Question

A Name the particle emitted in the following reaction.
4^9Be + alpha->6^ 12C+_​

Answer 1

Answer:

Neutron (₀n¹)

Explanation:

₄Be⁹ + $\propto$ -----> ₆C¹² + ₀n¹

⇒ ₄Be⁹ + ₂He⁴  -----> ₆C¹² + ₀n¹ ---------- alpha = ₂He⁴

Sum of masses (superscript numbers 'ⁿ') => 9 + 4 = 12 + 1

Sum of atomic numbers (subscript numbers 'ₙ') => 4 + 2 = 6 + 0

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