${_{92} X}^{238} ⟶ A + {_{2} H e}^{4}$
The number of neutrons in element ' $A$ ' will be-

146

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148

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242

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144

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Solution

The correct option is D $144$
No.of Neutrons = atomic mass -atomic number
No.of Neutrons in X is $= 238 - 92$
$= 146.$
No.of Neutrons in A is $= 146 - n o . o f n e u t r o n s i n H e .$
= 146- (4-2)
$= 146 - 2$
$= 144.$
So number of neutrons in A is $144$ .
So answer is D.

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92X238⟶A+2He4The number of neutrons in element 'A' will be-

The correct option is D 144No.of Neutrons = atomic mass -atomic numberNo.of Neutrons in X is =238−92 =146.No.of Neutrons in A is=146− no. of neutrons in He. = 146- (4-2) =146−2 =144.So number of neutrons in A is 144.So answer is D.

${_{92} X}^{238} ⟶ A + {_{2} H e}^{4}$
The number of neutrons in element ' $A$ ' will be-

The correct option is D $144$
No.of Neutrons = atomic mass -atomic number
No.of Neutrons in X is $= 238 - 92$
$= 146.$
No.of Neutrons in A is $= 146 - n o . o f n e u t r o n s i n H e .$
= 146- (4-2)
$= 146 - 2$
$= 144.$
So number of neutrons in A is $144$ .
So answer is D.