nuclear decay reactions worksheet answers

nuclear decay reactions worksheet answers

 A radioactive decay is a natural type of nuclear reaction involving the emission of alpha, beta, positrons or gamma rays,

 alpha decay create new atom, in two places above the parent nuclide in the periodic table of element, beta and positron create isobars  while gamma does cause any transmutations.

 Another type of radioactive decay is electron capture, this also create an isobars, below are some questions and  answers

                                 

describe what changes occur during alpha decay

 when an atom undergo an alpha decay,   it loses two from the atomic number an four from the mass umber, the atomic number of the daughter nucleus therefore decreases by two and the mass number decreases by four. The new atom formed is two places above the parent nucleus in te periodic table of element.

 For example

²³⁰ ₉₀Th → ²²⁶ ₈₈Ra  + ⁴₂He

 The atomic number of thorium is 88 , while the mass number is 226.

describe what changes occur during beta decay.

 When an atom undergo a beta emission, the atomic number increases by one while the mass number of the daughter nucleus remain unchanged. During beta decay,  a protons decays into a neutron, an electron and an electron antineutrino.  This result to increases in the total number of protons by one( an increased atomic number) while the total number of nucleon( sum of protons and neutron)  and hence the mass number remain unchanged.

²³⁴ ₉₀Th → ²³⁴ ₉₁Pa + ⁰₋₁e

 The atomic number of protactinium-234  is  91 while the that of the thorium is 90, the mass number of protactinium is 234 while that of protactinium is also 234.

describe what changes occur during gamma ray emission.

  When an atom undergo a gamma decay, the atomic number and the mass number of the daughter nucleus does  not change, in other word  no new nucleus is formed during a gamma decay.

 In a gamma decay, an excited nucleus, complex or unstable nucleus will emit a gamma ray to return to its stable state.

 Gamma decay is usually accompanied after a nuclear reaction.

describe what changes occur during electron capture

 when an atom undergo an  electron ,  one of its inner electron is absorbed by a proton, the protons, decays into an neutron and a neutrino, the  daughter atom formed has atomic number less than the parent nucleus by one.

 The resulting nucleus from an electron capture is a one position above the parent nucleus in the periodic table of element.

Electron capture and a positron decay has the same effect, the only difference is that in electron capture the atom  there is no emission of particle, in other words it is an internal affair within the atom.

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

 when carbon-11 undergo a positron emission, one of its proton decays into neutron, a positron and an electron neutrino.  The atomic number of the parent nuclide decreases by one while the mass number remain unchanged, 

a proton decays into a neutron, no change in the total number of nucleons ,  therefore the atomic mass does not changes, the atom that fit the above description is boron-11.

 Let us put down the decay equation to see if we are right or wrong.

¹¹ ₆C → ¹¹ ₅Pa + ⁰₊₁e

 let us balance tis equation to check if we  our answer is correct or not.

for the superscript, we have

11→11+0

 and for the subscript , we have

6 →5 +1

 therefore , since both sides are equal the equation is balanced.

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

 When americium undergo an alpha decay, it loses two protons and two neutron, thereby ejecting an alpha particle, the atomic number of the resulting nucleus decreases by two while the mass number decreases by four, the daughter nucleus formed is in two places below the  parent nucleus in the periodic table of element.

  The atom that fit the above description is neptunium-237

 Let us write the decay equation to verify.

²⁴¹ ₉₅Am → ²³⁷ ₉₃Pa + ⁴₂He

 Let us balance the above equation to check if we are wrong

 For the superscript, we have

241 → 237+4=241

And for the subscript

95 → 93+2=95.

 Therefore the above equation is balanced.

determine the identity of the daughter nuclide from the positron emission of 15 8 o

 when oxygen-15 undergo a positron emission, the atomic number of the daughter nucleus decreases by one while the mass number remain unchanged, the daughter nucleus formed will be in a one positron above the parent nucleus in the periodic table of element.  The nucleus that fit the above description is nitrogen-15.

