Question

A radioactive element belongs to $IIIB$ group; it emits one '$\alpha$' and one $\beta$-particle to form a daughter nuclide. The position of daughter nuclide will be in:

• A. $IIA$
• B. $IA$
• C. $IIB$
• D. $IVB$

Answer 1

Answer: (A)

$IIA$

$IIIB$ group is group 3, according to

modern periodic table.

Let radioactive element be x & has

atomic number z & mass number A.

It limits one $\alpha$ & mass number A.

It limits one $\alpha$ & one $\beta$ particle as

$z{X}^A\to z-2X^{A-4}+2H{e}^4$

$z-2X^{A-4}\to z-1X^{A-4}+-1{\beta }^0$

Here, atomic number of x decrease by 1

i.e, $(z-1)$ So, The daughter nuclide will

be in group $IIA$ (group 2).