Question

The element Lawrencium ${(}_{103}^{256}Lr)$ is named after Ernest Lawrence, whose invention - the cyclotron - helped discover many radioactive nuclei. Consider now, the alpha decay of ${}_{103}^{256}Lr$.
${}_{103}^{256}Lr{\to }_b^{a}Md+\alpha$
Where $\alpha$ represents the alpha particle given out. What are the values of a and b?

• A. a=252 ; b =105
• B. a=252 ; b =101
• C. a=260 ; b =101
• D. a=260 ; b =105

Answer 1

The correct option is B a=252 ; b =101

We have to find a and b in the following equation:
${}_{103}^{256}Lr{\to }_b^{a}Md+\alpha$
We know an $\alpha$-particle is just a helium-4 nucleus, ${}_2^{4}He$. We can write:
${}_{103}^{256}Lr{\to }_b^{a}Md{+}_2^{4}He.\left(1\right)$
Since the number of protons on both sides of the equation should be the same, and the lower index indicates the atomic number (Z ), from (1) we can say,
103=b +2
$\Rightarrow b=101$
If Z is the same, and the number of nucleons is same for the decaying nucleus and products, it means the mass number should be same too. We can write from (1),
256 = a + 4
$\Rightarrow a=252$