 Let us write the beta decay equation to find out

¹⁵ ₈O → ¹⁵ ₇N + ⁰₊₁e

In a positron emission, a proton decays into an neutron and a neutrino,  a positron is an antiparticle of electron, they both have same masses but opposite charges

determine the identity of the daughter nuclide from the alpha decay of 215 84po.

When polonium under an alpha decay, it loses two proton and two neutron, the atom formed or in other words the daughter nucleus formed has atomic number less by two to that of the parent nucleus while the mass number is less by four. The atom that will fit the above position will have an atomic number of 82 and a mass number of 211. And the atom that fit the above description is lead -211

 Let us write the decay equation to verify.

²¹⁵ ₈₄Po → ²¹¹ ₈₂Pb + ⁴₂He

 Let us balance the above equation to find out what is happening

 For the superscript, we have

215 → 211+4=215

 And for the subscript

 We have                                   

84 → 82+2=84

 Therefore since  both sides are balanced, we can confidently say that we are  not wrong.

 An alpha decay result in the emission of nucleus the size of a completely ionize helium-4 atom.

determine the identity of the daughter nuclide from the positron emission of 13 7 n

 when nitrogen-13 undergo a positron emission, one of its protons decays into neutron, apositron and an electron neutrino,  this result to the formation of daughter  nucleus with atomic number of 6 and a mass number of 13,  the reason why mass number doesn’t change in a positron emission is that , the proton lost is recounted as a neutron,  of which both are counted same when computing the mass number of atoms.

 The atom that fit the above description is carbon-13

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

¹³ ₇N → ¹³ ₆C + ⁰₊₁e

 Let us balance the above equation to find out what is happening

 For the superscript, we have

13 → 13+0=13

 And for the subscript

 We have                                   

7 → 6+1=7

 Therefore since  both sides are balanced, we can confidently say that we are  not wrong.

determine the identity of the daughter nuclide from the positron emission of 68 32 ge

 when germanium undergo a positron  emission, on of it proton decays into  a neutron, the atomic number decreases by one while the mass number  remain unchanged the nucleus formed  will have an atomic number of 31 and a mass number of 68, the atom that will fit the above description is  gallium-68.

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

⁶⁸ ₃₂Ge → ⁶⁸ ₃₁C + ⁰₊₁e

 Let us balance the above equation to find out what is happening

 For the superscript, we have

68 → 68+0=13

 And for the subscript

 We have                                   

32 → 31+1=32

 Therefore since  both sides are balanced, we can confidently say that we are  not wrong.

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

 During an electron capture , am inner electron is captures by the  electropositive nucleus, a proton decays into a neutron and a neutrino,   the atomic number of the daughter nucleus formed is less than that of the original element by one while the mass number remain unchanged.

 In the  above case, the daughter nucleus formed will have an atomic number of 36, and the atom with this identity is krypton-81

 Let us put down the equation to find out if we are wrong

⁸¹ ₃₇Rb + ⁰_-1e →X

⁸¹ ₃₇Rb + ⁰_-1e → ⁸¹ ₃₆Kr

 Let us balance the above equation to find out what is happening

 For the superscript, we have

81 → 81 +0 =81

 And for the subscript

 We have                                   

37 → 36+1=36

determine the identity of the daughter nuclide from the electron capture by 84be.

when a berium-8 nuclide under a an electron capture, the identity of the daughter atom can be found by finding the atomic number of the daughter nuclide.

 when an electron capture occur,  a proton decays into a neutron and a neutrino, the atomic number of the resulting nuclide is decreased by one while the mass number remain unchanged.

 The nuclide that will fit the above description is lithium-8

 Let us put down the decay equation to see if lithium -8 is the daughter nuclide.

⁸₄Be + ⁰_-1e →X

⁸ ₄Be + ⁰_-1e → ⁸ ₃Li

 Let us balance the above equation to find out what is happening

 For the superscript, we have

8 → 8 +0 =8

 And for the subscript

 We have                                   

4+(-1) → 3=3

 Therefore since the above equation is balanced , , the daughter nuclide formed is therefore lithium-8.