Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

Question: Assume that immediately after the fission of ${}^{236}U$ according to Eq. 43-1, the resulting ${}^{140}{Xe} and {}^{94}{Sr}$ nuclei are just touching at their surfaces. (a) Assuming the nuclei to be spherical, calculate the electric potential energy associated with the repulsion between the two fragments. (Hint: Use Eq. 42-3 to calculate the radii of the fragments.) (b) Compare this energy with the energy released in a typical fission event.

(a) The potential electrical energy is 251 MeV.

(b) The electrical potential energy is greater than typical fission energy.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

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

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

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

Step 2: Determine the formulas to calculate the electric potential energy

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

$r = r_{0} A^{1 / 3}$ ...(i)

Here, A is the mass number.

The expression to calculate the electrical potential energy is given as follows.

$W = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r}$ ...(ii)

Here, $q_{1} . q_{2}$ are the charges and r is the distance between the charges.

Step 3: (a) Calculate the electrical potential energy.

The atomic number of Strontium, $Z_{S r} = 38$

The atomic number of Xenon, $Z_{X e} = 54$

Calculate the radii of the Xenon,

Substitute 1.2 fm for $r_{0}$ and 140 for A into equation (i).

$r_{X e} = 1.2 fm \times {(140)}^{1 / 3} r_{X e} = 1.2 fm \times 5.1924 r_{X e} = 6.2 fm$

Calculate the radii of the strontium,

Substitute 1.2 fm for $r_{0}$ and 94 for A into equation (i).

$r_{S r} = 1.2 fm \times {(94)}^{1 / 3} r_{S r} = 1.2 fm \times 4.546 r_{S r} = 5.48 fm$

Calculate the electrical potential energy.

Substitute $Z_{X e}$ for $q_{1}$ , $Z_{S r}$ for $q_{2}$ , $r_{S r} + r_{X e}$ for r into above equation (i).

$W = \frac{1}{4 π \times ε_{0}} \times \frac{Z_{X e} Z_{S r}}{(r_{X e} + r_{S r})}$

Substitute 6.24 for $r_{X e}$ , 5.48 for $r_{S r}$ , 54e for $Z_{X e}$ and 38e for $Z_{X e}$ into above equation.

localid="1661927633337" $\begin{aligned} W = \frac{1}{4 π \times (8.85 \times 10^{- 12} C^{2} / {Nm}^{2})} \times \frac{54 e \times 38 e}{(6.24 + 5.48) \times 10^{- 15} m} \\ W = 8.99 \times 10^{9} \frac{{Nm}^{2}}{C^{2}} \times \frac{2052 e^{2}}{11.72 \times 10^{- 15} m} \end{aligned}$

Substitute $1.6 \times 10^{- 19} C$ for e into above equation.

$\begin{aligned} W = 8.99 \times 10^{9} \frac{{Nm}^{2}}{C^{2}} \times \frac{2052 {(1.6 \times 10^{- 19} C)}^{2}}{11.72 \times 10^{- 15} m} \\ W = \frac{18447.48 \times 2.56 \times 10^{24} \times 10^{- 38} J}{11.72} \\ W = \frac{47225.54 \times 10^{- 14} J}{11.72} \\ W = 4029 \times 10^{- 11} J \end{aligned}$

Convert the energy into MeV,

$W = \frac{4.033 \times 10^{- 11}}{1.602 \times 10^{- 19}} eV W = 2.515 \times 10^{8} eV W = 251.5 MeV$

By rounding down the value of the energy is 251 MeV.

Hence the potential electrical energy is 251 MeV.

Step 4: (b) Compare the electrical potential energy with typical fission energy.

The energy released in the typical fission energy is 200 MeV and the electrical potential energy from the part (a) is 251 MeV . By comparing the two energies, it is clear that the electrical potential energy is greater than the typical fission energy. This energy appears in the form of kinetic, beat and sound energy.

Hence the electrical potential energy is greater than typical fission energy.

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Given data

Step 2: Determine the formulas to calculate the electric potential energy

Step 3: (a) Calculate the electrical potential energy.

Step 4: (b) Compare the electrical potential energy with typical fission energy.

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