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Found in: Page 1305

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# Large radionuclides emit an alpha particle rather than other combinations of nucleons because the alpha particle has such a stable, tightly bound structure. To confirm this statement, calculate the disintegration energies for these hypothetical decay processes and discuss the meaning of your findings:$\left(a\right){\mathbf{}}{}^{{\mathbf{238}}}{\mathbf{U}}{\mathbf{\to }}{}^{{\mathbf{232}}}{\mathbf{Th}}{\mathbf{+}}{}^{{\mathbf{3}}}{\mathbf{He}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}\left(b\right){\mathbf{}}{}^{{\mathbf{235}}}{\mathbf{U}}{\mathbf{\to }}{}^{{\mathbf{231}}}{\mathbf{Th}}{\mathbf{+}}{}^{{\mathbf{4}}}{\mathbf{He}}{\mathbf{}}{\mathbf{}}\phantom{\rule{0ex}{0ex}}\left(c\right){\mathbf{}}{}^{{\mathbf{235}}}{\mathbf{U}}{\mathbf{\to }}{}^{{\mathbf{230}}}{\mathbf{Th}}{\mathbf{+}}{}^{{\mathbf{5}}}{\mathbf{He}}{\mathbf{}}{\mathbf{}}$The needed atomic masses arerole="math" localid="1661928659878" ${}^{{\mathbf{232}}}{\mathbf{Th}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{232}}{\mathbf{.}}{\mathbf{0381}}{\mathbf{}}{\mathbf{}}{\mathbf{u}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{}^{{\mathbf{3}}}{\mathbf{He}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{3}}{\mathbf{.}}{\mathbf{0160}}{\mathbf{}}{\mathbf{}}{\mathbf{u}}\phantom{\rule{0ex}{0ex}}{}^{{\mathbf{231}}}{\mathbf{Th}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{231}}{\mathbf{.}}{\mathbf{0363}}{\mathbf{}}{\mathbf{}}{\mathbf{u}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{}^{{\mathbf{4}}}{\mathbf{He}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{4}}{\mathbf{.}}{\mathbf{0026}}{\mathbf{}}{\mathbf{}}{\mathbf{u}}\phantom{\rule{0ex}{0ex}}{}^{{\mathbf{230}}}{\mathbf{Th}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{230}}{\mathbf{.}}{\mathbf{0331}}{\mathbf{}}{\mathbf{}}{\mathbf{u}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{}^{{\mathbf{5}}}{\mathbf{He}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{5}}{\mathbf{.}}{\mathbf{0122}}{\mathbf{}}{\mathbf{}}{\mathbf{u}}\phantom{\rule{0ex}{0ex}}{}^{{\mathbf{235}}}{\mathbit{U}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{235}}{\mathbf{.}}{\mathbf{0429}}{\mathbf{}}{\mathbf{}}{\mathbf{u}}$

1. The disintegration energy for the hypothetical decay ${}^{235}\mathrm{U}\to {}^{232}\mathrm{Th}+{}^{3}\mathrm{He}$ is - 9.50 MeV.
2. The disintegration energy for the hypothetical decay ${}^{235}\mathrm{U}\to {}^{231}\mathrm{Th}+{}^{4}\mathrm{He}$ is 4.66 MeV.
3. The disintegration energy for the hypothetical decay ${}^{235}\mathrm{U}\to {}^{230}\mathrm{Th}+{}^{5}\mathrm{He}$ is - 1.30 MeV.
See the step by step solution

## Step 1: Given data

The given atomic masses of the nuclides and alpha particles are:

${}^{232}\mathrm{Th}232.0381\mathrm{u}{}^{3}\mathrm{He}3.0160\mathrm{u}\phantom{\rule{0ex}{0ex}}{}^{231}\mathrm{Th}231.0363\mathrm{u}{}^{4}\mathrm{He}4.0026\mathrm{u}\phantom{\rule{0ex}{0ex}}{}^{230}\mathrm{Th}230.0331\mathrm{u}{}^{5}\mathrm{He}5.0122\mathrm{u}\phantom{\rule{0ex}{0ex}}{}^{235}\mathrm{U}235.0429\mathrm{u}$

## Step 2: Understanding the concept of decay

Massive nuclides tend to undergo alpha decay releasing disintegration energy. The disintegration energy, also known as the Q-value, is the energy that is absorbed or released when a nuclear reaction takes place. The Q-value is positive if the reaction is exothermic and negative if the reaction is endothermic. The potential barrier height of the nucleus indicates the energy it needs to overcome the internal forces and become an individual nucleus from the parent nucleus.

Formula:

The disintegration energy of a nuclear reaction,

…… (i)

## Step 3: a) Calculate the disintegration energy

The disintegration energy for uranium-235 “decaying” into thorium-232 is given using the atomic masses and equation (i) as follows:

$\mathrm{Q}=\left({\mathrm{m}}_{{235}_{\mathrm{U}}}-{\mathrm{m}}_{{232}_{\mathrm{Th}}}-{\mathrm{m}}_{{3}_{\mathrm{He}}}\right){\mathrm{c}}^{2}$

Substitute the values and solve as:

$Q=\left(235.0429u-232.0381u-3.0160u\right)\left(931.5\frac{MeV}{u}\right)\phantom{\rule{0ex}{0ex}}=-9.50MeV$

Hence, the disintegration energy is - 9.50 MeV.

## Step 4: b) Calculate the disintegration energy

The disintegration energy for uranium-235 decaying into thorium-231 is given using the atomic masses and equation (i) as follows:

$\mathrm{Q}=\left({\mathrm{m}}_{{235}_{\mathrm{U}}}-{\mathrm{m}}_{{231}_{\mathrm{Th}}}-{\mathrm{m}}_{{4}_{\mathrm{He}}}\right){\mathrm{c}}^{2}$

Substitute the values and solve as:

$Q=\left(235.0429u-231.0363u-4.0026u\right)\left(931.5\frac{MeV}{u}\right)\phantom{\rule{0ex}{0ex}}=4.66MeV$

Hence, the disintegration energy is 4.66 MeV.

## Step 5: c) Calculate the disintegration energy

The disintegration energy for uranium-235 decaying into thorium-230 is given using the atomic masses and equation (i) as follows:

$\mathrm{Q}=\left({\mathrm{m}}_{{235}_{\mathrm{U}}}-{\mathrm{m}}_{{230}_{\mathrm{Th}}}-{\mathrm{m}}_{{5}_{\mathrm{He}}}\right){\mathrm{c}}^{2}$

Substitute the values and solve as:

role="math" localid="1661929289112" $Q=\left(235.0429u-230.0331u-5.0122u\right)\left(931.5\frac{MeV}{u}\right)\phantom{\rule{0ex}{0ex}}=-1.30MeV$

Hence, the disintegration energy is - 1.30 MeV.

Only the second decay process (the $\alpha$ decay) is spontaneous, as it releases energy considering the positive sign of Q-value.