When a _4Be^9 atom is bombarded with alpha-pa | vedclass

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(D) The nuclear reaction is given by: $_4Be^9 + _2He^4 \rightarrow _6C^{12} + X$. To find $X$,we balance the mass number and atomic number on both sid

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$_0n^1$

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(D) The nuclear reaction is given by: $_4Be^9 + _2He^4 \rightarrow _6C^{12} + X$. To find $X$,we balance the mass number and atomic number on both sides. Sum of mass numbers on the left: $9 + 4 = 13$. Sum of mass numbers on the right: $12 + A = 13 \implies A = 1$. Sum of atomic numbers on the left: $4 + 2 = 6$. Sum of atomic numbers on the right: $6 + Z = 6 \implies Z = 0$. Since the particle has mass number $1$ and atomic number $0$,it is a neutron,denoted as $_0n^1$.

When a $_4Be^9$ atom is bombarded with $\alpha$-particles,one of the products of nuclear transmutation is $_6C^{12}$. The other product is:

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