COMEDK · Chemistry · 19. Chemical Kinetics
The number of \(\alpha\) and \(\beta\)-particles emitted during the transformation of \({ }_{90} \mathrm{Th}^{232}\) to \({ }_{82} \mathrm{~Pb}^{208}\) are respectively
- A \(4,2\)
- B \(2,2\)
- C \(8,6\)
- D \(6,4\)
Answer & Solution
Correct Answer
(D) \(6,4\)
Step-by-step Solution
Detailed explanation
\({ }_{90}^{232} \mathrm{Th} \longrightarrow{ }_{82}^{208} \mathrm{~Pb}+n \alpha+m \beta\) Number of \(\alpha\)-particles, \(n=\frac{232-208}{4}=6\) Now, for charge balanced \[ \begin{aligned} 90 &=82+2 n-m \\ m &=4 \end{aligned} \] Number of \(\beta\)-particles, \(m=4\)
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