Question
State Soddy $-$ Fajan’s displacement laws for radioactive transformations.
✓
Answer
The atoms of radioactive element are unstable. When an atom of a radioactive element disintegrates, an entirely new element is formed.
This new element possesses entirely new chemical and radioactive properties.
The disintegrating element is called the parent element and the resulting product after disintegration is called the daughter element.
Soddy and Fajan studied the successive product elements of disintegration of radioactive elements and gave the following conclusions:
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\alpha-\text{particle})$
Examples:
$\text{92U238 \ 90Th234 + _2He^4}$
$\text{80Ra226 \ 86Rn222 + _2He^4}$
where $\bar{\text{v}}$ is a fundamental particle called antineutrino which is massless and chargeless.
Example:
$_{90}\text{Th}^{228}\rightarrow_{89}\text{Ac}^{228}+_{-1}\text{b}^0+\bar{\text{v}}$
This new element possesses entirely new chemical and radioactive properties.
The disintegrating element is called the parent element and the resulting product after disintegration is called the daughter element.
Soddy and Fajan studied the successive product elements of disintegration of radioactive elements and gave the following conclusions:
- Alpha $-$ Emission:
- $a -$ particle is nucleus of a helium atom having atomic number $2$ and atomic weight $4$. It is denoted by $_2He^4$.
- Therefore when an $\alpha-$ particle is emitted from a radioactive parent atom $(X),$ its atomic number is reduced by $2$ and atomic weight is reduced by $4$.
- Thus the daughter element has its place two groups lower in the periodic table.
- Thus the process of $a-$ emission may be expressed as:
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\alpha-\text{particle})$
Examples:
$\text{92U238 \ 90Th234 + _2He^4}$
$\text{80Ra226 \ 86Rn222 + _2He^4}$
- Beta $-$ Emission:
- $\beta-$ particle is an electron $(e)$ and is denoted by $_{-1}\beta^0.$ When a $\beta-$particle is emitted from a parent atom $(X),$ it atomic number increases by $1$. while atomic weight remains unchanged.
- As a result the doughter element $(Y)$ has a place one group higher in the periodic table. Thus the process of $\beta-$ emission may be expressed as:
where $\bar{\text{v}}$ is a fundamental particle called antineutrino which is massless and chargeless.
Example:
$_{90}\text{Th}^{228}\rightarrow_{89}\text{Ac}^{228}+_{-1}\text{b}^0+\bar{\text{v}}$
- Gamma $-$ Emission:
- The emission of $\lambda-$ray from a radioactive atom neither chages its atomic number nor its atomic weight.
- Therefore its place in periodic table remains undisplaced. In natural radioactivity $\lambda-$ radiation is accompanied with either $\alpha$ or $\beta-$ emission.
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