solving alpha beta gamma decay equations

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 The following are some radioactive decays example, we wrote on the nuclear decay equations , solve it and  explain how we obtain our answers.

         y-90 beta decay equation in which zirconium-90 is produced

 When yttrium-90 decays into zirconium-90,   the mass number does not change, the atomic number also increases by one, this signifies a beta decay ,. Therefore we can have the decay equation as follows.

⁹⁰ ₃₉Yt → ⁹⁰ ₄₀Zr + ⁰ _-1e

 The atomic number increases by one, the daughter atom is more positive than the parent nucleus.

 Let us balance the equation to see if we are right or wrong.

We start with super-script on both side, the law required that they all be equal.

 Therefore

90  → 90  +0=0

39 → 40+(-1)=39

^balanced.

·       identify the nuclide produced when fluorine-18 decays by positron emission: 18 9f→01e + ?

 to identify the nuclide produced we have to start by finding what type nuclear decay has taken part.

 The fluorine decays by a positron decay, therefore from the known characteristics of a positron decay, the atomic number of the daughter nucleus will reduce by one while he atomic number will remain constant.

 If the atomic number will reduce by one, then the nucleus that fit this our description is oxygen-18.

 Therefore the decay equation will look like this

¹⁸ ₉F →¹⁸₈O + ⁰₁e

 Let us check this equation to see if is balanced

The superscript are 18 on the product side and 18+0=18 On the product side

 The subscript is 9 on the reactant side and 8 + 1 =9 on the product side

 Therefore the equation is balanced.

 When a nucleus undergo a positron decay, the new atom formed , gain one more proton .  a neutron decays into a proton, a positron and an electron neutrino.

·       identify the nuclide produced when carbon-14 decays by beta emission: 14 6c→ 0−1e + ?

  The decay process is beta. In a beta decay, a proton decays into a neutron and an electron, the atomic number increases by one while the atomic number remain the same,

 Why the atomic mass remain the same?

 A lost of protons result to a gain of neutron, both proton and neutron are nucleons, they are counted same when determining  the mass number of atoms, therefore  this is why the mass number does not change in a beta decay.

 Let us put down the equation for the beta decay.

¹⁴₆C →⁰_-1e + ?

 What will fit the above position is an atom with atomic number higher than carbon-14 with atomic number of one, the atom that fit this descrition is nitrogen-14

 Therefore

 The new equation is

¹⁴₆C →⁰_-1e + ¹⁴₇C

 Balancing this equation to determine if we are write or wrong

 For the superscript

 We have

14 → 0+14=4

Balanced

And  for subscript

6 → (-1)+ 7 =6

Balanced

 Therefore we are not wrong

·       identify the nuclide produced when phosphorus-32 decays by beta emission: 3215p→ 0−1e + ?

   phosphorus-32 has  an atomic number of 15, for it to undergo a beta emission, it mean one of its neutron decays into a proton , an electron and an electron  antineutrino.

 The daughter nucleus that will fit this description will be suphur-32,

 Therefore our decay equation will be

³²₁₅P→ ⁰₋₁e + ?

³²₁₅P→ ⁰₋₁e +  ³²₁₆S

 Let us check to see if this equation is balanced

For the super scrip part we have 32 →0+32 =32

For the subscript

15 →15+(-1)=15

 Everything is balanced , we are therefore  not wrong.

·       identify the nuclide produced when thorium-234 decays by beta emission:

 when a thorium nucleus  decays by a beta emission , one of its neutron decays into a proton, an electron and an electron antineutrino, the atomic number o the new nucleus formed or simply the atomic number of the daughter nucleus has one more charge than that of the parent nucleus and the mass number does not change.

  The atom that fit this description is protactinium-234

 let us write down the equation

²³⁴ ₉₀Th → X + ⁰₋₁e

 ²³⁴ ₉₀Th → ²³⁴ _9iPa + ⁰₋₁e

Let us balance this equation to see if we are wrong

 For the super script we have

234  →234+0 =234 balanced

 For the subscript , we have

90 → 91 +(-1) =90

Balanced

 Therefore we are not wrong

·       identify the nuclide produced when iodine-131 decays by beta emission: 131 53i→ 0−1e + ?

let us start by writing the nuclear decay equation

¹³¹ ₅₃I→ ⁰₋₁e + ?

