General Physics, 2nd Edition
General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
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Chapter 30, Problem 26E

(a)

To determine

The difference in mass defects of 82207Pb and 82208Pb.

(a)

Expert Solution
Check Mark

Answer to Problem 26E

The difference in mass defects of 82207Pb and 82208Pb is 0.00797u.

Explanation of Solution

Write the expression for the mass defect of a nucleus.

    Δm=n(mn)+p(mp)m        (I)

Here, Δm is the mass defect, n is the number of neutrons, p is the number of protons, mn is the mass of neutron, mp is the mass of proton and m is the atomic mass.

Consider that Δm1 is the mass defect of 82208Pb and Δm2 is the mass defect of 82207Pb.

Conclusion:

Substitute 1.008665u for mn, 1.007825u for mp and 207.9766u for m in expression (I).

    Δm1=(20882)(1.008665u)+82(1.007825u)207.9766u=1.75684u

Substitute 1.008665u for mn, 1.007825u for mp and 206.9759u for m in expression (I).

    Δm2=(20782)(1.008665u)+82(1.007825u)206.9759u=1.74887u

Calculate the difference between Δm1 and Δm2.

    Δm1Δm2=(1.75684u)(1.74887u)=0.00797u

Thus, the difference in mass defects of 82207Pb and 82208Pb is 0.00797u.

(b)

To determine

The minimum energy required to remove a neutron from 82208Pb.

(b)

Expert Solution
Check Mark

Answer to Problem 26E

The minimum energy required to remove a neutron from 82208Pb is 7.86MeV.

Explanation of Solution

Write the expression for the binding energy.

    E=Δm(931MeV1u)        (II)

Here, E is the total binding energy.

Write the expression of average binding energy.

    Eav=EA        (III)

Here, A is the mass number.

Conclusion:

Substitute 1.75684u for Δm in expression (II).

    E=(1.75684u)(931MeV1u)=1635.62MeV

Substitute 1635.62MeV for E and 208 for A in expression (III).

    Eav=1635.62MeV208=7.86MeV

Thus, the minimum energy required to remove a neutron from 82208Pb is 7.86MeV.

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