Question

Find the atomic number of $X_3$ in the following nuclear reaction,

$21485X\to X_1+\beta \to X_2+\gamma \to X_3+\alpha$

• A. $82$
• B. $86$
• C. $84$
• D. $80$

Answer 1

The correct option is C $84$

In a $\beta$-emission, atomic number increases by 1 and mass number remains same.

So after first emission we get,

$21485X\to 21486X1+\beta$
In gamma emission, there is no change in atomic number or, mass number. So, after second emission we get,

$21485X\to 21486X1+\beta \to 21486X2+\gamma$

In alpha emission, atomic number decreases by 2, and the mass number decreases by 4. So after the third emission we get,

$21485X\to 21486X1+\beta \to 21486X2+\gamma \to 21084X3+\alpha$

Hence, the atomic number of $X_3$, $\text{Z}=84$