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Q.Discuss the Beta+^{+} (β+\beta^{+}) decay process with an example.

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2024Subjective· 3mImportance★★★★★
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In β+\beta^+ decay a nuclear proton converts to a neutron, emitting a positron and a neutrino, decreasing the atomic number by 1 while the mass number stays the same; e.g. 1122Na→1022Ne+e++ν^{22}_{11}\text{Na}\to{}^{22}_{10}\text{Ne}+e^++\nu.

1. Process. Positron (β+\beta^+) decay occurs in proton-rich (neutron-deficient) nuclei. A proton inside the nucleus transforms into a neutron, with the emission of a positron (the electron's antiparticle, e+e^+, same mass as an electron but opposite/positive charge) and a neutrino ν\nu:

p⟶n+β++νp \longrightarrow n + \beta^+ + \nu

2. Nuclear equation. For a parent nucleus ZAX^A_ZX:

ZAX⟶Z−1AY+β++ν^A_ZX \longrightarrow {}^A_{Z-1}Y + \beta^+ + \nu

The mass number AA is unchanged (a nucleon is converted, not lost), while the atomic number decreases by 1 (one proton becomes a neutron), so the daughter YY lies one place to the left of XX in the periodic table.

3. Example.

1122Na⟶1022Ne+β++ν^{22}_{11}\text{Na} \longrightarrow {}^{22}_{10}\text{Ne} + \beta^+ + \nu

Sodium-22 (Z=11) decays by positron emission to neon-22 (Z=10).

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