A) \[8\alpha \]and \[4\beta \]
B) \[8\alpha \]and \[16\,\beta \]
C) \[4\alpha \]and \[2\beta \]
D) \[6\alpha \]and \[4\beta \]
Correct Answer: D
Solution :
Key Idea: During any change mass of reactants = mass of products. \[_{90}^{232}TH\xrightarrow{{}}_{82}^{208}Pb+x_{2}^{4}\alpha {{+}_{y-1}}{{\beta }^{0}}\] Suppose number of \[\alpha \] particles \[=x\] Suppose number of p particles \[=y\] Equating mass numbers of both sides, we get \[232=208+4x+0y\] or \[4x=24\] \[\therefore \] \[x=\frac{24}{4}=6\] Equating atomic number of both sides, we get \[90=82+2x-y\] or \[90=82+2\times 6-y\] \[\therefore \] \[y=4\] \[\therefore \]Th emits \[6\alpha \] particles and \[4\beta \] particles.
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