Suggested languages for you:

Americas

Europe

Q15P

Expert-verifiedFound in: Page 1331

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Question: A ${\mathbf{66}}$ kiloton atomic bomb is fueled with pure ${}^{{\mathbf{235}}}{\mathbf{U}}$ (Fig. 43-14), ${\mathbf{4}}{\mathbf{.}}{\mathbf{0}}{\mathbf{\%}}$ 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$\times {10}^{25}$.

(c) The number of the neutron released to the environment is $2.13\times {10}^{25}$.

The mass, m = 66kilotonn

The 4% of the total mass is actually fissionable.

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

**${\mathit{E}}{\mathbf{=}}{\mathit{m}}{{\mathit{c}}}^{{\mathbf{2}}}$ **** …… (1)**

** **

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

**${\mathit{n}}{\mathbf{=}}\frac{\mathbf{E}}{\mathbf{Q}}$ **** …… (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)**

The typical energy released per fission is Q = 200MeV

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

Calculate the released energy.

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

role="math" $E=m{c}^{2}\phantom{\rule{0ex}{0ex}}=66\times {10}^{-3}\times 2.6\times {10}^{28}\phantom{\rule{0ex}{0ex}}=171.6\times {10}^{25}\phantom{\rule{0ex}{0ex}}=1.716\times {10}^{27}\mathrm{MeV}$

Calculate the number of the fission.

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

$n=\frac{E}{Q}\phantom{\rule{0ex}{0ex}}=\frac{1.71\times {10}^{27}}{200}\phantom{\rule{0ex}{0ex}}=8.55\times {10}^{24}$

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.

${m}_{U}=2.15\times {10}^{26}\times 235\times u\phantom{\rule{0ex}{0ex}}=2.15\times {10}^{26}\times 235\times 1.66\times {10}^{-27}\phantom{\rule{0ex}{0ex}}=83.87kg$

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

Hence, the mass of the Uranium is 84kg.

Calculate the number of the fragment.

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

$N=2n\phantom{\rule{0ex}{0ex}}=2\times 8.55\times {10}^{24}\phantom{\rule{0ex}{0ex}}=17.1\times {10}^{24}\phantom{\rule{0ex}{0ex}}=1.71\times {10}^{25}$

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

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.

${N}_{n}=2.5\times 8.55\times {10}^{24}\phantom{\rule{0ex}{0ex}}=21.3\times {10}^{24}\phantom{\rule{0ex}{0ex}}=2.13\times {10}^{25}$

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

94% of StudySmarter users get better grades.

Sign up for free