Question: A U236nucleus undergoes fission and breaks into two middle-mass fragments, X140eandSr96. (a) By what percentage does the surface area of the fission products differ from that of the original U236nucleus? (b) By what percentage does the volume change? (c) By what percentage does the electric potential energy change? The electric potential energy of a uniformly charged sphere of radius r and charge Q is given by

U=35(Q24πε0r)

Short Answer

Expert verified

(a) The percentage change in the surface area is 25.5%.

(b)The percentage change in the volume is 0%.

(c) The percentage change in the electrical potential energy is -35.9%.

Step by step solution

01

Write the given data from the question.

The Uranium U236fission X140eand S96rbreaks into two fragments and .

The radius of the sphere is r.

The charge on the sphere is Q.

The mass number of Xenon, AXe=140

The mass number of Strontium, ASr=94

The mass number of Uranium,AU=236

The electric potential of the uniformly charged sphere, U=35(Q24πε0r)

02

Determine the formulas:

The expression to calculate the radii of the fragment is given as follows.

r=r0A1/3 …… (1)

Here, A is the mass number.

The expression for the surface area of the sphere is given as follows.

a=4πr2 …… (2)

The expression to calculate the volume of the sphere is given as follows.

V=43πr3 …… (3)

03

(a) Calculate the percentage change in the surface area of the sphere:

Derive the expression for the surface area in terms of the mass number.

Substitute r=r0A1/3for r into equation (2).

r=4πr0A1/32=4πr0A2/3

The surface area after the fission process would be the sum of the surface area of the Xenon and Strontium.

Calculate the surface area after the fission process.

ar=4πr0AXe2/3+ASr2/3

Calculate the surface area before the fission process.

ar=4πr0AU2/3

Calculate the percentage change in the surface.

aa=Af-AiAi

Substitute 4πr0AXe2/3+ASr2/3for afand 4πr0AU2/3for aiinto above equation.

Δaa=4πr0AXe2/β+ASr2/β4πr0AU2/34πr0AU2/3=4πr0Aχe2/β+ASr2/βAU2/34πr0AU2/β=AXe2/3+Asr2/3AU2/βAU2/3

Substitute 236 for Au,140 for AXeand 96 for ASrinto above equation.

aa=1402/3+962/3-2362/32362/3=26.9619+20.9659-38.189238.1892=9.738638.1892=0.255

aa%=0.255×100%=25.5%

Hence, the percentage change in the surface area is 25.5%.

04

(b) Calculate the percentage change in the volume of the sphere:

Derive the expression for the volume in terms of the mass number.

Substitute r0A1/3for r into equation (3).

V=43πr0A1/32=43πr0A

The volume after the fission process would be the sum of the volume of the Xenon and Strontium.

Calculate the volume after the fission process.

Vr=43πr0AXe+ASr

Calculate the volume before the fission process.

Vi=43πr0AU

Calculate the percentage change in the surface.

VV=Vf-ViVi

Substitute role="math" localid="1661930807586" 43πr0AXe+ASrfor Vfand 43πr0AUfor Viinto above equation.

role="math" localid="1661931138966" VV=43πr0AXe+ASr-43πr0AU43πr0AU=43πr0AXe+ASr-AU43πr0AU=AXe+ASr-AUAU

Substitute 236 for AU, 140 for AXeand 96 for ASrinto above equation.

vv=140+96-236236=236-236236=0

Hence the percentage change in the volume is 0%.

05

(c) Calculate the percentage change in electrical potential energy:

The charges are equal to the product of the atomic number and electron charge.

Therefore,

Qu=ZueQXe=ZXeeQ=ZSre

Derive the expression for electrical potential energy in terms of the mass number.

role="math" localid="1661931639748" U=3514πε0r0Ze2r0A1/3=3514πε0r0Ze2A1/3

Calculate the electrical potential energy after the fission process.

role="math" localid="1661931904914" Uf=3514πε0r0Zχee2Aχe1/3+ZSre2ASr1/3=35e24πε0r0ZXe2Aχe1/3+ZSr2ASr1/3

Calculate the volume before the fission process.

Uf=3514πε0r0ZUe2AU1/3=35e24πε0r0ZXU2AU1/3

Calculate the percentage change in the surface.

UU=Uf-UiUi

Substitute 35e24πε0r0Zχe2Aχe1/3+ZSr2ASr1/3for Uf and role="math" localid="1661932204802" 35e24πε0r0ZU2AU1/3for Ufinto above equation.

role="math" localid="1661932416965" UU=35e24πε0r0Zχe2Aχe1/3+ZSr2ASr1/3-35e24πε0r0ZU2AU1/335e24πε0r0ZU2AU1/3UU=35e24πε0r0Zχe2Aχe1/3+ZSr2ASr1/3-ZU2AU1/335e24πε0r0ZU2AU1/3=Zχe2Aχe1/3+ZSr2ASr1/3-ZU2AU1/3ZU2AU1/3

Substitute 236 for AU, 140 for AXe, 96 for ASr,54 for ZXe,38 for ZSrand 92 for ZUinto above equation.

UU=5421401/3+382961/3+9222361/39222361/3=561.5798+315.3625+1369.63541369.6354=-492.69311369.6354=-0.359UU%=-0.359×100%=-35.9%

Hence, the percentage change in the electrical potential energy is -35.9%.

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Most popular questions from this chapter

(a) Calculate the disintegration energy Q for the fission of the molybdenum isotope M98ointo two equal parts. The masses you will need are 97.90541u forM98o and 48.95002u for S49c. (b) If Q turns out to be positive, discuss why this process does not occur spontaneously.

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Calculate the height of the Coulomb barrier for the head-on collision of two deuterons, with effective radius2.1 fm.

The uncompressed radius of the fuel pellet of Sample Problem 43.05 is 20μm. Suppose that the compressed fuel pellet “burns” with an efficiency of 10%—that is, only 10% of the deuterons and 10% of the tritons participate in the fusion reaction of Eq. 43-15. (a) How much energy is released in each such micro explosion of a pellet? (b) To how much TNT is each such pellet equivalent? The heat of combustion of TNT is 4.6 MJ/kg . (c) If a fusion reactor is constructed on the basis of 100 micro explosions per second, what power would be generated? (Part of this power would be used to operate the lasers.)

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