Neutron decay in free space is given as follows
$_{0} n^{1} ⟶_{1} H^{1} +_{1} e^{0} +$ [ ]
Then the parenthesis [ ] represents a

neutrino

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photon

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antineutrino

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graviton

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Solution

The correct option is B antineutrino
Since the charge is already conserved in the decay process, proton cannot be ejected.
To conserve the spin angular momentum of the initial neutron particle, an antineutrino (chargeless) is ejected.

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_{0} n^{1} \to_{1} H^{1} +_{- 1} e^{0} + []

_{e} n^{1} \to_{0} H^{1} +_{- o} e^{0} + []

_{7} N^{14} +_{0} n^{1} \to

+_{1} H^{1}

β

β^{-}

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