Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

Question: A ${}^{236}U$ nucleus undergoes fission and breaks into two middle-mass fragments, ${}^{140}X e$ $and {}^{96}{Sr}$ . (a) By what percentage does the surface area of the fission products differ from that of the original ${}^{236}U$ nucleus? (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
$U = \frac{3}{5} (\frac{Q^{2}}{4 π ε_{0} r})$

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

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

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Write the given data from the question.

The Uranium ${}^{236}U$ fission ${}^{140}X e$ and ${}^{96}S r$ breaks into two fragments and .

The radius of the sphere is r.

The charge on the sphere is Q.

The mass number of Xenon, $A_{X e} = 140$

The mass number of Strontium, $A_{S r} = 94$

The mass number of Uranium, $A_{U} = 236$

The electric potential of the uniformly charged sphere, $U = \frac{3}{5} (\frac{Q^{2}}{4 {πε}_{0} r})$

Step 2: Determine the formulas:

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

$r = r_{0} A^{1 / 3}$ …… (1)

Here, A is the mass number.

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

$a = 4 π r^{2}$ …… (2)

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

$V = \frac{4}{3} {πr}^{3}$ …… (3)

Step 3: (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 $r = r_{0} A^{1 / 3}$ for r into equation (2).

$r = 4 π r_{0} {(A^{1 / 3})}^{2} = 4 π r_{0} A^{2 / 3}$

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.

$a_{r} = 4 π r_{0} (A_{X e}^{2 / 3} + A_{S r}^{2 / 3})$

Calculate the surface area before the fission process.

$a_{r} = 4 π r_{0} A_{U}^{2 / 3}$

Calculate the percentage change in the surface.

$\frac{∆ a}{a} = \frac{A_{f} - A_{i}}{A_{i}}$

Substitute $4 π r_{0} (A_{X e}^{2 / 3} + A_{S r}^{2 / 3})$ for $a_{f}$ and $4 π r_{0} A_{U}^{2 / 3}$ for $a_{i}$ into above equation.

$\begin{aligned} \frac{Δ a}{a} = \frac{4 π r_{0} (A_{X_{e}}^{2 / β} + A_{S r}^{2 / β}) - 4 π r_{0} A_{U}^{2 / 3}}{4 π r_{0} A_{U}^{2 / 3}} \\ = \frac{4 π r_{0} (A_{χ_{e}}^{2 / β} + A_{S r}^{2 / β} - A_{U}^{2 / 3})}{4 π r_{0} A_{U}^{2 / β}} \\ = \frac{(A_{X e}^{2 / 3} + A_{s r}^{2 / 3} - A_{U}^{2 / β})}{A_{U}^{2 / 3}} \end{aligned}$

Substitute 236 for $A_{u}$ ,140 for $A_{X e}$ and 96 for $A_{S r}$ into above equation.

$\frac{∆ a}{a} = (\frac{140^{2 / 3} + 96^{2 / 3} - 236^{2 / 3}}{236^{2 / 3}}) = \frac{26.9619 + 20.9659 - 38.1892}{38.1892} = \frac{9.7386}{38.1892} = 0.255$

$\frac{∆ a}{a} % = 0.255 \times 100 % = 25.5 %$

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

Step 4: (b) Calculate the percentage change in the volume of the sphere:

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

Substitute $r_{0} A^{1 / 3}$ for r into equation (3).

$V = \frac{4}{3} π r_{0} {(A^{1 / 3})}^{2} = \frac{4}{3} π r_{0} A$

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.

$V_{r} = \frac{4}{3} π r_{0} (A_{X e} + A_{S r})$

Calculate the volume before the fission process.

$V_{i} = \frac{4}{3} π r_{0} A_{U}$

Calculate the percentage change in the surface.

$\frac{∆ V}{V} = \frac{V_{f} - V_{i}}{V_{i}}$

Substitute role="math" localid="1661930807586" $\frac{4}{3} π r_{0} (A_{X e} + A_{S r})$ for $V_{f}$ and $\frac{4}{3} π r_{0} A_{U}$ for $V_{i}$ into above equation.

role="math" localid="1661931138966" $\frac{∆ V}{V} = \frac{\frac{4}{3} π r_{0} (A_{X e} + A_{S r}) - \frac{4}{3} π r_{0} A_{U}}{\frac{4}{3} π r_{0} A_{U}} = \frac{\frac{4}{3} π r_{0} (A_{X e} + A_{S r} - A_{U})}{\frac{4}{3} π r_{0} A_{U}} = \frac{(A_{X e} + A_{S r} - A_{U})}{A_{U}}$

Substitute 236 for $A_{U}$ , 140 for $A_{X e}$ and 96 for $A_{S r}$ into above equation.

$\frac{∆ v}{v} = \frac{(140 + 96 - 236)}{236} = \frac{236 - 236}{236} = 0$

Hence the percentage change in the volume is 0%.

