Consider the following nuclear reactions:_{92 | vedclass

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(B) In the first reaction:$_{92}^{238}M 	o _{y}^{x}N + 2_{2}^{4}He$Applying conservation of mass number: $x = 238 - (2 	imes 4) = 230$Applying conse

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$144$

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(B) In the first reaction:$_{92}^{238}M 	o _{y}^{x}N + 2_{2}^{4}He$Applying conservation of mass number: $x = 238 - (2 	imes 4) = 230$Applying conservation of atomic number: $y = 92 - (2 	imes 2) = 88$So,the element is $_{88}^{230}N$.In the second reaction:$_{88}^{230}N 	o _{B}^{A}L + 2\beta^{+}$For $\beta^{+}$ decay (positron emission),the mass number $A$ remains the same and the atomic number decreases by $1$ per particle.$A = 230$$B = 88 - (2 	imes 1) = 86$So,the element is $_{86}^{230}L$.Number of neutrons in $_{86}^{230}L = A - B = 230 - 86 = 144$.

Consider the following nuclear reactions:
$_{92}^{238}M \to _{y}^{x}N + 2_{2}^{4}He$
$_{y}^{x}N \to _{B}^{A}L + 2\beta^{+}$
The number of neutrons in the element $L$ is:

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Consider the following nuclear reactions:$_{92}^{238}M \to _{y}^{x}N + 2_{2}^{4}He$$_{y}^{x}N \to _{B}^{A}L + 2\beta^{+}$The number of neutrons in the element $L$ is: