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
E) \[5\alpha \]particles and\[2\beta \]particles
Correct Answer: C
Solution :
\[_{92}^{238}U\xrightarrow[{}]{{}}_{82}^{206}Pb+m_{2}^{4}He+n_{-1}^{0}e\] \[238=206+4m\] \[\Rightarrow \] \[m=8\] \[92=82+2m-n\] \[2m-n=10\] \[\Rightarrow \] \[n=6\] \[\therefore \] \[\alpha \]particles emitted, \[m=8\] \[\beta \]particles emitted, \[n=6\]
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