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Step-by-Step Solution
Step 1: Identify the Given Nuclear Reactions
We have two nuclear reactions provided:
1) $\,{}_{92}^{238}M \rightarrow \,{}_{Y}^{X}N + 2\,{}_{2}^{4}He$
2) $\,{}_{Y}^{X}N \rightarrow \,{}_{B}^{A}L + 2\,\beta^{+}$
The question asks for the number of neutrons in the element $L$.
Step 2: Analyze the Effect of Alpha ($\alpha$) Decay
An alpha particle ($\alpha$) has a mass number of 4 and an atomic number of 2. When an atom undergoes alpha decay, its mass number decreases by 4, and its atomic number decreases by 2:
$\displaystyle {}_{Z}^{A}X \xrightarrow{-\alpha} {}_{Z-2}^{A-4}Y$
In the given question, two alpha particles are emitted from $\,{}_{92}^{238}M$:
$\displaystyle {}_{92}^{238}M \xrightarrow{-2\alpha} {}_{92-(2\times2)}^{238-(2\times4)}N = {}_{88}^{230}N.$
Step 3: Analyze the Effect of Positron ($\beta^+$) Decay
A positron ($\beta^+$) is a positively charged electron. Each positron emission reduces the atomic number by 1 but does not change the mass number:
$\displaystyle {}_{Z}^{A}X \xrightarrow{-\beta^{+}} {}_{Z-1}^{A}Y.$
The question states that $N$ emits two positrons to form $L$:
$\displaystyle {}_{88}^{230}N \xrightarrow{-2\,\beta^{+}} {}_{86}^{230}L.$
Step 4: Determine the Atomic Number and Mass Number of $L$
From the above steps, the final nuclide $L$ has:
Atomic number ($Z_L$) = 86
Mass number ($A_L$) = 230
Step 5: Calculate the Number of Neutrons in $L$
The number of neutrons ($n$) in any atom is given by $n = A - Z,$ where $A$ is the mass number and $Z$ is the atomic number. For $L,$
$\displaystyle n_L = A_L - Z_L = 230 - 86 = 144.$
Final Answer
The number of neutrons in the element $L$ is $\boxed{144}.$
