Question

1. rewrite each nuclear reaction, then fill in the missing parts, represented by the blanks and question marks. you want to write again here with our mouse, or write in your coursepack or journal and upload photos.

a) 40? → β⁻ + 40 20ca
b) -? → 4 2he + 226 88ra
c) 35 14si → -1 e⁻ + -?
d) 110 53i → 4? + 106 51sb
e) 140 56ba →? + 140 57la

Explanation:

Step1: Recall nuclear - reaction conservation laws

In nuclear reactions, the sum of atomic numbers and the sum of mass numbers are conserved on both sides of the equation.

Step2: Solve for part a

The mass number on the left - hand side is 40. The mass number of $\beta^-$ is 0. The mass number of $_{20}^{40}Ca$ is 40. So the missing mass number is 40. The atomic number of $\beta^-$ is - 1, the atomic number of $_{20}^{40}Ca$ is 20. Let the atomic number of the missing nucleus be $Z$. Then $Z+( - 1)=20$, so $Z = 21$. The element with atomic number 21 is scandium ($Sc$). The reaction is $_{21}^{40}Sc
ightarrow_{ - 1}^0\beta+_{20}^{40}Ca$.

Step3: Solve for part b

The mass number of the product $_{88}^{226}Ra$ is 226 and the mass number of $_{2}^{4}He$ is 4. So the mass number of the missing nucleus is $226 + 4=230$. The atomic number of $_{88}^{226}Ra$ is 88 and the atomic number of $_{2}^{4}He$ is 2. So the atomic number of the missing nucleus is $88 + 2=90$. The element with atomic number 90 is thorium ($Th$). The reaction is $_{90}^{230}Th
ightarrow_{2}^{4}He+_{88}^{226}Ra$.

Step4: Solve for part c

The mass number of $_{14}^{35}Si$ is 35 and the mass number of $_{ - 1}^0e$ is 0. So the mass number of the missing nucleus is 35. The atomic number of $_{14}^{35}Si$ is 14 and the atomic number of $_{ - 1}^0e$ is - 1. Let the atomic number of the missing nucleus be $Z$. Then $14=( - 1)+Z$, so $Z = 15$. The element with atomic number 15 is phosphorus ($P$). The reaction is $_{14}^{35}Si
ightarrow_{ - 1}^0e+_{15}^{35}P$.

Step5: Solve for part d

The mass number of $_{53}^{110}I$ is 110 and the mass number of $_{2}^{4}He$ is 4. So the mass number of the missing nucleus is $110 - 4 = 106$. The atomic number of $_{53}^{110}I$ is 53 and the atomic number of $_{2}^{4}He$ is 2. So the atomic number of the missing nucleus is $53-2 = 51$. The element with atomic number 51 is antimony ($Sb$). The reaction is $_{53}^{110}I
ightarrow_{2}^{4}He+_{51}^{106}Sb$.

Step6: Solve for part e

The mass number of $_{56}^{140}Ba$ is 140 and the mass number of $_{57}^{140}La$ is 140. The mass number of the missing particle is 0. The atomic number of $_{56}^{140}Ba$ is 56 and the atomic number of $_{57}^{140}La$ is 57. The missing particle is a $\beta^-$ particle. The reaction is $_{56}^{140}Ba
ightarrow_{ - 1}^0\beta+_{57}^{140}La$.

Answer:

a. $_{21}^{40}Sc
ightarrow_{ - 1}^0\beta+_{20}^{40}Ca$
b. $_{90}^{230}Th
ightarrow_{2}^{4}He+_{88}^{226}Ra$
c. $_{14}^{35}Si
ightarrow_{ - 1}^0e+_{15}^{35}P$
d. $_{53}^{110}I
ightarrow_{2}^{4}He+_{51}^{106}Sb$
e. $_{56}^{140}Ba
ightarrow_{ - 1}^0\beta+_{57}^{140}La$