Which of the following particles is emitted in the nuclear reaction: 13Al27+2He4⟶14P30+…?
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a
0n1
b
-1e0
c
1H1
d
1H2
answer is C.
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Detailed Solution
13A27+2He4⟶14P30+ZXAEquating mass number on both sides27 + 4 = 30 + A∴A = 31 - 30 = 1Equating atomic number on both sides13 + 2 = 14 + Z∴ Z = 1∴ Particle is 1H1.