Question

review: what isotope is formed when u - 238 emits an alpha particle? your answer review: what isotope produces boron - 11 when it emits a positron? your answer

Explanation:

Step1: Recall alpha - decay equation

An alpha particle is $_2^4He$. Uranium - 238 has atomic number 92 and mass number 238, written as $_{92}^{238}U$. In alpha - decay, the mass number decreases by 4 and the atomic number decreases by 2.
$_{92}^{238}U
ightarrow _{90}^{234}Th + _{2}^{4}He$

Step2: Recall positron - decay equation

A positron is $_1^0e$. Boron - 11 has atomic number 5 and mass number 11, written as $_{5}^{11}B$. In positron - decay, the mass number remains the same and the atomic number decreases by 1. Let the parent isotope be $_{Z}^{A}X$. Then $_{Z}^{A}X
ightarrow _{5}^{11}B+_{1}^{0}e$. So $A = 11$ and $Z=6$. The isotope is carbon - 11, $_{6}^{11}C$.

Answer:

When U - 238 emits an alpha particle, thorium - 234 ($_{90}^{234}Th$) is formed.
When an isotope emits a positron to produce boron - 11, the isotope is carbon - 11 ($_{6}^{11}C$).