Solve 3215p→ 0−1e+?: physics tutorial

Solve 3215p→ 0−1e+?

 To  solve the above equation, let me rewrite the equation

³²₁₅P→ ⁰₋₁e+?

we can balance , the above equation, by using the law of conservation of masses and charges, the law requires all mass number on both side must be equal and all atomic number for both side must be equal

let me rewrite the above equation

³²₁₅P→ ⁰₋₁e+^a_zX

Where a I the mass number, z is the atomic number and X is the unknow nuclide

For the mass number

We can have

32= 0+a

32-0=a

32=a

a =32

and for the atomic number.

15=(-1)+z

15+1 =z

z= 16

 therefore the nuclide formed has an atomic number of 16 and a mass number of 32.

 The atom with the above property is sulphur-32

 The above equation, will then be written as

³²₁₅P→ ⁰₋₁e+³²₁₆S

write the identity of the missing nucleus for the following nuclear decay reaction:?→84be+01e

?→⁸₄Be+⁰₁e

let us balance the above equation to see

 let me start by rewriting the above equation

^a_zX →⁸₄Be+⁰₁e

a=8+0

a= 8

and z= 4+1

z=5

 the original nuclide have an atomic number  5 an a mass number  8, the atom hat fit the above description is boron-8, the above decay characterize a positron decay, where a proton decays into a neutron and an electron anitneutrino. The atomic number of the parent nuclide decreases by one while the mass number did not change, this is why the uclide beryllium is formed.

 The above equation can be written as follows

⁸₅B →⁸₄Be+⁰₁e

provide a symbol for the isotope of bismuth that contains 124 neutrons.

 The atom bismuth has an  atomic number of

?→237 93np+42he+00γ

 The above equation can be balanced and we can therefore find the missing atom,

 Let us start by rewriting the  decay equation

?→²³⁷₉₃Np+⁴₂He+ ⁰₀γ

Let us start by rewriting the equation

^a_zX →²³⁷₉₃Np+⁴₂He+ ⁰₀γ

a=237+4+0 =241

and

z =93+2+0=95

 the nuclide that decayed has an atomic number of 5 and a mass number of 241, the nuclide this property is americium-241, the atom decay is two, alpha decay and gamma decay , in alpha decay the atomic number of the parent atom decreases by 2 while the mass number decreases by 4, the nuclide formed will therefore have an atomic number of 95 and a mass number of 241, the atom with the above characteristics is americium-241

 the above decay equation can be rewritten as follows

²⁴¹₉₅Am →²³⁷₉₃Np+⁴₂He+ ⁰₀γ

Solve 5927co+ → 5625mn 42he

 When manganese undergo an alpha decay , the atomic number decreass by 2 while the mass number decreases by 4, the nuclide formed  will therefore have an atomic number of 25 an a mass number of 55, the atom that correpon to this is manganese-55

 Let us write the decay equation

⁵⁹₂₇Co→ ⁵⁶₂₅Mn +⁴₂He

write the identity of the missing nucleus for the following nuclear decay reaction: 224 88ra→42he+?

Let us rewrite the above nuclear equations

²²⁴ ₈₈Ra→⁴₂He+?

LET US BALANCE THE equation

 We will have to rewrite the equation again

²²⁴₈₈Ra→⁴₂He+ ^a_zX

224 = 4+a

A=224-4

A=220

And

For the subscript  we have

88= 2+z

Z=86

 Therefore the nuclide formed will have an atomic number of 86 and a mass number of  220.

 the above equation will therefore be

²²⁴₈₈Ra→⁴₂He+ ²²⁰₈₆Rn

  The mass number of radium decreases by four while the atomic number decreases by 2,  this osa characteristics of an alpha decay.