\( ^{235} _{92}U + ^1 _0n → ^{144} _{56}Ba + ^{90} _{36}Kr +2X \)
In the reaction above, X is
\( ^{235} _{92}U + ^1 _0n → ^{144} _{56}Ba + ^{90} _{36}Kr +2^1_0X \)
In the reaction above, balancing the left with the right shows that X is a neutron
There is an explanation video available below.
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