Question:
When \({}^{228}_{90}Th\) transforms to \({}^{212}_{83}Bi\), find number of \(\alpha\) and \(\beta\) particles emitted.
When \({}^{228}_{90}Th\) transforms to \({}^{212}_{83}Bi\), find number of \(\alpha\) and \(\beta\) particles emitted.
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\(\alpha\): \(-4,-2\), \(\beta\): \(+1\) in atomic number.
Updated On: Apr 23, 2026
- \(8\alpha,\ 7\beta\)
- \(4\alpha,\ 7\beta\)
- \(4\alpha,\ 4\beta\)
- \(4\alpha,\ 1\beta\)
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The Correct Option is D
Solution and Explanation
Step 1: Mass number change
\[ 228 \to 212 \Rightarrow \Delta A = 16 \] Each \(\alpha\) reduces mass by 4: \[ \frac{16}{4} = 4\alpha \]
Step 2: Atomic number change
\[ 90 \to 83 \Rightarrow \Delta Z = 7 \] 4 \(\alpha\) reduce Z by \(8\): \[ 90 - 8 = 82 \] To reach 83: \[ +1 \Rightarrow 1\beta \] Conclusion: \[ 4\alpha,\ 1\beta \]
\[ 228 \to 212 \Rightarrow \Delta A = 16 \] Each \(\alpha\) reduces mass by 4: \[ \frac{16}{4} = 4\alpha \]
Step 2: Atomic number change
\[ 90 \to 83 \Rightarrow \Delta Z = 7 \] 4 \(\alpha\) reduce Z by \(8\): \[ 90 - 8 = 82 \] To reach 83: \[ +1 \Rightarrow 1\beta \] Conclusion: \[ 4\alpha,\ 1\beta \]
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