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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 # Calculate the disintegration energy Q for the fission of ${}^{{\mathbf{52}}}{\mathbit{C}}{\mathbit{r}}$ into two equal fragments. The masses you will need are role="math" localid="1661753124790" ${}^{{\mathbf{52}}}{\mathbit{C}}{\mathbit{r}}{\mathbf{}}{\mathbf{}}{\mathbf{51}}{\mathbf{.}}{\mathbf{94051}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{}^{{\mathbf{26}}}{\mathbit{M}}{\mathbit{g}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{25}}{\mathbf{.}}{\mathbf{98259}}{\mathbit{u}}{\mathbf{}}{\mathbf{}}$

The disintegrated energy is $-23\mathrm{MeV}$.

See the step by step solution

## Step 1: Given data

The mass of ${}^{52}Cr,{m}_{cr}=51.94051u$

The mass of ${}^{26}Mg,{m}_{Mg}=25.98259u$

## Step 2: Determine the formula to calculate the disintegrated energy.

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

${\mathbit{Q}}{\mathbf{=}}{\mathbf{‐}}{\mathbf{∆}}{\mathbit{m}}{{\mathbit{c}}}^{{\mathbf{2}}}\phantom{\rule{0ex}{0ex}}{\mathbit{Q}}{\mathbf{=}}\left({m}_{Cr}‐2{m}_{Mg}\right){{\mathbit{c}}}^{{\mathbf{2}}}$ ...(i)

## Step 3: Calculate the value of disintegrated energy.

Consider the fission reaction as given follow.

${}^{52}Cr\to {}^{26}Mg+{}^{26}Mg$

Calculate the disintegrated energy.

Substitute $51.94051u$ for ${m}_{Cr}$ , $25.98259u$ for ${M}_{Mg}$ and $931.5MeV/u$ for ${c}^{2}$ into equation (i).

$Q=\left(51.9405\mathrm{u}-2×25.98259\mathrm{u}\right)931.5\mathrm{MeV}/\mathrm{u}\phantom{\rule{0ex}{0ex}}Q=-0.02467×931.5\mathrm{MeV}\phantom{\rule{0ex}{0ex}}Q=-23\mathrm{MeV}$

Hence the disintegrated energy is $-23\mathrm{MeV}$. ### Want to see more solutions like these? 