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