Nuclei
Question
What is the respective number of α and β particles emitted in the following radioactive decay 90X200→80Y168
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Answer: Option D
: D
nα=A−A′4=200−1684=8
nβ=2nα−Z+Z′=2×8−90+80=6
: D
nα=A−A′4=200−1684=8
nβ=2nα−Z+Z′=2×8−90+80=6
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