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Question: When \(90Th^{228}\) transforms to \(83Bi^{212}\), then the number of the emitted a and b particles i...

When 90Th22890Th^{228} transforms to 83Bi21283Bi^{212}, then the number of the emitted a and b particles is, respectively

A

8a, 7b

B

4a, 7b

C

4a, 4b

D

4a, 1b

Answer

4a, 1b

Explanation

Solution

Z=90ThA=228→Z′=83BiA′=212Z = 90Th^{A = 228} \rightarrow_{Z' = 83}Bi^{A' = 212}

Number of α-particles emitted

nα=A−A′4=228−2124=4n_{\alpha} = \frac{A - A'}{4} = \frac{228 - 212}{4} = 4

Number of *β-*particles emitted

nβ=2nα−Z+Z′=2×4−90+83=1.n_{\beta} = 2n_{\alpha} - Z + Z' = 2 \times 4 - 90 + 83 = 1.