balancing nuclear equations worksheet with answers : nuclear physics

 Alpha, beta , gamma and positron decays are types of are types of nuclear decay that create new atoms.

 When an atoms undergo and alpha decay it create a daughter nucleus in two places above the parent nucleus in the periodic table, when an atom undergo a beta decay it created an a nucleus in one place above the parent nucleus in a periodic table, when an atom undergo a positron decay it created a daughter nucleus in one place above the parent nucleus in the periodic table and when an atom undergo a gamma decay does not created a new atom.

 Below are some example of radioactive decays and the resulting nucleus formed.

 

23491pa undergoes beta decay. what is the mass number of the resulting element?

 The mass number is 234, how?

 LET US START BY WRITING THE DECAY EQUATION

 ²³⁴₉₁Pa → X + ⁰_-1e

 When protactinium undergo a beta decay, the daughter nucleus formed is will have an atomic number which +1  to that of the parent nucleus. In other word the atomic number of the new nucleus formed will increase by one while the mass number will remain the same. The atom that fit the above description is uranium-234

²³⁴₉₁Pa → ²³⁴₉₂U + ⁰_-1e

 In a  beta emission , a proton decays into a neutron, electron and an electron antineutrino, the mass number does not change because the total number of nucleon did not change.

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

 For the superscript

234  → 234+0= 234

For the sub-script

91 → 92+(-1)= 91.

 The above equation is therefore balanced.

23490th undergoes beta decay. what is the atomic number of the resulting element?

 The atomic number of the resulting element is 91, and the element is protactinium, how?

 When a thorium undergo a beta decay, the atomic number, a protons decays into a neutron and an electron, the atomic number therefore increases by one, because the atomic number is the sum of the protons in an atom while the mass number is the sum of the neutrons in the atom, why does that the mass does doesn’t change after a beta decay.

 The mass number is the sum of the protons and the neutron, when counting for the mass number we count both the neutrons and the protons as nucleon, therefore, even though a neutron changes to a proton , it is still a nucleon.

 Let us consider this decay equation

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

part b 23490th undergoes beta decay. what is the atomic number of the resulting element?

The atomic number of the resulting nucleus will be the atomic number of the parent nucleus plus one, that is Z+ 1, but the Z = 90, which is equal to 90 +1=91

 Therefore the atomic number  of the resulting element is 91 and this correspond to protactinium-234. Consider the equations below

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

²³⁴ ₉₀Th → ²³⁴ _9iPa + ⁰₋₁e

23090th undergoes alpha decay. what is the mass number of the resulting element?

 When thorium undergo an alpha decay,  two proton is lost and two neutrons is also lose an a particle , the size of a helium nucleus is emitted,  this result to the lost of two from the atomic number and four from the mass number, the resulting atom will therefore have a mass number of 230-4= 226 and an atomic number of 90-2=88.

 Le us see the alpha decay equation below

²³⁰ ₉₀Th → X + ⁴₂He

  The atom that will fit the above equation must have an atomic of 88 and a mass of 226, the atom that fit the above description is radium-226.

 The above nuclear equation will therefore be

²³⁰ ₉₀Th → ²²⁶ ₈₈Ra  + ⁴₂He

 Let us check to see if we are wrong.

 Balancing the above equation , we have

 For the superscript

230  → (226+ 4)=230 balanced

 And the

 For the sub-script , we have

90 → (88+2)= 90

234 92 u undergoes alpha decay what is the atomic number of the resulting element

 The atomic umber of the daughter nucleus from an alpha decay of uranium-234 will be  ( 90-2=88),  ,

From the decay equation

²³⁴ ₉₂U → ²³⁰ ₉₀Th  + ⁴₂He

  When uranium-234 undergo an alpha decay, the atomic number decreases by two while the atomic mass decreases by four.   We can check the above  equation to fin if it is balanced or not.

 To balance a nuclear equation, the law of conservation of matter requires the , the superscripts on the left hand side must be equal to the superscript on the right hand side and the sub-script on the left hand side must be equal to the sub-script on the right han side.

