Physics for Scientists and Engineers with Modern Physics
Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 43, Problem 62AP

(a)

To determine

The edge dimension of the cube.

(a)

Expert Solution
Check Mark

Answer to Problem 62AP

The edge dimension of the cube is 15.4cm.

Explanation of Solution

Write the equation for volume of a cube.

    V=l3                                                                                                                  (I)

Here, V is the volume of a cube and l is the edge dimension of the cube.

Write the equation connecting mass, density and volume.

    V=mρ                                                                                                                (II)

Here, m is the mass and ρ is the density.

Write the equation for the edge dimension of the cube from equations (I) and (II).

    l3=mρl=(mρ)1/3                                                                                                         (III)

Conclusion:

Substitute 70.0kg for m  and 19.1×103kg/m3 for ρ in equation (III) to find l.

    l=(70.0kg19.1×103kg/m3)1/3=(0.00366)1/3=(0.154m)(102cm1m)=15.4cm

Thus, the edge dimension of the cube is 15.4cm.

(b)

To determine

The net decay energy.

(b)

Expert Solution
Check Mark

Answer to Problem 62AP

The net decay energy is 51.7MeV.

Explanation of Solution

Write the given reaction.

    U922388(H24e)+6(e10)+P82206b+6ν¯+Qnet

Add 92 electrons on both sides of the reaction to obtain the new nuclear reaction.

    U92238atom8H24eatom+P82206batom                                                          (IV)

Write the equation for Q value.

    Q=[MU922388MH24eMP82206b]c2                                                                        (V)

Conclusion:

Substitute 238.050783u for MU92238, 4.002603u for MH24e and 205.974449u for MP82206b in equation (V) to find Q.

    Q=[238.050783u8(4.002603u)205.974449u]c2=(0.05551u)c2(931.5MeV/c2u)=51.7MeV

Thus, the net decay of energy is 51.7MeV.

(c)

To determine

Prove that the power output P=QR.

(c)

Expert Solution
Check Mark

Answer to Problem 62AP

The power output is P=QR.

Explanation of Solution

The number of decays per second is equal to the decay rate given by R.

The energy released in each decay is Q.

Conclusion:

The energy released per unit time interval is the power output.

Write the equation for the power output.

    P=QR

Thus, the power output is p=QR.

(d)

To determine

The power output due to the radioactivity of the uranium and its daughters.

(d)

Expert Solution
Check Mark

Answer to Problem 62AP

The power output due to the radioactivity of the uranium and its daughters is 2.27×105J/yr.

Explanation of Solution

Find the amount of the nuclei.

    N=(70.0kg238g/mol)=((70.0kg)(103g1kg)238g/mol)(6.02×1023nuclei/mol)=1.77×1026nuclei

Write the equation for the decay constant.

    λ=ln2T1/2                                                                                                           (VI)

Here, λ is the decay constant and T1/2 is the half-life period.

Write the equation for the rate of decay.

    R=λN                                                                                                          (VII)

Here, R is the rate of decay.

Write the equation for the power output.

    P=QR                                                                                                         (VIII)

Here, P is the power output and Q is the Q factor.

Conclusion:

Substitute 4.47×109yr for T1/2 in equation (VI) to find λ.

    λ=ln24.47×109yr=0.6934.47×109yr=1.55×10101yr

Substitute 1.55×10101yr for λ and 1.77×1026nuclei for N in equation (VII) to find R.

    R=(1.55×10101yr)(1.77×1026nuclei)=2.75×1016decays/yr

Substitute 2.75×1016decays/yr for R and 51.7MeV for Q in equation (VIII) to find P.

    P=(51.7MeV)(2.75×1016decays/yr)=(51.7MeV)(2.75×1016decays/yr)(1.60×1013J1MeV)=2.27×105J/yr

Thus, the power output due to the radioactivity of the uranium and its daughters is 2.27×105J/yr.

(e)

To determine

The rate per year at which the person absorb the energy of radiation.

(e)

Expert Solution
Check Mark

Answer to Problem 62AP

The rate per year at which the person absorb the energy of radiation is 3.18J/yr.

Explanation of Solution

Find the dose in rad/yr.

    dose in rad/yr=dose in rem/yrRBE                                                                     (IX)

Here, RBE is the relative biological effectiveness.

Write the equation for the allowed whole-body dose.

    dose=m(dose in rad/yr)                                                                                 (X)

Conclusion:

Substitute 5.00rem/yr for dose in rem/yr and 1.10 for RBE in equation (IX) to find dose in rad/yrλ.

    dose in rad/yr=5.00rem/yr1.10=4.55rad/yr

Substitute 4.55rad/yr for dose in rad/yr and 70.0kg for m in equation (X) to find the dose.

    dose=(70.0kg)(4.55rad/yr)=(70.0kg)(4.55rad/yr)(102J/yr1rad)=3.18J/yr

Thus, the rate per year at which the person absorb the energy of radiation is 3.18J/yr.

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