A) \[2\alpha \] and\[6\beta \]particles
B) \[5\alpha \] and\[6\beta \]particles
C) \[6\alpha \] and\[2\beta \]particles
D) \[10\alpha \] and\[6\beta \]particles
Correct Answer: C
Solution :
: \[_{92}^{235}U\to _{82}^{211}Pb+x_{2}^{4}He+y_{-1}^{0}e\] Equating atomic mass, we get \[235=211+4x\Rightarrow 4x=24\] \[\Rightarrow \]\[x=6\] Equating atomic number, we get \[92=82+2x-y\] \[\Rightarrow \]\[92=82+12-y\Rightarrow y=2\] It involves emission of\[6\alpha \]and\[6\beta \]particles.
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