A) \[_{90}^{233}Th\xrightarrow{{}}_{90}^{233}Pa{{+}_{-2}}{{e}^{0}}\]
B) \[_{90}^{233}Th\xrightarrow{{}}_{91}^{233}Pa{{+}_{-1}}{{e}^{0}}\]
C) \[_{90}^{233}Th\xrightarrow{{}}_{90}^{233}Pa{{+}_{0}}{{e}^{4}}\]
D) \[_{90}^{233}Th\xrightarrow{{}}_{91}^{233}Pa{{+}_{-1}}{{e}^{4}}\]
Correct Answer: B
Solution :
Key Idea The \[\beta \]-particles are electrons \[_{-1}^{0}e\](charge -1 and mass number zero} Therefore the emission of a \[\beta \]-particle will not cause any change in mass number of the daughter element but the nucleus of daughter element will have atomic number more by one unit. \[\underset{\begin{smallmatrix} radioactive \\ element \end{smallmatrix}}{\mathop{_{Z}^{A}X}}\,\xrightarrow{{}}\underset{\begin{smallmatrix} daughter \\ element \end{smallmatrix}}{\mathop{\,_{Z+1}^{A}X}}\,\,\,\,\,\,+\,\,\,\underset{\beta -particle}{\mathop{_{-1}^{0}\,e}}\,\] When \[_{90}^{233}Th\] undergoes one \[\beta \] -emission, the the following nuclear reaction occur \[_{90}^{233}Th\xrightarrow{{}}_{91}^{233}Pa+_{-1}^{a}e\]
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