solving electron capture vs positron emission: physics

 electron capture occur when an inner electron is captured by a nucleus , a proton decays in a neuton and the atomic number decreases while the mass number remain constant.

 in a positron decay,  a proton decays in a positron and a neutron, the atomic number decreases by one. while the mass number remain constant.

  They both have same effect on the  atomic number and mass number of atom, they only difference is that they occur in different ways,

Example of electron capture

¹⁹⁶₈₂Pb+⁰_-1e → ¹⁹⁶₈₁Tl

 Example of positron decay

⁵⁰₂₅Mn→ ⁵⁰₂₄Cr +⁰₊₁e

 differences

As we can see the electron capture  dos involve emission of particles  while positron involves  emission of positron

 Electron capture is an internal affair while positron is not.

what is the daughter nucleus produced when 123in undergoes electron capture?

When indium -123 undergo  an electron capture, one of its proton decay into a neutron, in the process an inner electron is captured by the nucleus, and this result to a decreases in atomic number by one while the mass number remain unchanged. The atom that fit the above description is  cadmium-123.

 Let us write the nuclear decay equation for the above reaction.

¹²³₄₉ In +⁰_-1e →  ¹²³₄₈Cd

 From the above reaction we can  balance it , to see if we are right or wrong.

to balance a nuclear equation we have to ensure that the total sum of the supercritp on both side must be equal as well the total sum of the subscript,

 for the superscript

 we have

123 + → 0 123

And for the subscript we have 49+(-1)= 48

 Therefore we are not wrong.

                                                                                             

 what is the daughter nucleus produced when 63zn undergoes electron capture?

Let us start  by writing the decay equation

Therefore

⁶³₃₀ Zn +⁰_-1e →  ⁶³₂₉Cd

                                                                     Wen zinc-63 undergo an electron capture, one of its inner electron is captures  by the nucleus, a proton decays into a neutron and neutrino, the atomic number of the  daughter nuclide decreases by one while the mass number remain the same.

 The nuclide formed will therefore have an atomic number of 29 and a mass number of 63.

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

For the subscript ,we have

63+0 → 63

 And for the subscript, we have

30+(-1) →30

 Therefore the above equation is balanced, since both sides are the same.

what is the daughter nucleus produced when 167tm undergoes electron capture?

 when thulium with atomic number of 69 and mass number of 167 under an electron capture,  the  one of the inner electron of thulium decay into a neutron and neutrino. The atomic number of thulium decreases by one while that of the daughter nuclide atomic number  increases by one. The  daughter atom formed  will have an atomic number of 68 and mass number of 167. The atom that fit the above description is Erbium-167.

 let us put down the decay equation

¹⁶⁷₆₉ Tm +⁰_-1e → ¹⁶⁷₆₈Er

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

To balance a nuclear decay equation, we start by checking the atomic number of the nuclide and mass number, or simply the subscript and the super superscript.

 For the superscript, we have;

167 +0 → 167

And for the subscript, we have

69+(-1) → 68

 The above equation is therefore balanced.

 what is the daughter nucleus produced when 195au undergoes electron capture?

 When  gold with a mass number of 195 and atomic number of 79 undergo electron capture, one of its inner electron is absorb by the nucleus,  the number of protons in gold decreases by one while the total number of nucleons remain unchanged.

 The  daughter nuclide formed will have an atomic number of 78 and a mass number of 195, the atom that fit the above description is platinum-195.

 Let us write the decay equation now.

¹⁹⁵₇₉Au +⁰_-1e → ¹⁹⁵₇₈Pt

Let us check to see if the above equation is correct.

 The atomic number on both side must be equal, therefore

We have 79+(-1) =78 on the left hand side and 78 on the right hand side,

 Secondly mass numbers on both side must be equal

 We have195+0=195 on the left hand side and 195 on the right hand side

 Therefore the above equation is balanced.

what is the daughter nucleus produced when 83 sr undergoes positron emission

when strontium-83 with atomic number 38 undergo an electron capture, the atomic number of the daughter nuclide formed will be 37 and the mass number will be 83.

 Let us put down the decay equation first

⁸³₃₈Au → ⁸³₃₇Pt +⁰₊₁e

During an positron emission, a proton decays into a neutron, a positron and an electron neutrino in the process, the atomic number of the daughter nuclide decreases by one while the mass number did not change.