Step 5: (c) Calculate the percentage change in electrical potential energy:

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

Therefore,

$Q_{u} = Z_{u} e Q_{X e} = Z_{X e} e Q = Z_{S r} e$

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

role="math" localid="1661931639748" $U = \frac{3}{5} \frac{1}{4 π ε_{0} r_{0}} \frac{{(Z e)}^{2}}{r_{0} A^{1 / 3}} = \frac{3}{5} \frac{1}{4 π ε_{0} r_{0}} \frac{{(Z e)}^{2}}{A^{1 / 3}}$

Calculate the electrical potential energy after the fission process.

role="math" localid="1661931904914" $\begin{aligned} U_{f} = \frac{3}{5} \frac{1}{4 π ε_{0} r_{0}} [\frac{{(Z_{χ_{e} e})}^{2}}{A_{χ_{e}}^{1 / 3}} + \frac{{(Z_{S r} e)}^{2}}{A_{S r}^{1 / 3}}] \\ = \frac{3}{5} \frac{e^{2}}{4 π ε_{0} r_{0}} [\frac{{(Z_{X_{e}})}^{2}}{A_{χ_{e}}^{1 / 3}} + \frac{{(Z_{S r})}^{2}}{A_{S r}^{1 / 3}}] \end{aligned}$

Calculate the volume before the fission process.

$\begin{aligned} U_{f} = \frac{3}{5} \frac{1}{4 π ε_{0} r_{0}} \frac{{(Z_{_{U} e})}^{2}}{A_{U}^{1 / 3}} \\ = \frac{3}{5} \frac{e^{2}}{4 π ε_{0} r_{0}} \frac{{(Z_{X_{U}})}^{2}}{A_{U}^{1 / 3}} \end{aligned}$

Calculate the percentage change in the surface.

$\frac{∆ U}{U} = \frac{U_{f} - U_{i}}{U_{i}}$

Substitute $\begin{array}{r} \frac{3}{5} \frac{e^{2}}{4 π ε_{0} r_{0}} [\frac{{(Z_{χ_{e}})}^{2}}{A_{χ_{e}}^{1 / 3}} + \frac{{(Z_{S r})}^{2}}{A_{S r}^{1 / 3}}] \end{array}$ for $U_{f}$ and role="math" localid="1661932204802" $\frac{3}{5} \frac{e^{2}}{4 π ε_{0} r_{0}} \frac{{(Z_{U})}^{2}}{A_{U}^{1 / 3}}$ for $U_{f}$ into above equation.

role="math" localid="1661932416965" $\frac{∆ U}{U} = \frac{\frac{3}{5} \frac{e^{2}}{4 π ε_{0} r_{0}} [\frac{{(Z_{χ_{e}})}^{2}}{A_{χ_{e}}^{1 / 3}} + \frac{{(Z_{S r})}^{2}}{A_{S r}^{1 / 3}}] - \frac{3}{5} \frac{e^{2}}{4 π ε_{0} r_{0}} \frac{{(Z_{U})}^{2}}{A_{U}^{1 / 3}}}{\frac{3}{5} \frac{e^{2}}{4 π ε_{0} r_{0}} \frac{{(Z_{U})}^{2}}{A_{U}^{1 / 3}}} \frac{∆ U}{U} = \frac{\frac{3}{5} \frac{e^{2}}{4 π ε_{0} r_{0}} [\frac{{(Z_{χ_{e}})}^{2}}{A_{χ_{e}}^{1 / 3}} + \frac{{(Z_{S r})}^{2}}{A_{S r}^{1 / 3}} - \frac{{(Z_{U})}^{2}}{A_{U}^{1 / 3}}]}{\frac{3}{5} \frac{e^{2}}{4 π ε_{0} r_{0}} \frac{{(Z_{U})}^{2}}{A_{U}^{1 / 3}}} = \frac{\frac{{(Z_{χ_{e}})}^{2}}{A_{χ_{e}}^{1 / 3}} + \frac{{(Z_{S r})}^{2}}{A_{S r}^{1 / 3}} - \frac{{(Z_{U})}^{2}}{A_{U}^{1 / 3}}}{\frac{{(Z_{U})}^{2}}{A_{U}^{1 / 3}}}$

Substitute 236 for $A_{U}$ , 140 for $A_{X e}$ , 96 for $A_{S r}$ ,54 for $Z_{X e}$ ,38 for $Z_{S r}$ and 92 for $Z_{U}$ into above equation.

$\frac{∆ U}{U} = \frac{\frac{{(54)}^{2}}{140^{1 / 3}} + \frac{{(38)}^{2}}{96^{1 / 3}} + \frac{{(92)}^{2}}{236^{1 / 3}}}{\frac{{(92)}^{2}}{236^{1 / 3}}} = \frac{561.5798 + 315.3625 + 1369.6354}{1369.6354} = \frac{- 492.6931}{1369.6354} = - 0.359 \frac{∆ U}{U} % = - 0.359 \times 100 % = - 35.9 %$

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

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Short Answer

Step by Step Solution

Step 1: Write the given data from the question.

Step 2: Determine the formulas:

Step 3: (a) Calculate the percentage change in the surface area of the sphere:

Step 4: (b) Calculate the percentage change in the volume of the sphere:

Step 5: (c) Calculate the percentage change in electrical potential energy:

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Step by Step Solution

Step 1: Write the given data from the question.

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Step 3: (a) Calculate the percentage change in the surface area of the sphere:

Step 4: (b) Calculate the percentage change in the volume of the sphere:

Step 5: (c) Calculate the percentage change in electrical potential energy:

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