balancingnuclear equations worksheet with answers: physics

Beta decay is a type of nuclear decay, where an isobar  is formed, a more positive nuclide is formed after every beta decay,  the atomic number of the resulting nuclide increases while the mass number remain constant.

alpha decreases atomic number by two and mass number by four, positron and electron capture decreases atomic number  while gamma decay has no effect on the number of nucleons

 To balance a nuclear  decay equation, as required by law of conservation of matter,  the sum of superscript  and subscript on both side must be equal, otherwise the equation is must balanced.

 Below are some example :

determine the identity of the daughter nuclide from the beta decay of 9943tc.

When technetium undergo a beta emission, one of its neutron decays into a proton, an electron and an electron antineutrino, the atomic number of the daughter nuclide formed creases by one while the mass number remained unchanged. The nuclide that fit the above description will behave an atomic number of 44 and the atom is ruthenium.

 Let us write the decay equation

⁹⁹₄₃Tc → ⁹⁹₄₄Ru + ⁰_-1e

Let us write the decay equation to see if we are wrong.

99→99+0 =99

 And

44 → 44+(-1+)

determine the identity of the daughter nuclide from the beta decay of 100 43tc.

When technetium-100 undergo a beta decay,  one of its neutron decays into a proton, an electron and an electron antineutrino,  the atomic number of the daughter nuclide formed will be 44 whilethe mass number will remain the same.

 The atom the fit the description above is ruthenium-100.

 Let us write an equation for the above decay.

¹⁰⁰₄₃Tc → ¹⁰⁰₄₄Ru + ⁰_-1e

Let us check to see, if the above equation is balanced or  not.

 100 →100+0=100

For the superscript

And for the subscript we have

43 →44+(-1)=43

 Therefore since both sides are balanced, then the above equation is balanced and we are not wrong.

 When different isotopes of an atom undergo beta decay they will produce the same aom but will different mass .

determine the identity of the daughter nuclide from the electron capture by 8137rb.

 When rubidium-81 undergo an electron capture, one of its inner electron is capture by the electro-positive nucleus, a proton is converted to a neutron and a neutrino in the process. The atomic number of the resulting nucleus is decreased by one while the mass number remain unchained.

 The  daughter nuclide that will be formed will therefore have an atomic number of 36 and a mass number of 81

 Let us write the decay equation below

⁸¹₃₇Rb + ⁰_-1e → ⁸¹₃₆Kr

 Let  us check to see if the above equation is balanced or not,

 81 +0→ 81 , for the superscript

And 37+(-1) = 36

 Balanced , therefore , the above equation is balanced and we are not wrong.

determine the identity of the daughter nuclide from the beta decay of 89 38 sr

 when strontium-89 undergo a beta decay, one of its neutron  decays into a proton and an electron and an neutrino, the total number of the protons increases by one while the total number of nucleons remain  the same. The atomic number of the resulting nuclide increases by one while the mass number remain unchanged. The new nuclide formed above  will therefore have an atomic number of 39  and mass number of 89.

 The nuclide that fit the above condition is  Yttrium-89

 Let us write down t decay equation also

⁸⁹₃₈Sr → ⁸⁹₃₉Y + ⁰_-1e

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

The law of conservation of matter requires that the total sum of the super script on the left hand side must the equal to the total sum of the superscript on the right hand side.

 Also the total sum of the subscript on the left hand side must be equal to the total sum of the subscript on the right hand side.

 For the superscript we have

89 → 89+0

And  for the subscript , we have

38 → 39+(-1)

 Balanced, since everything is balanced, the above equation is therefore balanced.

determine the identity of the daughter nuclide from the alpha decay of 22286rn

 when radon undergo an alpha decay, it loses two protons and two neutrons, the atomic number of the daughter nuclide decrease by two while the mass number decreases by four the daughter nuclide formed is  will have an atomic number of 84 and a mass number of 218, the atom  that fit the above description is polonium-218

 let us rite the equation first.

²²²₈₆Rn → ²¹⁸₈₄Po + ⁴₂He

 Let us check if our answer is correct by checking the above equation .

 To balance an equation, the law of conservation of matter requires that the total sum of superscript on the left and right hand side must be equal and the total sum of subscript on the left and right hand side must be equal.

 We have 222 → 218 +4=222

 And we also have

86 → 84+2=86.

 Therefore the above equation is balanced .

determine the identity of the daughter nuclide from the beta decay of 210 82 pb

 when lead-210 under a beta decay, one of its neutron changes into a proton, an electron and an electron neutrino, the atomic number of the daughter atom formed increases by one while the mass number remain unchanged.

 The atom that fit the above description is bismuth-210

 Let us write down the decay equations

²¹⁰₈₂Pb → ²⁰⁶₈₃Bi + ⁰_-1e

 Let us balance the above equation to find out if we are wrong

 The law of conservation of matter requires that the total sum of superscript on both side must be equal as well as the total sum of subscript on both side must be equal, the superscript represent the atomic mass while the subscript represent the nuclear charge.

 Therefore now, for the superscript , we have

 210 →83+(-1)

And for the subscript, we have

82 → 83+(-1)

 Therefore, the above equation is balanced.

the following reaction represents what nuclear process? 24195am→42he+25793np

 let us write  down the nuclear decay equation first,

²⁴¹₉₅Am → ⁴₂He + ²³⁷₉₃Np

 balanced nuclear equation for beta decay

 The above equation is an alpha decay, during an alpha decay the atomic number of new nuclide decreases by two while the mass number decreases by four.

The law of conservation of matter for a nuclear decay equation requires that the  total sum of subscript on the left hand side must be equal to the total sum of subscript on the right hand side must be equal to the total sum of the superscript on the left hand side must the equal to total sum of the super script on the right hand side. In other word the total nuclear charge on the reactant side must be equal to the total nuclear charge on the product side and the total nuclear mass on the reactant side must be equal to the total sum of nuclear mass on the product side.