Question

plutonium - 238 undergoes alpha decay as shown: what is the the appropriate values for a and z and the chemical symbol for x.
$_{94}^{238}pu
ightarrow _{z}^{a}x + _{2}^{4}he$
234 238 242 pu u cm 94 92 96
x
a
z
what isotope produces boron - 11 when it emits a positron? *
carbon - 11
magnesium - 23
sodium - 23

Explanation:

Step1: Balance mass - number (A)

In alpha - decay, the mass - number of the parent nucleus is equal to the sum of the mass - numbers of the daughter nucleus and the alpha particle. Given the parent nucleus $\ce{^{238}_{94}Pu}$ with mass - number $A = 238$ and the alpha particle $\ce{^{4}_{2}He}$ with mass - number 4. So, $A$ of $X$ is $238−4 = 234$.

Step2: Balance atomic - number (Z)

The atomic - number of the parent nucleus is equal to the sum of the atomic - numbers of the daughter nucleus and the alpha particle. The atomic - number of $\ce{Pu}$ is 94 and that of $\ce{He}$ is 2. So, $Z$ of $X$ is $94 - 2=92$. The element with atomic - number 92 is uranium, symbol $\ce{U}$.

Step3: Positron decay for the second part

In positron decay, the atomic - number of the parent nucleus is one more than that of the daughter nucleus, and the mass - number remains the same. Boron ($\ce{B}$) has atomic - number $Z = 5$. The parent nucleus for positron decay to form $\ce{^{11}_{5}B}$ should have $Z=5 + 1=6$ (carbon) and mass - number $A = 11$. So the parent isotope is $\ce{^{11}_{6}C}$.

Answer:

X: U
A: 234
Z: 92
What isotope produces boron - 11 when it emits a positron: Carbon - 11