iii. practice with equations\ntry these.\n1. $_{27}^{60}co\rightarrow\boxed{_{28}^{60}ni}+_{ - 1}^{0}e$\n2…

iii. practice with equations\ntry these.\n1. $_{27}^{60}co\rightarrow\boxed{_{28}^{60}ni}+_{ - 1}^{0}e$\n2. $_{43}^{99}tc\rightarrow\boxed{_{44}^{99}?}+_{1}^{0}\beta$\n3. $_{53}^{131}i\rightarrow\boxed{_{54}^{131}?}+_{ - 1}^{0}e$\n4. $\boxed{_{55}^{137}?}\rightarrow_{56}^{137}ba+_{ - 1}^{0}e$\n5. $\boxed{_{10}^{23}?}\rightarrow_{11}^{23}na+_{1}^{0}\beta$\n6. $_{90}^{229}th\rightarrow\boxed{_{88}^{225}?}+_{2}^{4}he$

iii. practice with equations\ntry these.\n1. $_{27}^{60}co\rightarrow\boxed{_{28}^{60}ni}+_{ - 1}^{0}e$\n2. $_{43}^{99}tc\rightarrow\boxed{_{44}^{99}?}+_{1}^{0}\beta$\n3. $_{53}^{131}i\rightarrow\boxed{_{54}^{131}?}+_{ - 1}^{0}e$\n4. $\boxed{_{55}^{137}?}\rightarrow_{56}^{137}ba+_{ - 1}^{0}e$\n5. $\boxed{_{10}^{23}?}\rightarrow_{11}^{23}na+_{1}^{0}\beta$\n6. $_{90}^{229}th\rightarrow\boxed{_{88}^{225}?}+_{2}^{4}he$

Answer

Explanation:

Step1: Apply conservation of mass - number and atomic - number

In a nuclear reaction, the sum of mass - numbers and atomic - numbers on the left - hand side must equal the sum of mass - numbers and atomic - numbers on the right - hand side respectively.

  1. For $\ce{^{60}{27}Co}\to X+\ce{^{0}{ - 1}e}$: The mass - number of $\ce{Co}$ is $60$ and atomic - number is $27$. For the electron $\ce{^{0}{ - 1}e}$, mass - number is $0$ and atomic - number is $-1$. Let the unknown nucleus be $\ce{^{A}{Z}X}$. Then $A = 60$ and $Z=27+1 = 28$. The element with atomic - number $28$ is $\ce{Ni}$, so the unknown is $\ce{^{60}_{28}Ni}$.

  2. For $\ce{^{99}{43}Tc}\to X+\ce{^{0}{1}\beta}$: The mass - number of $\ce{Tc}$ is $99$ and atomic - number is $43$. For the positron $\ce{^{0}{1}\beta}$, mass - number is $0$ and atomic - number is $1$. Let the unknown nucleus be $\ce{^{A}{Z}X}$. Then $A = 99$ and $Z=43 - 1=42$. The element with atomic - number $42$ is $\ce{Mo}$, so the unknown is $\ce{^{99}_{42}Mo}$.

  3. For $\ce{^{131}{53}I}\to X+\ce{^{0}{ - 1}e}$: The mass - number of $\ce{I}$ is $131$ and atomic - number is $53$. For the electron $\ce{^{0}{ - 1}e}$, mass - number is $0$ and atomic - number is $-1$. Let the unknown nucleus be $\ce{^{A}{Z}X}$. Then $A = 131$ and $Z=53 + 1=54$. The element with atomic - number $54$ is $\ce{Xe}$, so the unknown is $\ce{^{131}_{54}Xe}$.

  4. For $X\to\ce{^{137}{56}Ba}+\ce{^{0}{ - 1}e}$: The mass - number of $\ce{Ba}$ is $137$ and atomic - number is $56$. For the electron $\ce{^{0}{ - 1}e}$, mass - number is $0$ and atomic - number is $-1$. Let the unknown nucleus be $\ce{^{A}{Z}X}$. Then $A = 137$ and $Z=56-1 = 55$. The element with atomic - number $55$ is $\ce{Cs}$, so the unknown is $\ce{^{137}_{55}Cs}$.

  5. For $X\to\ce{^{23}{11}Na}+\ce{^{0}{1}\beta}$: The mass - number of $\ce{Na}$ is $23$ and atomic - number is $11$. For the positron $\ce{^{0}{1}\beta}$, mass - number is $0$ and atomic - number is $1$. Let the unknown nucleus be $\ce{^{A}{Z}X}$. Then $A = 23$ and $Z=11 + 1=12$. The element with atomic - number $12$ is $\ce{Mg}$, so the unknown is $\ce{^{23}_{12}Mg}$.

  6. For $\ce{^{229}{90}Th}\to X+\ce{^{4}{2}He}$: The mass - number of $\ce{Th}$ is $229$ and atomic - number is $90$. For the alpha - particle $\ce{^{4}{2}He}$, mass - number is $4$ and atomic - number is $2$. Let the unknown nucleus be $\ce{^{A}{Z}X}$. Then $A = 229 - 4=225$ and $Z=90 - 2=88$. The element with atomic - number $88$ is $\ce{Ra}$, so the unknown is $\ce{^{225}_{88}Ra}$.

Answer:

  1. $\ce{^{60}_{28}Ni}$
  2. $\ce{^{99}_{42}Mo}$
  3. $\ce{^{131}_{54}Xe}$
  4. $\ce{^{137}_{55}Cs}$
  5. $\ce{^{23}_{12}Mg}$
  6. $\ce{^{225}_{88}Ra}$