A) 10
B) 5
C) 8
D) 32
Correct Answer: C
Solution :
Suppose the no. of a-particles emitted = x and the no. of b-particles emitted = y, then \[_{92}{{U}^{238}}{{\to }_{82}}P{{b}^{206}}+x{{\,}_{+2}}{{\alpha }^{4}}+y{{\,}_{-1}}{{\beta }^{0}}\] Equating the mass number on both sides, we get 238 = 206 + 4x + 0y or 4x = 32 or \[x=\frac{32}{4}=8\] Hence 8 a-particles will be emitted.You need to login to perform this action.
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