how to balance radioactive decay equations

 

 To balance  a nuclear equation, the atomic number and the mass number on both side must be equal, in other words the sum of  the superscript on the left hand side must be equal to the sum of the superscript on the right hand side, the sum of the subscript on the left hand side must be equal to the sum of the subscript on the right hand side

 For example

 ²³⁷₉₃Np→⁴₂He + ²³³₉₁Pa

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

 To balance a nuclear equation, 

All the reds must be equal on both side

 237 =4+233=237 balanced

 All the greens must be equal on both side

93 = 91+2=93

If the above condition is satisfied, then an equation is balanced

enter the isotope symbol for the strontium (sr) nucleus in this reaction. 235 92u+10n→azsr+143 54xe+310n

Let us write the equation for the decay first

²³⁵₉₂U+ ¹₀n →^a_zSr + ¹⁴³₅₄Xe + 3¹₀n

 For a nuclear equation to be balanced , the sum of the superscript on the left hand side must be equal to te sum of the superscript on the right hand side and the sum of the subscript on the left han side must be equal to the sum of the subscript on the right hand side, in other word the sum of the mass numbers on the left hand side must be equal to the sum of the mass numbers on the right hand side and the sum of the atomic numbers on the left hand side must be equal to the sum of the atomic numbers on the right hand side.

Therefore for the superscript , we have

235 +1 =a+143 +3(1)

236=147 +a

A= 236-147 = 89

 An for the subscript, we have

92+0 =z+54+3(0)

92= 54 +z

Z= 92-54

Z = 38

 Therefore the above equation can be written as follows

²³⁵₉₂U+ ¹₀n →⁸⁹₃₈Sr + ¹⁴³₅₄Xe + 3¹₀n

what is the value of a in the following nuclear reaction?237 93np→233 91pa+azx

let us rewrite the equation first

²³⁷₉₃ Np→²³³ ₉₁Pa+^a_zX

 From the above equation

 We can use two method to find the value of a,

 1^st method

 In a balanced nuclear equation, the atomic number and mass numbers on both sides are equal , in other words the sum of the superscript on the left hand side must be equal to the sum of the superscript on the right hand side  and the sum of subscript on the left hand side must also be equal to the sum of subscript on the left hand side.

 To find the value of a

 We say

237 =233+a

A =237-233 =4

 And for the value of z

We say

93= 91+z

Z =93-91

Z=2

  We can rewrite the above equation

²³⁷₉₃ Np→²³³ ₉₁Pa+⁴₂X

 And nuclide has an atomic number of 2 and a mass of 4, it is helium nucleus, and an alpha particle is a helium nucleus.

 Therefore the above decay is alpha decay.

2^nd method

²³⁷₉₃ Np→²³³ ₉₁Pa+^a_zX

  From the  above equation , the atomic number of the parent atom is 93 and that of the daughter is 91 , likewise the mass number of the parent nucleus is 237 and that of the daughter is 233

 a process that lead to the loss of 4 from the atomic number and 2 from the mass number has occurred, the process is called alpha decay.

what is the value of a in the following nuclear reaction 212 84 po → 208 82 pb+ azx

let us rewrite the equation

²¹²₈₄Po → ²⁰⁸₈₂Pb+ ^a_zX

From the superscript, the atomic number has decreased by 2 while the mass number has decreased by 4,  the nuclear process that has the above characteristics is alpha decay

 Therefore the particle X is alpha particle and therefore the  value a is 4

 We can find it analytically as follows

212 = 208+a

A = 212-208=4

And

The value of  z

84=82+ z

Z= 84-82 = 2

We can rewrite the above equation as follow

²¹²₈₄Po → ²⁰⁸₈₂Pb+ ⁴₂He

Ra−210 (alpha) express your answer as a nuclear equation.

 The equation for the decay is as follows

²¹⁰₈₈Ra →? + ⁴₂He

 When radium undergo  an alpha decay, the atomic number and the mass number changes, the atomic number decreases by 2 while the mass number decreases by 4, the new daughter nucleus formed will have an atomic number of 82 and a mass number of 86.