 Therefore for the above equations , we have:

234 → 230+ 4=234

And

92 →90+2= 92

 Therefore ,  the above equation is balanced.

describe what changes occur during alpha decay

when an atom under an alpha decay,  it lose  a particle  the size of  a helium  nucleus, it lose two protons and two neutron, the atom number of the resulting nucleus is less by two while the mass number is less by four the daughter nucleus formed is therefore A-4 and Z-2.

See the equation below

^A_Zx → ^A-4 _Z-2 Y + ⁴₂He

 For example ,

 When thorium  undergo an alpha emission, the equation will be

²³⁰ ₉₀Th → ²²⁶ ₈₈Ra  + ⁴₂He

 the daughter nucleus formed, radium-226, has an atom ic number of 88 and mass number of 226.

describe what changes occur during beta decay.

when an atom undergoes a beta decay, one of its neutron decays into a proton,  an electron and an electron antineutrino and  energy.

 When this occur, the atomic number increases by one while the mass number remain constant.

 The mass number does not change because, the neutron lost is recounted as the new protons formed, ( they are both nucleons).

 Beta decay create isobars, nucleus with the same atomic  mass but different atomic number or in other words, different atoms with the same  masses,

 Isobars are completely different atoms, because they exhibit entirely different chemical and nuclear properties.

  For  example, when protactinium, undergo beta decay, we have a uranium-234 formed.

 See the equation below

²³⁴₉₁Pa → ²³⁴₉₂U + ⁰_-1e

 From the above we can see that the mass number  does not change while the atomic number increases by one.

write a nuclear equation for the alpha decay of 23892u.

²³⁴ ₉₂U → ²³⁰ ₉₀Th  + ⁴₂He

 

.

 Therefore for the above equations , we have:

234 → 230+ 4=234

And

92 →90+2= 92

 Therefore ,  the above equation is balanced

describe what changes occur during gamma ray emission.

During a gamma ray emission, an excited atom , return to its ground state by emitting  an energy in form of electromagnet wave, hence the gamma ray, this is why gamma has all the properties of an e.m waves.

Gamma ray emission usually occur after a nuclear decay or  nuclear reaction, when an atom has already absorb an excess energy, the electrons have jumps from lower energy levels to higher ones.

 For this  atoms to return to its round state( the unexcited state) the bohr’s theory requires that some packets of radiations to be emitted before the atom will stabilize.

 The  atoms will do so through gamma emission.

determine the identity of the daughter nuclide from the positron emission of 116c.

 let us start by writing the nuclear equations

¹¹ ₆C → ¹¹ ₅B + ⁰₊₁e

 the atom formed is boron-11

 when an tom undergo a positron one of its protons  decays into a neutron, a positron and an electron neutrino, the atomic number of the daughter nucleus decreases by one while mass number  remain unchanged.

 The new atom formed is one position below the parent nucleus in the periodic table of element.

Positron decay create an isobars that is more negative than  the parent atoms,

 One similarity between  positron emissions and beta decay is that they all create isobars, whereas one is higher up  and the other is lower in the periodic table of element.

write a nuclear equation for the alpha decay of 24195am.

 ²⁴¹ ₉₅Am → ²³⁷₉₃Th + ⁴₂He

for americium-241 to undergo  an alpha decay, it has to lose two protrons and two neutron from the nucleus, this result to decreases of two from the atomic number and four from the atomic mass.

We  balance the above equation as follows

Therefore for the above equations , we have:

241 → 237+ 4=241

And

95 →93+2= 95

 Therefore ,  the above equation is balanced

determine the identity of the daughter nuclide from the positron emission of  

158O

¹⁵₈O → ¹⁵₇N + ⁰₊₁e

 From the above nuclear equation we can see that the atomic number of the daughter nucleus nitrogen-15 is 7 while the mass number remain 15. The nucleus formed is nitrogen 15

 When an atom undergo , a positron emission, a proton changes to an electron and  a neutron and an electron neutrino. The atomic number of the daughter nucleus decreases by one while the mass number remain unchanged.  The nucleus formed is an isobar but one place above the parent nucleus in the periodic table.