A radioactive element \({}^{242}_{92}X\) emits two \(\alpha\)-particles, one electron, and two positrons. The product nucleus is represented by \({}^{234}_{P}Y.\) The value of \(P\) is _______.
A radioactive element \({}^{242}_{92}X\) emits two \(\alpha\)-particles, one electron, and two positrons. The product nucleus is represented by \({}^{234}_{P}Y.\) The value of \(P\) is _______.
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For radioactive decay:
- \(\alpha\)-particle emission decreases atomic number by 2 and mass number by 4.
- \(\beta^-\) emission increases atomic number by 1.
- \(\beta^+\) emission decreases atomic number by 1.
Correct Answer: 87
Solution and Explanation
- Effect of Emission:
- Emission of an \(\alpha\)-particle reduces the mass number by 4 and the atomic number by 2.
- Emission of an electron (\(\beta^-\)) increases the atomic number by 1.
- Emission of a positron (\(\beta^+\)) decreases the atomic number by 1.
- Calculate Atomic Number (\(P\)):
- After two \(\alpha\)-particles:
\[ \text{Atomic number} = 92 - 2 \times 2 = 88, \quad \text{Mass number} = 242 - 2 \times 4 = 234. \]
- After one electron (\(\beta^-\)):
\[ \text{Atomic number} = 88 + 1 = 89. \]
- After two positrons (\(2\beta^+\)):
\[ \text{Atomic number} = 89 - 2 = 87. \]
Final Answer: 87
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