Question

In the equation \(\rm{_{13}^{27}Al} \ + \ {_2^4He} \ \longrightarrow \ {_{15}^{30}P} \ + \ X\), The correct symbol for X is

1. \(\rm {_{-1}^{ \ \ \ 0} e}\)
2. \(\rm {_1^1 H}\)
3. \(\rm {_2^4 He}\)
4. \(\rm {_0^1 n}\)

Answer 1

Correct Answer - Option 4 : \(\rm {_0^1 n}\)

CONCEPT: 

• Nuclear reactions are the processes in which one or more nuclides are produced from a collision between two atomic nuclei or one atomic nucleus and subatomic particle
• All the nuclear reactions must obey all kind of conservation laws like conservation of charge and mass 

EXPLANATION:

• The reaction between Aluminium and Helium is given by

\(\Rightarrow \rm{_{13}^{27}Al} \ + \ {_2^4He} \ \longrightarrow \ {_{15}^{30}P} \ + \ X\)

• In an equation, the right-hand side of the equation is always equal to the left-hand side.
• The mass number and atomic number on the left side of the reaction is 

⇒ A = 27 + 4 = 31 

⇒ Z = 13 + 2 = 15

• The mass number and atomic number on the right side of the reaction is 

⇒ A = 30

⇒ Z = 15

• Right now the mass number is not the same on both sides and differ by 1, therefore in order to balance out the equation place 1n₀ in the equation.

\(\Rightarrow \rm{_{13}^{27}Al} \ + \ {_2^4He} \ \longrightarrow \ {_{15}^{30}P} \ + \ \rm {_0^1 n}\)

• Hence, option 4 is the answer