¹³¹ ₅₃I→ ⁰₋₁e + ¹³¹₅₄Xe

 For iodine -131 to undergo a beta emission, one of it proton has to decay into neutron, an electron ( beta particle) and an electron antineutrino.   The atomic number of the daughter atom formed increases by one while the mass number remain unchanged. The  atom that fit the description above is xen0n-131.

 Let us check if we are wrong

 Starting with the super-script

131 → 0 +131

For the subscript

53 → (-1) + 54 =53

Therefore the equation is balanced.

·       identify the nuclide produced when thallium-201 decays by electron capture: 201 81tl+ 0−1e→00γ + ?

 let us put own the nuclear decay  equation  first.

 ²⁰¹ ₈₁Tl+ ⁰_−1e→⁰₀γ + ?

²⁰¹ ₈₁Tl+ ⁰_−1e→⁰₀γ + ²⁰¹ ₈₀Hg

 When an electron capture occur, the nucleus of the parent atom  captures one of the inner electrons  and emits a antineutrino,  the process lead to decreases in atomic number while the mass number remain the same. An isobar is created, an isobar is any two or more atoms of different element with the same mass number.

 During an electron capture a protons decays into a neutron to and gamma ray is emitted. It is is similar to positron emission in many ways but do only occur n nucleus with insufficient energy to emit a positron.

·       identify the nuclide produced when americium 241 decays by alpha emission

 when americium emits an alpha particle, the  atomic number decreases by two while the mass number decreases by four, the daughter atom  formed is found in two position less than the parent nucleus in the periodic table . the atom that fit the description above is neptunium-237

 let us write the nuclear decay equation below

²⁴¹ ₉₅Am → X + ⁴₂He

²⁴¹₉₅Am → ²³⁷₉₃Np + ⁴₂He

 Let us balance this equation to find out if we are wrong

 For the superscript,

 We have   241 → ( 237+4) =241 balanced

For the subscript

We have

95 →  (93+ 2) =95 balanced

 Therefore we are not wrong. The daughter nucleus is neptunium-237

·       what particle is produced by the decay of thorium-214 to radium-210? 21490th→21088ra+?

Let us put the equation 1^st

²¹⁴₉₀Th→²¹⁰₈₈Ra+?

 When a thorium nucleus decays to radium,  we can see that  the atomic number of radium   is less than that of   by two while the mass number is less by four. In other word the daughter nucleus formed is less by 2 and 4 in atomic number an mass number respectively( 214-210=4 and 90-88. The type of process that exhibit this characteristics is alpha particle emission.  Therefore the particle product is  alpha particles.

 Let us write own the nuclear decay equation.

²¹⁴₉₀Th→²¹⁰₈₈Ra+ ⁴₂He

 LET US BALANCE THE equation

For the superscript

 We have 214  → 210+4 =214 Balanced

 For the subscript

 We have

90 → 88+ 2 =90  balanced

 Therefore the left and right  hand side has equal atomic masses and nuclear charges . hence we are not wrong.

·       uranium-238 decays by alpha emission. what is the mass number of the resulting element?

  When uranium-238 undergo a beta decay , it loses four from the atomic number and two from the mass number,  the mass number of the atom formed will be ( 234-4)=4.  The atom that fit this description is thorium-234.

 Let us put down the decay equation first.

²³⁸ ₉₂U → ²³⁴ ₉₀U + ⁴₂He

  Let us balance this equation  to see if we are wrong

 For the superscript, we have

238  → (234+ 4) =238

 For  the subscript

 92 → (90+ 2)= 92 balanced

 Therefore, we are not wrong.

 When atom undergo an alpha emission , I atomic number decreases by two while the mass number increases by four.

·       10947ag+42he→? express your answer as an isotope.

¹⁰⁹₄₇Ag→?  + ⁴₂He

 When silver undergo an alpha decay, it losses four from it mass number and two from it atomic number, the daughter nucleus formed will have an atomic number of 45 and an atomic mass of 107, the atom that fit this our description is rhodium-105

 The above nuclear equation will therefore be

¹⁰⁹₄₇Ag → ¹⁰⁵₄₅Rh + ⁴₂He

 Balancing this equation, we have

 109  → 107+ 5 =109

For the superscript  a,

And for the subscript

 We have

47 → 45+ 2 =47.

 The above equation is balanced.