8O16+1H1⟶2X4+2He4 Use conservation of charge and mass The conservation of mass number gives 16+2=A+4⇒A=14 The conservation of atomic number gives. 8+1=2+2⇒2=7 Therefore, 2X4 is 7 N14
A deuteron is bombard on 8O16 nuclei than α-particle is emitted then the product nuclei is:
Held on 30 Apr 2002 · Verified 9 Jul 2026.
7 N13
5 B10
4Be9
7 N14
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