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Q11P

Expert-verifiedFound in: Page 1331

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Calculate the disintegration energy Q for the fission of ${}^{{\mathbf{52}}}{\mathit{C}}{\mathit{r}}$ into two equal fragments. The masses you will need are **

**role="math" localid="1661753124790" ${}^{{\mathbf{52}}}{\mathit{C}}{\mathit{r}}{\mathbf{}}{\mathbf{}}{\mathbf{51}}{\mathbf{.}}{\mathbf{94051}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{}^{{\mathbf{26}}}{\mathit{M}}{\mathit{g}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{25}}{\mathbf{.}}{\mathbf{98259}}{\mathit{u}}{\mathbf{}}{\mathbf{}}$**

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

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

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

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

${\mathit{Q}}{\mathbf{=}}{\mathbf{\u2010}}{\mathbf{\u2206}}{\mathit{m}}{{\mathit{c}}}^{{\mathbf{2}}}\phantom{\rule{0ex}{0ex}}{\mathit{Q}}{\mathbf{=}}{\left({m}_{Cr}\u20102{m}_{Mg}\right)}{{\mathit{c}}}^{{\mathbf{2}}}$ ...(i)

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).

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

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

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