Question

nuclear chemistry worksheet- show \k - u - e - s\ where necessary, otherwise answer completely. work that does not fit in the provided space needs to be completed on your own paper. part a: completing nuclear decay reactions: 1 - 10 for each of the atoms listed below, complete the decay reaction by solving for $_z^a x$ or other missing information. remember that the mass and protons on each side of the arrow need to equal each other. $_{103}^{256}lr
ightarrow _{2}^{4}he + _{z}^{a}x$, $_{z}^{247}am
ightarrow _{ - 1}^{0}e+_{z}^{a}x$, $_{z}^{a}x
ightarrow _{87}^{211}fr + _{2}^{4}he$, $_{93}^{175}np
ightarrow _{2}^{4}he+_{z}^{a}x$, $_{2}^{6}he
ightarrow _{ - 1}^{0}e+_{z}^{a}x$, $_{5}^{13}b
ightarrow _{ - 1}^{0}e+_{z}^{a}x$, $_{79}^{211}au
ightarrow _{ - 1}^{0}e+_{z}^{a}x$, $_{67}^{151}ho
ightarrow _{2}^{4}he+_{z}^{a}x$, $_{z}^{a}x+_{ - 1}^{0}e+_{z}^{213}po$, $_{57}^{148}la
ightarrow _{2}^{4}he+_{z}^{a}x$

Explanation:

Step1: Recall nuclear - decay rules

In alpha - decay ($\alpha$ - decay), a helium nucleus ($_{2}^{4}He$) is emitted. In beta - decay ($\beta^-$ - decay), an electron ($_{ - 1}^{0}e$) is emitted. The sum of mass numbers (top numbers) and atomic numbers (bottom numbers) on both sides of the nuclear - decay equation must be equal.

Step2: Solve for the first reaction $_{103}^{256}Lr

ightarrow_{2}^{4}He + _{Z}^{A}X$
For the mass number $A$: $256=4 + A$, so $A = 252$. For the atomic number $Z$: $103=2 + Z$, so $Z = 101$. The element with atomic number $Z = 101$ is $Md$ (Mendelevium). So the product is $_{101}^{252}Md$.

Step3: Solve for the second reaction $_{Z}^{247}Am

ightarrow_{ - 1}^{0}e+_{Z}^{A}X$
For the mass number $A$: $247=0 + A$, so $A = 247$. For the atomic number $Z$: $Z=-1 + Z_{new}$, so $Z_{new}=Z + 1$. Americium ($Am$) has $Z = 95$, so the new element has $Z = 96$, which is $Cm$ (Curium). The product is $_{96}^{247}Cm$.

Step4: Solve for the third reaction $_{Z}^{A}X

ightarrow_{87}^{211}Fr+_{2}^{4}He$
For the mass number $A$: $A=211 + 4=215$. For the atomic number $Z$: $Z=87 + 2=89$, which is $Ac$ (Actinium). The reactant is $_{89}^{215}Ac$.

Step5: Solve for the fourth reaction $_{93}^{175}Np

ightarrow_{2}^{4}He+_{Z}^{A}X$
For the mass number $A$: $175=4 + A$, so $A = 171$. For the atomic number $Z$: $93=2 + Z$, so $Z = 91$, which is $Pa$ (Protactinium). The product is $_{91}^{171}Pa$.

Step6: Solve for the fifth reaction $_{2}^{6}He

ightarrow_{ - 1}^{0}e+_{Z}^{A}X$
For the mass number $A$: $6=0 + A$, so $A = 6$. For the atomic number $Z$: $2=-1 + Z$, so $Z = 3$, which is $Li$ (Lithium). The product is $_{3}^{6}Li$.

Step7: Solve for the sixth reaction $_{5}^{13}B

ightarrow_{ - 1}^{0}e+_{Z}^{A}X$
For the mass number $A$: $13=0 + A$, so $A = 13$. For the atomic number $Z$: $5=-1 + Z$, so $Z = 6$, which is $C$ (Carbon). The product is $_{6}^{13}C$.

Step8: Solve for the seventh reaction $_{79}^{211}Au

ightarrow_{ - 1}^{0}e+_{Z}^{A}X$
For the mass number $A$: $211=0 + A$, so $A = 211$. For the atomic number $Z$: $79=-1 + Z$, so $Z = 80$, which is $Hg$ (Mercury). The product is $_{80}^{211}Hg$.

Step9: Solve for the eighth reaction $_{67}^{151}Ho

ightarrow_{2}^{4}He+_{Z}^{A}X$
For the mass number $A$: $151=4 + A$, so $A = 147$. For the atomic number $Z$: $67=2 + Z$, so $Z = 65$, which is $Tb$ (Terbium). The product is $_{65}^{147}Tb$.

Step10: Solve for the ninth reaction $_{Z}^{A}X+_{ - 1}^{0}e

ightarrow_{Z}^{213}Po$
For the mass number $A$: $A=213$. For the atomic number $Z$: $Z-1 = 84$ (since $Po$ has $Z = 84$), so $Z = 85$, which is $At$ (Astatine). The reactant is $_{85}^{213}At$.

Step11: Solve for the tenth reaction $_{57}^{148}La

ightarrow_{2}^{4}He+_{Z}^{A}X$
For the mass number $A$: $148=4 + A$, so $A = 144$. For the atomic number $Z$: $57=2 + Z$, so $Z = 55$, which is $Cs$ (Cesium). The product is $_{55}^{144}Cs$.

Answer:

$_{101}^{252}Md$; $_{96}^{247}Cm$; $_{89}^{215}Ac$; $_{91}^{171}Pa$; $_{3}^{6}Li$; $_{6}^{13}C$; $_{80}^{211}Hg$; $_{65}^{147}Tb$; $_{85}^{213}At$; $_{55}^{144}Cs$