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Expert-verified Found in: Page 1331 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # In an atomic bomb, energy release is due to the uncontrolled fission of plutonium ${}^{{\mathbf{239}}}{\mathbf{Pu}}$ (or ${}^{{\mathbf{235}}}{\mathbf{U}}$ ). The bomb’s rating is the magnitude of the released energy, specified in terms of the mass of TNT required to produce the same energy release. One megaton of TNT releases ${\mathbf{2}}{\mathbf{.}}{\mathbf{6}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{28}}}{\mathbf{}}{\mathbf{MeV}}$ of energy. (a) Calculate the rating, in tons of TNT, of an atomic bomb containing 95 kg of ${}^{{\mathbf{239}}}{\mathbf{Pu}}$, of which 2.5 kg actually undergoes fission. (See Problem 4.) (b) Why is the other 92.5 kg of ${}^{{\mathbf{239}}}{\mathbf{Pu}}$ needed if it does not fission?

(a) The rating of the atomic bomb is $4.36×{10}^{4}\mathrm{ton}$.

(b) The other 92.5 kg of plutonium is used to increase the process of fission.

See the step by step solution

## Step 1: Write the given data from the question:

The total mass of the atomic bomb, ${M}_{t}=95\mathrm{kg}$

The mass of atomic bomb that actually undergoes fission, $m=2.5\mathrm{kg}$

The molar mass of plutonium, $M=239\mathrm{g}/\mathrm{mol}$

## Step 2: Determine the formulas to calculate the rating of the bomb:

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

${\mathbf{E}}{\mathbf{=}}\frac{{\mathbf{mN}}_{\mathbf{A}}\mathbf{Q}}{\mathbf{M}}$ …… (1)

Here, ${{\mathbf{N}}}_{{\mathbf{A}}}$ is the Avogadro number $\left(6.022×{10}^{23}{\mathrm{mol}}^{-1}\right)$, and Q is the average energy released in per fission.

## Step 3: Calculate the rating of the atomic bomb:

The average energy released in the per fission is $\mathrm{Q}=180\mathrm{MeV}$.

Calculate the fission energy.

Substitute 2.5 kg for m, $6.022×{10}^{23}{\mathrm{mol}}^{-1}\mathrm{for}{\mathrm{N}}_{\mathrm{A}},239\mathrm{g}/\mathrm{mol}$ for M and 180MeV for Q into equation (1).

$E=\frac{n{M}_{A}Q}{M}\phantom{\rule{0ex}{0ex}}=\frac{2.5×1000×6.022×{10}^{23}×180}{239}\phantom{\rule{0ex}{0ex}}=\frac{2709.0×{10}^{26}}{239}\phantom{\rule{0ex}{0ex}}=11.338×{10}^{26}\mathrm{MeV}$

As, 1 of TNT releases $2.6×{10}^{28}\mathrm{MeV}$ .

The rating of the atomic bomb is given by,

$\mathrm{Rating}=\frac{11.338×{10}^{26}}{2.6×{10}^{28}}\phantom{\rule{0ex}{0ex}}=4.36×{10}^{-2}\mathrm{megaton}\phantom{\rule{0ex}{0ex}}=4.36×{10}^{4}\mathrm{ton}$

Hence, the rating of the atomic bomb is $4.36×{10}^{4}\mathrm{ton}$.

## Step 4: Why is the other 92.5 kg of Pu239 needed if it does not fission?

In the atomic bomb, only 2.5 kg of plutonium undergoes the fission process. Due to the fission process, many Neutrons release and collide with the other 92.5kg of plutonium, increasing the fission process and producing a large amount of energy.

Hence, the other 92.5 kg of plutonium is used to increase the process of fission. ### Want to see more solutions like these? 