Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

In an atomic bomb, energy release is due to the uncontrolled fission of plutonium ${}^{239}{Pu}$ (or ${}^{235}U$ ). The bomb’s rating is the magnitude of the released energy, specified in terms of the mass of TNT required to produce the same energy release. One megaton of TNT releases $2.6 \times 10^{28} MeV$ of energy. (a) Calculate the rating, in tons of TNT, of an atomic bomb containing 95 kg of ${}^{239}{Pu}$ , of which 2.5 kg actually undergoes fission. (See Problem 4.) (b) Why is the other 92.5 kg of ${}^{239}{Pu}$ needed if it does not fission?

(a) The rating of the atomic bomb is $4.36 \times 10^{4} ton$ .

(b) The other 92.5 kg of plutonium is used to increase the process of fission.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Write the given data from the question:

The total mass of the atomic bomb, $M_{t} = 95 kg$

The mass of atomic bomb that actually undergoes fission, $m = 2.5 kg$

The molar mass of plutonium, $M = 239 g / mol$

Step 2: Determine the formulas to calculate the rating of the bomb:

The expression to calculate the fission energy is given as follows.

$E = \frac{{mN}_{A} Q}{M}$ …… (1)

Here, $N_{A}$ is the Avogadro number $(6.022 \times 10^{23} {mol}^{- 1})$ , and Q is the average energy released in per fission.

Step 3: Calculate the rating of the atomic bomb:

The average energy released in the per fission is $Q = 180 MeV$ .

Calculate the fission energy.

Substitute 2.5 kg for m, $6.022 \times 10^{23} {mol}^{- 1} for N_{A}, 239 g / mol$ for M and 180MeV for Q into equation (1).

$E = \frac{n M_{A} Q}{M} = \frac{2.5 \times 1000 \times 6.022 \times 10^{23} \times 180}{239} = \frac{2709.0 \times 10^{26}}{239} = 11.338 \times 10^{26} MeV$

As, 1 of TNT releases $2.6 \times 10^{28} MeV$ .

The rating of the atomic bomb is given by,

$Rating = \frac{11.338 \times 10^{26}}{2.6 \times 10^{28}} = 4.36 \times 10^{- 2} megaton = 4.36 \times 10^{4} ton$

Hence, the rating of the atomic bomb is $4.36 \times 10^{4} ton$ .

Step 4: Why is the other 92.5 kg of Pu239 needed if it does not fission?

In the atomic bomb, only 2.5 kg of plutonium undergoes the fission process. Due to the fission process, many Neutrons release and collide with the other 92.5kg of plutonium, increasing the fission process and producing a large amount of energy.

Hence, the other 92.5 kg of plutonium is used to increase the process of fission.

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Step by Step Solution

Step 1: Write the given data from the question:

Step 2: Determine the formulas to calculate the rating of the bomb:

Step 3: Calculate the rating of the atomic bomb:

Step 4: Why is the other 92.5 kg of Pu239 needed if it does not fission?

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Fundamentals Of Physics

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Step by Step Solution

Step 1: Write the given data from the question:

Step 2: Determine the formulas to calculate the rating of the bomb:

Step 3: Calculate the rating of the atomic bomb:

Step 4: Why is the other 92.5 kg of Pu239 needed if it does not fission?

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