A) \[12p.13n.13{{e}^{-}}\]
B) \[13p,\text{ }12n,\text{ }13{{e}^{-}}\]
C) \[12p,13n,12{{e}^{-}}\]
D) \[14p,\text{ }11n,14{{e}^{-}}\]
Correct Answer: A
Solution :
Positron is emitted with proton changes to neutron \[_{1}^{1}H\xrightarrow[{}]{{}}_{0}^{1}n+_{+1}^{0}e\] \[_{13}^{25}Al\xrightarrow[{}]{{}}_{12}^{25}X+_{0}^{1}e\] Number of electrons remained = 13 (as in\[Al\]) Therefore,\[_{12}^{25}X\]has\[12p,\text{ }13n,\text{ }13\text{ }{{e}^{-}}\].
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