Suggested languages for you:

Americas

Europe

Q8P

Expert-verified
Found in: Page 1331

### Fundamentals Of Physics

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 ${}^{{\mathbit{98}}}{\mathbit{M}}{\mathbit{o}}$ into two equal parts. The masses you will need are 97.90541u for ${}^{{\mathbf{98}}}{\mathbit{M}}{\mathbit{o}}$ and 48.95002u for ${}^{{\mathbf{49}}}{\mathbit{S}}{\mathbit{c}}$. (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.

See the step by step solution

## Step 1: Given data

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

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

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

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

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

## Step 3: (a) Calculate the value of disintegrated energy.

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).

$\mathrm{Q}=\left(97.90541\mathrm{u}-2×48.95002\mathrm{u}\right)931.5\mathrm{MeV}/\mathrm{u}\phantom{\rule{0ex}{0ex}}\mathrm{Q}=0.00537×931.5\mathrm{MeV}\phantom{\rule{0ex}{0ex}}\mathrm{Q}=5 \mathrm{MeV}$

Hence the disintegrated energy is 5 MeV.

## Step 4: (b) Determine the why the process does not occur spontaneously.

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.