Question

types of nuclear radiation - practice
instructions: write the particle (alpha, beta, or gamma) that completes the equation below.

1. $_{27}^{54}co\

ightarrow_{25}^{50}mn +$

1. $_{3}^{8}li\

ightarrow_{4}^{8}be +$

1. $_{88}^{222}ra\

ightarrow_{86}^{218}rn +$

1. $_{94}^{236}pu\

ightarrow_{92}^{232}u +$

1. $_{27}^{60}co^{*}\

ightarrow_{27}^{60}co +$

1. $_{90}^{234}th\

ightarrow_{91}^{234}pa + _{-1}^{0}e$
instructions: complete the following alpha decay equations.

1. $_{106}^{263}sg\

ightarrow$ +

1. $_{103}^{256}lr\

ightarrow$ +

1. $_{64}^{149}gd\

ightarrow_{62}^{145}sm + _{2}^{4}he$

1. $_{93}^{237}np\

ightarrow$ +
instructions: complete the following beta decay equations.

1. $_{26}^{52}fe\

ightarrow_{27}^{52}co + _{-1}^{0}e$

1. $_{20}^{45}ca\

ightarrow$ +

1. $_{11}^{24}n\

ightarrow$ +

1. $_{16}^{35}s\

ightarrow$ +

Explanation:

Step1: Recall alpha - decay rules

In alpha - decay, an alpha particle ($\alpha$ or $_2^4He$) is emitted. The mass number of the parent nucleus decreases by 4 and the atomic number decreases by 2.

Step2: Recall beta - decay rules

In beta - decay, a beta particle ($\beta^-$ or $_{- 1}^0e$) is emitted. The mass number remains the same and the atomic number increases by 1. In some cases of beta - plus decay, a positron ($\beta^+$ or $_{+1}^0e$) is emitted, mass number remains the same and atomic number decreases by 1. In gamma decay, a gamma ray ($\gamma$) is emitted, and there is no change in mass number and atomic number.

1. For $_{27}^{54}Co

ightarrow_{25}^{50}Mn +$ ___, the mass number changes by $54 - 50=4$ and the atomic number changes by $27 - 25 = 2$. So the particle is an alpha particle ($_{2}^{4}He$).

1. For $_{3}^{8}Li

ightarrow_{4}^{8}Be+ $ ___, the mass number is the same and the atomic number increases by 1. So the particle is a beta particle ($_{-1}^{0}e$).

1. For $_{88}^{222}Ra

ightarrow_{86}^{218}Rn +$ ___, the mass number changes by $222-218 = 4$ and the atomic number changes by $88 - 86=2$. So the particle is an alpha particle ($_{2}^{4}He$).

1. For $_{94}^{236}Pu

ightarrow_{92}^{232}U +$ ___, the mass number changes by $236-232 = 4$ and the atomic number changes by $94 - 92=2$. So the particle is an alpha particle ($_{2}^{4}He$).

1. For $_{27}^{60}Co^*

ightarrow_{27}^{60}Co +$ ___, since there is no change in mass number and atomic number, the particle is a gamma ray ($\gamma$).

1. Given $_{90}^{234}Th

ightarrow_{91}^{234}Pa+_{-1}^{0}e$, it is a beta - decay.

1. For $_{106}^{263}Sg

ightarrow$ __ $+$ __, in alpha - decay, the daughter nucleus has mass number $263 - 4=259$ and atomic number $106 - 2 = 104$. So the equation is $_{106}^{263}Sg
ightarrow_{104}^{259}Rf+_{2}^{4}He$.

1. For $_{103}^{256}Lr

ightarrow$ __ $+$ __, in alpha - decay, the daughter nucleus has mass number $256-4 = 252$ and atomic number $103 - 2=101$. So the equation is $_{103}^{256}Lr
ightarrow_{101}^{252}Md+_{2}^{4}He$.

1. Given $_{64}^{149}Gd

ightarrow_{62}^{145}Sm+_{2}^{4}He$, it is an alpha - decay.

1. For $_{93}^{237}Np

ightarrow$ __ $+$ __, in alpha - decay, the daughter nucleus has mass number $237-4 = 233$ and atomic number $93 - 2=91$. So the equation is $_{93}^{237}Np
ightarrow_{91}^{233}Pa+_{2}^{4}He$.

1. Given $_{26}^{52}Fe

ightarrow_{27}^{52}Co+_{-1}^{0}e$, it is a beta - decay.

1. For $_{20}^{45}Ca

ightarrow$ __ $+$ __, in beta - decay, the daughter nucleus has mass number 45 and atomic number $20 + 1=21$. So the equation is $_{20}^{45}Ca
ightarrow_{21}^{45}Sc+_{-1}^{0}e$.

1. For $_{11}^{24}N

ightarrow$ __ $+$ __, in beta - decay, the daughter nucleus has mass number 24 and atomic number $11+1 = 12$. So the equation is $_{11}^{24}N
ightarrow_{12}^{24}Mg+_{-1}^{0}e$.

1. For $_{16}^{35}S

ightarrow$ __ $+$ __, in beta - decay, the daughter nucleus has mass number 35 and atomic number $16 + 1=17$. So the equation is $_{16}^{35}S
ightarrow_{17}^{35}Cl+_{-1}^{0}e$.

Answer:

1. $_{2}^{4}He$
2. $_{-1}^{0}e$
3. $_{2}^{4}He$
4. $_{2}^{4}He$
5. $\gamma$
6. Already complete with $_{-1}^{0}e$
7. $_{104}^{259}Rf$, $_{2}^{4}He$
8. $_{101}^{252}Md$, $_{2}^{4}He$
9. Already complete as shown
10. $_{91}^{233}Pa$, $_{2}^{4}He$
11. Already complete as shown
12. $_{21}^{45}Sc$, $_{-1}^{0}e$
13. $_{12}^{24}Mg$, $_{-1}^{0}e$
14. $_{17}^{35}Cl$, $_{-1}^{0}e$