A nucleus X initially at rest, undergoes alpha decay according to the equation.
$_{92} X^{A} \to_{Z} Y^{228} + α$ .
Then, the value of A and Z are.

94, 230

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232, 90

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190, 32

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230, 94

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Solution

The correct option is A $232, 90$
The decay equation is
$_{92} X^{A} \to_{Z} Y^{228} + α$
$α$ -particle is nucleus of $_{2} H e^{4}$
$_{92} X^{A} \to_{Z} Y^{228} +_{2} H e^{4}$
$A = 228 + 4 = 232$
$Z = 92 - 2 = 90$ .

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A nucleus X initially at rest, undergoes alpha decay according to the equation.92XA→ZY228+α.Then, the value of A and Z are.

The correct option is A 232,90The decay equation is92XA→ZY228+αα-particle is nucleus of 2He492XA→ZY228+2He4A=228+4=232Z=92−2=90.

A nucleus X initially at rest, undergoes alpha decay according to the equation.
$_{92} X^{A} \to_{Z} Y^{228} + α$ .
Then, the value of A and Z are.

The correct option is A $232, 90$
The decay equation is
$_{92} X^{A} \to_{Z} Y^{228} + α$
$α$ -particle is nucleus of $_{2} H e^{4}$
$_{92} X^{A} \to_{Z} Y^{228} +_{2} H e^{4}$
$A = 228 + 4 = 232$
$Z = 92 - 2 = 90$ .