Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 40, Problem 15P

(a)

To determine

The binding energy and the binding energy per nucleon for 12C .

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The mass of one proton is 1.007825u .

The mass of one neutron is 1.008665u .

The mass of 12C is 12u .

Formula used:

Write the expression for the masses of protons and neutrons in 12C .

  mp+a=6mp+6ma

Here, mp is the mass of one proton, ma is the mass of one neutron and mp+a is the total mass of both proton and neutron in 12C .

Write the expression for the mass difference between total mass of both proton and neutron in 12C and mass of 12C alone.

  Δm=(6mp+6ma)m12C   ........ (1)

Here, Δm is the mass difference between total mass of both proton and neutron in 12C and m12C is the mass of 12C .

Write the expression for the binding energy.

  Eb=(Δm)c2   ........ (2)

Here, Eb is the binding energy and c is the speed of light.

Write the expression for the binding energy per nucleon.

  Eb=EbA   ........ (3)

Here, Eb is the binding energy per nucleon and A is the number of neutrons.

Calculation:

Substitute 1.007825u for mp , 1.008665u for ma and 12u for m12C in equation (1).

  Δm=(6( 1.007825u)+6( 1.008665u))(12u)=0.098940u

Substitute 0.098940u for Δm in equation (2).

  Eb=(0.098940u)c2( 931.5MeV c 2 u)=92.2MeV

Substitute 92.2MeV for Eb and 12 for A in equation (3).

  Eb=( 92.2MeV)12=7.68MeV

Conclusion:

Thus, the binding energy and the binding energy per nucleon for 12C is 92.2MeV and 7.68MeV .

(b)

To determine

The binding energy and the binding energy per nucleon for 56F .

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The mass of one proton is 1.007825u .

The mass of one neutron is 1.008665u .

The mass of 56F is 56u .

Formula used:

Write the expression for the masses of protons and neutrons in 56F .

  mp+a=26mp+30ma

Here, mp is the mass of one proton, ma is the mass of one neutron and mp+a is the total mass of both proton and neutron in 56F .

Write the expression for the mass difference between total mass of both proton and neutron in 56F and mass of 56F alone.

  Δm=(26mp+30ma)m56F   ........ (4)

Here, Δm is the mass difference between total mass of both proton and neutron in 56F and m56F is the mass of 56F .

Write the expression for the binding energy.

  Eb=(Δm)c2   ........ (2)

Here, Eb is the binding energy and c is the speed of light.

Write the expression for the binding energy per nucleon.

  Eb=EbA   ........ (3)

Here, Eb is the binding energy per nucleon and A is the number of neutrons.

Calculation:

Substitute 1.007825u for mp , 1.008665u for ma and 56u for m56F in equation (4).

  Δm=(26( 1.007825u)+30( 1.008665u))(56u)=0.528458u

Substitute 0.528458u for Δm in equation (2).

  Eb=(0.528458u)c2( 931.5MeV c 2 u)=492MeV

Substitute 92.2MeV for Eb and 56 for A in equation (3).

  Eb=( 492MeV)56=8.79MeV

Conclusion:

Thus, the binding energy and the binding energy per nucleon for 56F is 492MeV and 8.79MeV .

(c)

To determine

The binding energy and the binding energy per nucleon for 238U .

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

The mass of one proton is 1.007825u .

The mass of one neutron is 1.008665u .

The mass of 238U is 238u .

Formula used:

Write the expression for the masses of protons and neutrons in 238U .

  mp+a=92mp+146ma

Here, mp is the mass of one proton, ma is the mass of one neutron and mp+a is the total mass of both proton and neutron in 238U .

Write the expression for the mass difference between total mass of both proton and neutron in 238U and mass of 238U alone.

  Δm=(92mp+146ma)m238U   ........ (5)

Here, Δm is the mass difference between total mass of both proton and neutron in 238U and m238U is the mass of 238U .

Write the expression for the binding energy.

  Eb=(Δm)c2   ........ (2)

Here, Eb is the binding energy and c is the speed of light.

Write the expression for the binding energy per nucleon.

  Eb=EbA   ........ (3)

Here, Eb is the binding energy per nucleon and A is the number of neutrons.

Calculation:

Substitute 1.007825u for mp , 1.008665u for ma and 238u for m238U in equation (5).

  Δm=(92( 1.007825u)+146( 1.008665u))(238u)=1.934207u

Substitute 1.934207u for Δm in equation (2).

  Eb=(1.934207u)c2( 931.5MeV c 2 u)=1802MeV

Substitute 1802MeV for Eb and 238 for A in equation (3).

  Eb=( 1802MeV)238=7.57MeV

Conclusion:

Thus, the binding energy and the binding energy per nucleon for 238U is 1802MeV and 7.57MeV .

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