Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

Question: A $66$ kiloton atomic bomb is fueled with pure ${}^{235}U$ (Fig. 43-14), $4.0 %$ of which actually undergoes fission. (a) What is the mass of the uranium in the bomb? (It is not 66 kilotons—that is the amount of released energy specified in terms of the mass of TNT required to produce the same amount of energy.) (b) How many primary fission fragments are produced? (c) How many fission neutrons generated are released to the environment? (On average, each fission produces 2.5 neutrons.)

(a) The mass of the Uranium is 84kg.

(b) The number of the produced fragment is 1.71 $\times 10^{25}$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Write the given data from the question:

The mass, m = 66kilotonn

The 4% of the total mass is actually fissionable.

Step 2: Determine the formulas:

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

$E = m c^{2}$ …… (1)

The expression to calculate the number of fission is given as follows.

$n = \frac{E}{Q}$ …… (2)

Here, Q is the energy per fission.

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

n = 2n …… (3)

Step 3: (a) Calculate the mass of the uranium in the bomb:

The typical energy released per fission is Q = 200MeV

The mass in megaton is $m = 66 \times 10^{- 3} Megatron$ .

Calculate the released energy.

Substitute $66 \times 10^{- 3} Megatron$ for m and $2.6 \times 10^{28} Mev / Meagatron$ for $c^{2}$ into equation(1).

role="math" $E = m c^{2} = 66 \times 10^{- 3} \times 2.6 \times 10^{28} = 171.6 \times 10^{25} = 1.716 \times 10^{27} MeV$

Calculate the number of the fission.

Substitute $1.716 \times 10^{27} MeV$ for E and 200MeV for Q into equation (2).

$n = \frac{E}{Q} = \frac{1.71 \times 10^{27}}{200} = 8.55 \times 10^{24}$

Since the 4% of the total mass is actually fissionable. Therefore, the original nuclei present is,

$\frac{8.6 \times 10^{24}}{0.04} = 2.15 \times 10^{26}$

Calculate the mass of the uranium.

$m_{U} = 2.15 \times 10^{26} \times 235 \times u = 2.15 \times 10^{26} \times 235 \times 1.66 \times 10^{- 27} = 83.87 k g$

By rounding up the value of the mass of the uranium is 84kg.

Hence, the mass of the Uranium is 84kg.

Step 4: (b) Calculate the number of the produce fragment:

Calculate the number of the fragment.

Substitute $8.55 \times 10^{24}$ for n into equation (3).

$N = 2 n = 2 \times 8.55 \times 10^{24} = 17.1 \times 10^{24} = 1.71 \times 10^{25}$

Hence, the number of the produce fragment is $1.71 \times 10^{25}$ .

Step 5: (c) Calculate the number of fission neutrons generated are released to the environment:

It is given that, on average, each fission produces 2.5 neutrons. Therefore, number of the neutron released to the environment would be the 2.5 time the number of the fission.

$N_{n} = 2.5 \times 8.55 \times 10^{24} = 21.3 \times 10^{24} = 2.13 \times 10^{25}$

Hence, the number of the neutron released to the environment is $2.13 \times 10^{25}$ .

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Fundamentals Of Physics

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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 mass of the uranium in the bomb:

Step 4: (b) Calculate the number of the produce fragment:

Step 5: (c) Calculate the number of fission neutrons generated are released to the environment:

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Fundamentals Of Physics

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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 mass of the uranium in the bomb:

Step 4: (b) Calculate the number of the produce fragment:

Step 5: (c) Calculate the number of fission neutrons generated are released to the environment:

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