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 α-particle is just a helium-4 nucleus, ${}_2^{4}He.$ We can write:
${}_{103}^{256}Lr{\to }_b^{a}Md{+}_2^{4}He.$ (1)
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
$\Rightarrow \begin{array}{c}b=101\end{array}$.
If $Z$ is the same, and the number of nucleons is same for the decaying nucleus and products, it means the mass number $A$ should be the same too, no? We can write from(1),
256 = a + 4
$\Rightarrow \begin{array}{c}a=252\end{array}$.