A) \[_{n}{{X}^{m-4}}\]
B) \[_{n-2}{{Y}^{m-4}}\]
C) \[_{n-4}{{Z}^{m-4}}\]
D) none of these
Correct Answer: A
Solution :
Key Idea: In \[\alpha \]-particle emission, atomic mass decreases by 4 unit and atomic number decreases by 2 unit. \[\beta \]-particle emission, atomic mass remains unchanged and atomic number increases by 1 unit. The reaction can be shown as \[_{n}{{X}^{m}}\xrightarrow[{}]{\alpha }{{\,}_{n-2}}{{Y}^{m-4}}\] \[_{n-2}{{Y}^{m-4}}\xrightarrow[\,]{2\beta }{{\,}_{n}}{{X}^{m-4}}\] Thus, the resulting nucleus is the isotope of parent nucleus and is \[_{n}{{X}^{m-4}}\]
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