identify the nuclide produced when americium-241 decays by alpha emission: 241 95am→42he + ?

identify the nuclide produced when americium-241 decays by alpha emission: 241 95am→42he + ?

let us write the decay  equation 1^st

²⁴¹₉₅Am→⁴₂He + ²³⁷₉₃Np

When americium undergo an alpha decay, it atomic number decreases by two while the mass number decreases by 4, the new atom formed will have an atomic number of 93 and a mass number of 237, the nuclide that fit the above description is neptunium-237,

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

²⁴¹₉₅Am→⁴₂He + ²³⁷₉₃Np

All the number with red on the left hand side must be equal to all the numbers with red on the right hand side.

241→4+237=241

And all the numbers with green on the left hand side must be equal to all the number with green on the right hand side.

95 →2+93=95

Since the above condition is satisfied  , the above equation is balanced.

identify the nuclide produced when iodine-131 decays by beta emission: 131 53i→ 0−1e + ?

 when iodine-131 undergo a beta decay, the  atomic number increases by one while the mass number remain unchanged, the nuclide formed will therefore have an atomic number of  54 and a mass number of 131. The nuclide that fits the above description is xenon-131.

 Let us write the above equation

¹³¹ ₅₃I→ ⁰₋₁e + ?

¹³¹₅₃I→ ⁰₋₁e + ¹³¹₅₄Xe

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

To balance an equation all the superscript on the reactant side must be equal to the sum of the superscript on the product side, and all the subscript on the reactant side must be equal to the sum of the subscript on the product side.

 Therefore

131 =0+131=131

And

53= -1+54=53

 Balanced.

identify the nuclide produced when thorium-234 decays by beta emission: 234 90th→ 0−1e + ?

²³⁴₉₀Th→ ⁰₋₁e +?

A beta decay will cause the neutron to decay to  proton and an electron, the atomic number increases by one while the mass number remain unchanged. The  daughter nuclide formed will then have an atomic number of  91 and  mass number of 234,  the atom that fit the above condition is protactinium-234.

²³⁴₉₀Th→ ⁰₋₁e +²³⁴ ₉₁Pa

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

 For the superscript, we have

234  →0+234=234

 N for the subscript, we have

90 → (-1)+91 =90

 Since both sides are equal for both atomic numbers and mass number , the above equation is said to be balanced.

identify the nuclide produced when thallium-201 decays by electron capture: 201 81tl+ 0−1e→00γ + ?

 when thallium-201 undergo an electron capture, one of its inner electron is capture by  the nucleus, protons decays into a neutron and a neutrino, the atomic number decreases by one while the mass number remain unchanged. The nuclide formed will then have an atomic number of 80 and a mass number of 201, the nucleus that will fit the above description is  mercury-201.

 We can write the decay equation s follow

²⁰¹₈₁Tl + ⁰₋₁e → ²⁰¹₈₀Hg

 Let us check this equation to see if it is balanced or not

For the superscript, we have

201 +0  →201

81+(-1) →80

 SINCE, it is balanced, the above equation is correct and our atom formed is mercury.

identify the nuclide produced when fluorine-18 decays by positron emission:

 when  fluorine-18 undergo a positron emission,  one of it protons decays into a neutron, a positron and an electron neutrino, the nuclide formed, will then have an atomic number of  8 and  mass number of 19. The nuclide formed is therefore oxygen-19, because it has an atomic number of  9 and a mass number of 19.

 We can write the decay equation as follows

¹⁹₉F → ¹⁹₈O + ⁰₁e

 But, why does  the mass number does not change.

 The proton decays in the process result to the creation of neutron, the total number of nucleons did not changed, and mass number of any tom is the sum of it nucleons.

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

 For the superscript, we have

19 →19+0 =19

And for the subscript, we have

9 →8+1=9

what particle is produced by the decay of thorium-214 to radium-210? 21490th→21088ra+?

 Let us write the nuclear equation 1^st

²¹⁴₉₀Th→²¹⁰₈₈Ra+?

 From the above equation, we can see that the atomic number has decreases by one while the mass number has also decreases by 4 for the parent nuclide. Any nuclear decay that can result to a change in mass number is an alph decay , bet, gamm,positron and electron catures do not Cause Ny change in the mass number of atoms. They only create isobars.

 Therefore the particle produced is alpha particle.

 Let us put alph particle into the equation to see

²¹⁴₉₀Th→²¹⁰₈₈Ra+⁴₂He

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

for the superscript, we have

214 →210”+4=214 balanced

and for the subscript, we have

90 →88+2=90 balanced,

 the above equation is balanced

identify the nuclide produced when phosphorus-32 decays by beta emission: 3215p→ 0−1e + ?

When an atom undergo a beta emission, one of its neutrons decays into  proton and  an electron, the atomic number increases by one while the mass number decreases by zero, therefore the new atom formed will have an atomic number of +1 to that of the parent nuclide,

 For the case of phosphorus-32, the daughter nuclide will be sulphur-32

Let us put down the decay equation

³²₁₅P→ ⁰₋₁e + ?

 ³²₁₅P→ ⁰₋₁e + ³²₁₆S

identify the nuclide produced when iodine-131 decays by beta emission: 131 53i→ 0−1e + ?

when iodine-131 decays by beta decay one of its neutron decays into a proton and an electron, the atomic number increases by one while the mass number remain unchanged.  The nuclide formed will have an atomic number of 54 and a mass number of 131.

 The nuclear decay equation will therefore be

¹³¹₅₃P→ ⁰₋₁e + ¹³¹₅₄S

Let us balance the above equation to see, if

To balance a nuclear equation the  superscript on both side must be equal and likewise the subscript on both side must be equal.

 For the above equation, we have

131  →0+131=131

 For the subscript

53 → (-1)+54=53

 Since both sides are equal, then the equation is balanced.

?→5927co+ 0−1e, what is the parent nuclide

 The parent nuclide is

Let us write the above equation first,

?→⁵⁹₂₇Co+ ⁰₋₁e

 The above decay is a beta decay, the atomic number increases by one while the mass number reamin constant, the daughter nuclide formed will have an atomic number of 26 and  a mass number of  59. The nuclide that fit the above description is iron-57. He above equation will therefore be

⁵⁹₂₆Fe →⁵⁹₂₇Co+ ⁰₋₁e