²¹⁰₈₈Ra →²⁰⁶ ₈₆Rn + ⁴₂He

identify the nuclide produced when plutonium-239 decays by alpha emission:239 94pu→42he + ?

 let us rewrite the  alpha decay equation

²³⁹ ₉₄Pu→⁴₂He + ?

 From the above equation , it is an alpha decay, and in alpha decay the atomic number decreases by 2 while the atomic mass decreases by 4, the new daughter nuclide from the decay of plutonium-23 will be have an atomic number of 92 and a mass number of 235.

 The above decay equation will therefore be

²³⁹₉₄Pu→⁴₂He + ²³⁵₉₂U

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

For nuclear equation to be balanced , the  superscript on both side of the equation must be equal and likewise the subscript on both side must also be balanced.

 for the super script , we have

239→4+235 =239

For the subscript, we have

94 →92+2 =94

 Since both the two side are equal the above equation  is balanced and we are not wrong.

write a partial decay series for rn-220 undergoing the following sequential decays: α, α, β, α. po−210 (alpha)

 let us write the decay equations 1^st

 after the 1^st alpha decay, we will have

 ²²⁰ ₈₆Rn → ⁴₂He + ²¹⁶₈₄Po

 After a beta decay,  the atomic number decreases by two while the mass number deceases by 4.

 ²¹⁶ ₈₄Po → ⁴₂He + ²¹²₈₂Pb

 After a beta decay,  the atomic number decreases by two while the mass number deceases by 4.

²¹²₈₂Pb → ⁰_-1e + ²¹²₈₃Bi

 After a beta decay, the atomic number increases by 1 while the mass number remain unchanged.

After another alpha decay , the atomic number of bismuth will decreases by two while the mass number will be by 4.

  ²¹²₈₃Bi → ⁴₂He + ²⁰⁸₈₁ Tl

 Therefore the decay series are as follow

²²⁰ ₈₆Rn → ⁴₂He + ²¹⁶₈₄Po

²¹⁶ ₈₄Po → ⁴₂He + ²¹²₈₂Pb

²¹²₈₂Pb → ⁰_-1e + ²¹²₈₃Bi

²¹²₈₃Bi → ⁴₂He + ²⁰⁸₈₁ Tl

 Summary, when  radon-220 undergo an  alpha decay, it atomic number decreases by 2 while the mass number decreases by 4, the  new atom formed will have an atomic number of 82 and a mass number of   216, the atom that fit the above  description is polonium-216.

 When the polonium-216 undergo another alpha decay, it atomic number will go by 2 while the mass number  will decreases by 4, the new daughter nucleus formed will have an atomic number of  84 while the mass number will be 212, the atom that fit the above description is lead-212.

 When the lead-212 undergo a bet decay, its atomic number  will increases by one while the mass number remain constant, when an atom undergo a beta decay, a nucleus decayed into a proton, an electron and an electron  antineutrino. The mass number remain unchanged because the new proton formed is counted as a nucleon like the neutron destroyed. The new daughter atom formed will have an atomic number of 83 while a mass number of 212. The atom that fit this description is bismuth-212.

 When bismuth-212 undergo an alpha decay, the atomic number  decreases by 2 while the mass number decreases by four, the new nuclide formed will have an atomic number of  81 and a mass number of 210. The nuclide that fit the above description is thallium-210.

 The sequence of the atom formed will be

 ²²⁰₈₆Rn  →²¹⁶₈₄Po → ²¹²₈₂Pb→ ²¹²₈₃Bi→ ²⁰⁸₈₁Tl

identify the nuclide produced when neptunium-237 decays by alpha emission: 237 93np→42he + ?

²³⁷ ₉₃Np→⁴₂He + ?

 When neptunium undergo an alpha decay, the atomic number decreases by two while the mass number decreases by 4, the daughter nuclide formed will have an atomic number  of 91 and a mass number of 233,  the nuclide that fit the above description is protactinium-233.

Let us write the decay equation

²³⁷₉₃Np→⁴₂He + ²³³₉₁Pa

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

 To balance a nuclear equation, 

For the super script

 237 =4+233=237 balanced

 For subscript, we will have

93 = 91+2=93