Question 13 Marks
An element X has the symbol $_{84}X^{202}.$ It emits an alpha-particle and then a beta-particle. The final nucleus is $_bY^a.$ Find a and b.
Answer
View full question & answer→( $\beta$-emission)
$
{ }_{84} X ^{202} \longrightarrow{ }_{82} X _1{ }^{198}+{ }_2 He ^4
$
( $\alpha$ - particles)
$
{ }_{82} X _1{ }^{198} \longrightarrow{ }_{83} Y ^{198}+{ }_{-1} e ^0
$
( $\beta$-emission)
$
{ }_b Y ^a={ }_{83} Y ^{198}
$
$\therefore a=198$ and $b=83$
$
{ }_{84} X ^{202} \longrightarrow{ }_{82} X _1{ }^{198}+{ }_2 He ^4
$
( $\alpha$ - particles)
$
{ }_{82} X _1{ }^{198} \longrightarrow{ }_{83} Y ^{198}+{ }_{-1} e ^0
$
( $\beta$-emission)
$
{ }_b Y ^a={ }_{83} Y ^{198}
$
$\therefore a=198$ and $b=83$
