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TS EAMCET · Physics · Nuclear Physics

When an element \({ }_{90}^{232} \mathrm{Th}\) decays into \({ }_{82}^{208} \mathrm{~Pb}\), the number of \(\alpha\) and \(\beta^{-}\)particles emitted respectively are

  1. A 4,8
  2. B 8,2
  3. C 6,2
  4. D 6,4
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Answer & Solution

Correct Answer

(D) 6,4

Step-by-step Solution

Detailed explanation

\( \Delta A = 232 - 208 = 24 \) \( n_\alpha = \Delta A / 4 = 24 / 4 = 6 \) \( \Delta Z = Z_{initial} - (Z_{final} + 2n_\alpha) = 90 - (82 + 2 \times 6) = 90 - (82 + 12) = 90 - 94 = -4 \) \( n_\beta = -\Delta Z = -(-4) = 4 \) Thus, \( \alpha = 6, \beta^{-} = 4 \)
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