Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

Calculate and compare the energy released by (a) the fusion of 1.0 kg of hydrogen deep within the Sun and (b) the fission of 1.0 kg of ${}^{235}U$ in a fission reactor.

The energy released by the fusion of 1.0 kg hydrogen deep within the Sun is $6.4 \times 10^{14} J$ .
The energy released by the fission of 1.0 kg of ${}^{235}U$ in a fission reactor is $8.2 \times 10^{13} J$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given data

Mass of hydrogen, $m_{H} = 1.0 k g o r 1000 g$
Mass of ${}^{235}U$ , $m_{235_{U}} = 1.0 k g o r 1000 g$

Step 2: Understanding the concept of fusion and fission

In a fusion reaction, two or more light nuclei react with each other giving a heavier nucleus as the product releasing some energy. While in a fission reaction, a heavier unstable nucleus breaks down into two or daughter nuclei giving out some energy with it. In the hydrogen deep reaction, 4 protons undergo a fusion reaction. Similarly, if the uranium-235 nucleus undergoes a fission reaction, then the energy released is calculated as the total energy released with the number of particles being released.

Formula:

The number of particles in an atom is as follows

$N = \frac{m}{M} N_{A}$ …… (i):

Here, $N_{A} = 6.022 \times 10^{23} / m o l$

Here, m is the given mass and M is the molar mass of the atom.

Step 3: a) Calculate the energy released by the fusion process

Given the energy release per fusion in the overall fusion cycle $(Q = 26.7 MeV or 4.28 \times 10^{- 12} J)$ and also four protons are consumed in each fusion event. Now, the number of particles released in the reaction can be found using the given data in equation (i) for the four protons as follows:

$N = \frac{m_{H}}{4 (M_{H})} N_{A} = \frac{1000 g}{4 (1.0 \frac{g}{m o l})} (6.022 \times 10^{23} m o l^{- 1}) = 1.5 \times 10^{26}$

Now, the total energy released by the fusion reaction can be given as follows:

$Q_{t o t a l} = (1.5 \times 10^{2}) (4.28 \times 10^{- 12} J) = 6.4 \times 10^{14} J$

Hence, the amount of energy released is $6.4 \times 10^{14} J$ .

Step 4: b) Calculate the energy released by the fission process

Now, the number of particles in the uranium-235 nuclei can be calculated using the given data in equation (i) as follows:

$N = \frac{1000 g}{(235.0 g / m o l)} (6.022 \times 10^{23} / m o l) = 2.56 \times 10^{24}$

If all the U-235 nuclei fission, the total energy released in the fission process (using the result of Eq. 43-6, $Q_{f i s s o n} = 200 M e V$ ) is given as follows:

$Q_{t o t a l} = (2.56 \times 10^{24}) (200 M e V) = 5.1 \times 10^{26} M e V = 8.2 \times 10^{13} J$

Hence, the amount of energy released is $8.2 \times 10^{13} J$ .

Consider the fusion process (with regard to a unit mass of fuel) produces a larger amount of energy (despite the fact that the Q value per event is smaller).

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Fundamentals Of Physics

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Short Answer

Calculate and compare the energy released by (a) the fusion of 1.0 kg of hydrogen deep within the Sun and (b) the fission of 1.0 kg of ${}^{235}U$ in a fission reactor.

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of fusion and fission

Step 3: a) Calculate the energy released by the fusion process

Step 4: b) Calculate the energy released by the fission process

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Fundamentals Of Physics

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Short Answer

Calculate and compare the energy released by (a) the fusion of 1.0 kg of hydrogen deep within the Sun and (b) the fission of 1.0 kg of U235 in a fission reactor.

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of fusion and fission

Step 3: a) Calculate the energy released by the fusion process

Step 4: b) Calculate the energy released by the fission process

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Calculate and compare the energy released by (a) the fusion of 1.0 kg of hydrogen deep within the Sun and (b) the fission of 1.0 kg of ${}^{235}U$ in a fission reactor.