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. \[_{92}^{238}U\xrightarrow{{}}_{82}^{206}Pb+x(\alpha )+y(\beta )\] 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, \[\alpha =8,\,\beta =6.\]
You need to login to perform this action.
You will be redirected in
3 sec
