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Q40P

Expert-verifiedFound in: Page 1332

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**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 ${}^{{\mathbf{235}}}{\mathit{U}}$**** in a fission reactor.**

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

- Mass of hydrogen, ${m}_{H}=1.0kgor1000g$
- Mass of ${}^{235}U$, ${m}_{{235}_{U}}=1.0kgor1000g$

**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, ${N}_{A}=6.022\times {10}^{23}/mol$

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

Given the energy release per fusion in the overall fusion cycle $\left(\mathrm{Q}=26.7\mathrm{MeV}\mathrm{or}4.28\times {10}^{-12}\mathrm{J}\right)$ 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:

$N=\frac{{m}_{H}}{4\left({M}_{H}\right)}{N}_{A}\phantom{\rule{0ex}{0ex}}=\frac{1000g}{4\left(1.0\frac{g}{mol}\right)}\left(6.022\times {10}^{23}mo{l}^{-1}\right)\phantom{\rule{0ex}{0ex}}=1.5\times {10}^{26}$

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

${Q}_{total}=(1.5\times {10}^{2})(4.28\times {10}^{-12}J)\phantom{\rule{0ex}{0ex}}=6.4\times {10}^{14}J$

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

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

$N=\frac{1000g}{\left(235.0g/mol\right)}\left(6.022\times {10}^{23}/mol\right)\phantom{\rule{0ex}{0ex}}=2.56\times {10}^{24}$

If all the U-235 nuclei fission, the total energy released in the fission process (using the result of Eq. 43-6, ${Q}_{fisson}=200MeV$) is given as follows:

${Q}_{total}=\left(2.56\times {10}^{24}\right)\left(200MeV\right)\phantom{\rule{0ex}{0ex}}=5.1\times {10}^{26}MeV\phantom{\rule{0ex}{0ex}}=8.2\times {10}^{13}J$

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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