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 \[\alpha \] 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\[=6\]
You need to login to perform this action.
You will be redirected in
3 sec
