A) \[\alpha \]= 6, \[\beta \]= 8
B) \[\alpha \]=10,\[\beta \]=8
C) \[\alpha \]= 8, \[\beta \]= 10
D) \[\alpha \]= 8,\[\beta \] =6
Correct Answer: D
Solution :
Let\[x\]number of\[\alpha -\]particles and y number of \[\beta -\]particles be emitted, when uranium nucleus decays to lead nucleus. \[\therefore \] \[_{92}^{238}U\xrightarrow[{}]{{}}_{82}^{206}Pb+x(\alpha )+y(\beta )\] Since,\[\alpha -\]particle is doubly ionized helium atom and\[\beta -\]particles are fast moving electrons, also mass number and atomic number remains conserved in the reaction. \[\therefore \] \[_{92}^{238}U\xrightarrow[{}]{{}}_{82}^{206}Pb+x{{(}_{2}}H{{e}^{4}})+y{{(}_{-1}}{{\beta }^{0}})\] Equating mass number, we have \[238=206+4x\] \[\Rightarrow \] \[x=8\] Equating atomic number, we have \[92=82+2x-y=82+(2\times 8)-y\] \[\Rightarrow \] \[y=6\] Hence, \[\alpha =8,\text{ }\beta =6\].
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