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### Fundamentals Of Physics

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$×{10}^{25}$.

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

See the step by step solution

## 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.

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

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

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

## 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 $m=66×{10}^{-3}\mathrm{Megatron}$.

Calculate the released energy.

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

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

Calculate the number of the fission.

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

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

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

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

Calculate the mass of the uranium.

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

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

Hence, the mass of the Uranium is 84kg.

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

Calculate the number of the fragment.

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

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

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

## 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.

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

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