A) 2
B) 4
C) 6
D) 10
Correct Answer: C
Solution :
Suppose the no. of a-particles emitted \[=x\]and the no. of \[\beta \]-particles emitted =y. Then \[_{92}{{U}^{238}}{{\to }_{82}}P{{b}^{206}}+x_{+2}^{4}\alpha +y\,\,_{-1}^{0}\beta \] equating the mass number on both sides, we get 238 = 206 + 4x + 0 y or 4x = 32, x = 8 equating the atomic number on both sides, we get 92 = 82 + 2x ? y 92 = 82 + 2× 8 ?y y = 6 Hence 8\[\alpha \]and \[6\beta \] are emitted.You need to login to perform this action.
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