An element { }_{x} A^{y} emits 5 alpha and 4 | vedclass

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(B) The nuclear reaction is given by: ${ }_{x} A^{y} \longrightarrow { }_{82} B^{207} + 5 { }_{2} \alpha^{4} + 4 { }_{-1} \beta^{0}$ By balancing the

Vedclass · Accepted Answer

$88, 139$

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(B) The nuclear reaction is given by: ${ }_{x} A^{y} \longrightarrow { }_{82} B^{207} + 5 { }_{2} \alpha^{4} + 4 { }_{-1} \beta^{0}$ By balancing the atomic number $(x)$: $x = 82 + (5 	imes 2) + (4 	imes -1) = 82 + 10 - 4 = 88$ By balancing the mass number $(y)$: $y = 207 + (5 	imes 4) + (4 	imes 0) = 207 + 20 = 227$ Thus,the element $A$ is ${ }_{88} A^{227}$. Number of protons = $88$ Number of neutrons = $227 - 88 = 139$

An element ${ }_{x} A^{y}$ emits $5$ $\alpha$ and $4$ $\beta$ particles to give ${ }_{82} B^{207}$. The number of protons and neutrons in $A$ are respectively

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