What is the particle $X$ in the following nuclear reaction?
$_{4}^{9} B e +_{2}^{4} H e ⟶_{6}^{12} C + X$

Electron

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Proton

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Photon

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Neutron

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Solution

The correct option is C Neutron
$_{4}^{9} B e +_{2}^{4} H e ⟶_{6}^{12} C +_{Z}^{A} X$
Comparing sum of mass numbers and atomic numbers on both sides, we get
$A = 1, Z = 0$
Hence, $X$ represents $_{0}^{1} n$ (neutron).

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