  We can check the above equation see if we are right or wrong.

 For the super script we have

83 →83+0=83

And for the subscript, we have

38 →37+1= 38

 The above equation is balanced.

what is the daughter nucleus nuclide produced when 50mn undergoes positron emission

 when manganese undergo a positron decay, one of its proton decays into an a positron, a neutron and an electron neutrino .  the atomic  number of the daughter nuclide formed decreases by one while the mass number remain the same.

  Thee daughter nuclide that will be formed will have atomic number of  24 and mass number of 50. The atom that will fit the above description is chromium-50

 Let us write the above in nuclear equation

            ⁵⁰₂₅Mn→ ⁵⁰₂₄Cr +⁰₊₁e

what is the daughter nucleus produced when 196pb undergoes electron capture?

 When lead undergo an electron capture, the atomic number decreases by one while the atomic mass remain unchanged. The daughter nuclide formed will have an atomic number of 81 and a mass number of 196

 We can write this in a nuclear equation

¹⁹⁶₈₂Pb+⁰_-1e → ¹⁹⁶₈₁Tl

 During an electron capture,  an inner electron of  lead is captures by the nucleus, a proton decays into a neutron in the process, the atomic number of the daughter nuclide decreases by one while the mass number did not change

what nuclide is formed when 90sr undergoes β decay?

 When strontium undergo a beta decay, the atomic number of the new atom increases by one while the mass number remain the same, during a beta decay a neutron decays into a proton, an electron and an electron antineutrino.

 The daughter nuclide formed will then have an atomic number of 51 and a mass number of 90, the nuclide that will satisfied the above condition is antimony-90

⁹⁰₅₀Sr→ ⁹⁰₅₁Tl +⁰_-1e

We can balance the above equation to see if we are right or wrong.

 To balance a nuclear equation the law of conservation of mass requires that total mass and charge on the reactant side must be equal to the  total mass on the product side and the total charge on the product side must be equal to  the total charge on the product side.

 In order word, the sum of the mass numbers on the reactant side must be equal to the sum of the mass number on the product side and the sum nuclear charge on the product side must be equal to the sum of the nuclear charge on the product side.

 Therefore for the left hand side, we have a total of  90 on the left hand side and a total of 90+0 on the right hand side.  And for the nuclear charge we have a total of 50 on the reactant side and a total of 51+(-1) on the product side.

 calculate the mass defect for u-238 which has a mass of 238.050784 amu.

 mass defect is the difference between the calculated mass an the measured mass, the calculated mass is obtained by  multiplying the total number of nucleons  with their respective masses.

 For the above example, the measured mass is  238.050784 amu. While the calculated mass can be calculated as

The total number of nucleon in uranium is 238,

 The number of protons is 92

The number of neutrons is 238-92= 146

 The mass of a proton is 1.0072766 a.m.u

The mass of neutron is 1.008701 amu

 

Mass defects  is 239.9397932-238.050784

 1.8890092 amu

calculate the mass defect for u-235 which has a mass of 235.04392 amu.

mass defect is the difference between the calculated mass an the measured mass, the calculated mass is obtained by  multiplying the total number of nucleons  with their respective masses.

 For the above example, the measured mass is  235.04392amu. While the calculated mass can be calculated as

The total number of nucleon in uranium is 238,

 The number of protons is 92

The number of neutrons is 235-92= 143

 The mass of a proton is 1.0072766 a.m.u

The mass of neutron is 1.008701 amu

Mass defects  is 236.9136902-235.04392amu

1.8697702amu

                                                                                                                                                                                                                                                                                                                                            

  write a nuclear reaction for the neutron-induced fission of u−235 to produce te−137 and zr−97.

 Let us  write  the decay equation,

¹₀ n + ²³⁵₉₂U→  ¹³⁷₅₂Te +⁹⁷₄₀Zr  +2 ¹₀ n

The above equation is for nuclear fission decay equation,

 Uranium-235 is bombarded with a neutron and tellurium-137 and zirconium-97 is formed.

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

  The above equation is for nuclear fission decay equation,

 For the super script, we have,

1+235 → 97 +137+ 2= 236

And for the subscript

0+92 = 52+40+0=92

 Therefore, the equation is balanced.