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Q8P

Expert-verifiedFound in: Page 1331

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**(a) Calculate the disintegration energy Q for the fission of the molybdenum isotope ${}^{{\mathit{98}}}{\mathit{M}}{\mathit{o}}$ into two equal parts. The masses you will need are 97.90541u for ${}^{{\mathbf{98}}}{\mathit{M}}{\mathit{o}}$ and 48.95002u for ${}^{{\mathbf{49}}}{\mathit{S}}{\mathit{c}}$. (b) If Q turns out to be positive, discuss why this process does not occur spontaneously.**

(a) The disintegrated energy is 5 MeV.

(b) Due to the energy barrier the process is not spontaneously.

The mass of ${}^{98}Mo,{m}_{Mo}=97.90541u$,

The mass of ${}^{49}Sc,{m}_{SC}=48.95002u$,

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

${\mathit{Q}}{\mathbf{=}}{\mathbf{-}}{\mathbf{\u2206}}{\mathit{m}}{{\mathit{c}}}^{{\mathbf{2}}}\phantom{\rule{0ex}{0ex}}{\mathit{Q}}{\mathbf{=}}{\left({m}_{Mo}-2{m}_{SC}\right)}{{\mathit{c}}}^{{\mathbf{2}}}$ ...(i)

Consider the fission reaction as given follow.

${}^{98}Mo\to {}^{49}Sc+{}^{49}Sc$

Calculate the disintegrated energy.

Substitute 97.90514u for ${m}_{Mo}$,48.95002u for ${m}_{Sc}$ and 931.5 MeV/u for ${c}^{2}$ into equation (i).

$\mathrm{Q}=(97.90541\mathrm{u}-2\times 48.95002\mathrm{u})931.5\mathrm{MeV}/\mathrm{u}\phantom{\rule{0ex}{0ex}}\mathrm{Q}=0.00537\times 931.5\mathrm{MeV}\phantom{\rule{0ex}{0ex}}\mathrm{Q}=5\hspace{0.33em}\mathrm{MeV}$

Hence the disintegrated energy is 5 MeV.

From the part (a), the disintegrated energy is +5Mev. The nucleus must overcome the barrier of the energy which is larger than the +5MeV. Therefore, the positive sign of disintegrated energy does not indicate that the process is spontaneously. Therefore, due to the energy barrier the process is not spontaneously.

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