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+azX
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
azX →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
azX →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+ azX
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.
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