The number of neutrons in the element L in the following nuclear changes is
$_{92}^{238} M \to_{y}^{x} N +_{2}^{4} H e$
$_{Y}^{X} N \to_{B}^{A} L + 2 β^{+}$

146

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144

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140

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142

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Solution

The correct option is A $146$
$_{92}^{238} M \to_{y}^{x} N +_{2}^{4} H e$
$y = 92 - 2 = 90 x = 238 - 4 = 234$
$_{90}^{234} N \to_{B}^{A} L + 2_{+ 1}^{o} β B = 90 - 2 = 88 A = 234$
no. of neutrons $= 234 - 88 = 146$
So, $(A)$ is correct option.

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