Nuclei — Physics STD 12 Science — Question

1. Alpha-Emission: a-particle is nucleus of a helium atom having atomic number 2 and atomic weight 4. It is denoted by ₂He⁴. 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:

$_\text{Z}\text{X}^\text{A}\ \rightarrow \ _{\text{z}-2}\text{Y}^{\text{A}-4}\ +\ _2\text{He}^4\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\alpha-\text{particle})$

Examples:

92U238 90Th234 + ₂He⁴

80Ra226 86Rn222 + ₂He⁴

2. 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:

$_\text{Z}\text{X}^\text{A}\rightarrow_{\text{Z}+1}\text{Y}^\text{A}+_{-1}\beta^0+\bar{\text{v}}$

where $\bar{\text{v}}$ is a fundamental particle called antineutrino which is massless and chargeless.

Example: $_{90}\text{Th}^{228}\rightarrow_{89}\text{Ac}^{228}+_{-1}\text{b}^0+\bar{\text{v}}$

3. 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.