Question

these practice problems below are optional. i think youve had enough practice at this point, but please try these below to study, or if youd like to have more experience. ill post a key in a few days so you can check your work: 1. write the nuclear equation for the beta decay of cerium - 141. 2. write the nuclear equation for the alpha decay of platinum - 191. 3. write the nuclear equation for the formation of iron - 54 through fusion.

Explanation:

Step1: Recall beta - decay formula

In beta - decay, a neutron in the nucleus is converted into a proton, an electron (beta - particle), and an antineutrino. The general form is $_{Z}^{A}X
ightarrow_{Z + 1}^{A}Y+_{- 1}^{0}e+\bar{
u}$. For cerium - 141 ($_{58}^{141}Ce$), the beta - decay equation is $_{58}^{141}Ce
ightarrow_{59}^{141}Pr+_{-1}^{0}e+\bar{
u}$.

Step2: Recall alpha - decay formula

In alpha - decay, the nucleus emits an alpha - particle ($_{2}^{4}He$). The general form is $_{Z}^{A}X
ightarrow_{Z - 2}^{A - 4}Y+_{2}^{4}He$. For platinum - 191 ($_{78}^{191}Pt$), the alpha - decay equation is $_{78}^{191}Pt
ightarrow_{76}^{187}Os+_{2}^{4}He$.

Step3: Recall nuclear fusion formula

Nuclear fusion is the combination of two or more light nuclei to form a heavier nucleus. For the formation of iron - 54, a possible fusion reaction could be the combination of lighter nuclei. For example, if we consider the fusion of four helium - 4 nuclei ($_{2}^{4}He$) to form iron - 54 ($_{26}^{54}Fe$), the equation is $13_{2}^{4}He
ightarrow_{26}^{54}Fe + 2_{0}^{0}\gamma$ (gamma rays are also emitted in the process to conserve energy).

Answer:

1. $_{58}^{141}Ce

ightarrow_{59}^{141}Pr+_{-1}^{0}e+\bar{
u}$

1. $_{78}^{191}Pt

ightarrow_{76}^{187}Os+_{2}^{4}He$

1. $13_{2}^{4}He

ightarrow_{26}^{54}Fe + 2_{0}^{0}\gamma$