Nd (Z = 60) is a member of group 3 in the per | vedclass

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(B) The radioactive decay of a nucleus by $\beta$-emission involves the conversion of a neutron into a proton,which increases the atomic number $(Z)$

Vedclass · Accepted Answer

$4$

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(B) The radioactive decay of a nucleus by $\beta$-emission involves the conversion of a neutron into a proton,which increases the atomic number $(Z)$ by $1$. According to the Group Displacement Law,when an element undergoes $\beta$-decay,its atomic number increases by $1$,causing the element to shift one group to the right in the periodic table. Since $Nd$ $(Z = 60)$ belongs to Group $3$,the daughter nucleus will have an atomic number of $61$ $(Pm)$ and will belong to Group $4$.

$Nd$ $(Z = 60)$ is a member of group $3$ in the periodic table. An isotope of it is $\beta$-active. The daughter nuclei will be a member of Group $...$ ...

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