A) 14
B) 5
C) 8
D) 32
Correct Answer: A
Solution :
\[_{92}{{U}^{238}}\to {{\,}_{82}}P{{b}^{206}}+x{{\,}_{+2}}{{\alpha }^{4}}+y{{\,}_{-1}}{{\beta }^{0}}\] no. of a-particles = \[\frac{238-206}{4}=8\] no. of b-particles = \[92-82-2\times 8=6\] Total no. of particles = \[8+6=14\].You need to login to perform this action.
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