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Number of alpha (\(\alpha\)) and beta (\(\beta\)) particles emitted in the conversion of \(\mathrm{^{232}_{90}Th}\) to \(\mathrm{^{216}_{84}Po}\) are

Question and options

Number of alpha (α\alpha) and beta (β\beta) particles emitted in the conversion of 90232Th\mathrm{^{232}_{90}Th} to 84216Po\mathrm{^{216}_{84}Po} are
  1. α=3, β=3\alpha = 3,\ \beta = 3
  2. α=2, β=4\alpha = 2,\ \beta = 4
  3. α=4, β=2\alpha = 4,\ \beta = 2
  4. α=4, β=6\alpha = 4,\ \beta = 6

Answer and explanation

Correct answer: C - \(\alpha = 4,\ \beta = 2\)
Mass drop 232−216=16232 - 216 = 16, so 4 alpha (ZZ falls by 8 to 82). To reach Z=84Z = 84, 2 beta emissions are needed.

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