A) 8 beta particles and 6 alpha particles
B) 5 alpha particles and 0 beta particles
C) 8 alpha and 6 beta particles
D) 10 alpha particles and 10 beta particles
Correct Answer: C
Solution :
Key Idea: (i) Sum of atomic mass of reactants = sum of atomic mass of products. (ii) Sum of atomic number of reactants = sum of atomic number of products. \[_{92}{{U}^{238}}\xrightarrow{{}}{{\,}_{82}}P{{b}^{206}}+{{m}_{2}}H{{e}^{4}}+n{{\,}_{-1}}{{e}^{0}}\] where \[m=\]number of a particles \[n=\]number of \[\beta \] particles \[\therefore \] \[238=206+4\,m\] or \[m=\frac{238-206}{4}\] \[=\frac{32}{4}=8\] \[\therefore \] \[92=82+2m-n\] or \[92-82=2\times 8-n\] or \[10=16-n\] \[\therefore \] \[n=16-10=6\] \[\therefore \] \[\alpha -\]particles emitted \[=8\] \[\beta \]particles emitted \[=6\]
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