⋅m2/C2 for k , 92 for Z , 1.60×10−19 C for e and 7.44×10−15 m for R in equation (II) to find F .

F=(9×109 N⋅m2/C2)(92−1)(1.60×10−19 C)2(7.44×10−15 m)2=379 N

Substitute 9×109 N⋅m2/C2 for k , 92 for Z , 1.60×10−19 C for e and 7.44×10−15 m for R in equation (III) to find U .

U=(9×109 N⋅m2/C2)(92−1)(1.60×10−19 C)2(7.44×10−15 m)=2.82×10−12 J1 eV1.60×10−19 J1 MeV106 eV=17.6 MeV

Therefore, the radius of the U92238 nucleus is 7.44×10−15 m , the force of repulsion between a proton at the surface of a nucleus U92238 and the remaining protons is 379 N and the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 17.6 MeV .

Answer 28

(d)

To determine

The radius of the U92238 nucleus, force of repulsion between a proton at the surface of a U92238 nucleus and the remaining protons and the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus.

Answer 29

(d)

Expert Solution

Answer to Problem 10P

The radius of the U92238 nucleus is 7.44×10−15 m , the force of repulsion between a proton at the surface of a nucleus U92238 and the remaining protons is 379 N and the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 17.6 MeV .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

The mass number of U92238 is 238 and the atomic number is 92 .

Conclusion:

Substitute 1.2×10−15 m for R0 and 238 for A in equation (I) to find R .

R=(1.2×10−15 m)(238)1/3=7.44×10−15 m

Substitute 9×109 N⋅m2/C2 for k , 92 for Z , 1.60×10−19 C for e and 7.44×10−15 m for R in equation (II) to find F .

F=(9×109 N⋅m2/C2)(92−1)(1.60×10−19 C)2(7.44×10−15 m)2=379 N

Substitute 9×109 N⋅m2/C2 for k , 92 for Z , 1.60×10−19 C for e and 7.44×10−15 m for R in equation (III) to find U .

U=(9×109 N⋅m2/C2)(92−1)(1.60×10−19 C)2(7.44×10−15 m)=2.82×10−12 J1 eV1.60×10−19 J1 MeV106 eV=17.6 MeV

Therefore, the radius of the U92238 nucleus is 7.44×10−15 m , the force of repulsion between a proton at the surface of a nucleus U92238 and the remaining protons is 379 N and the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is 17.6 MeV .