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
- A 4,8
- B 8,2
- C 6,2
- D 6,4
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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