identify the nuclide produced when thallium-201 decays by electron capture:
201 81tl+ 0−1e→00γ + ?
When thallium
undergo a n electron capture, the atomic number of the daughter nuclide
decreases by one while the mass number remain unchanged. This is because during
an electron capture an inner electron is captured by the nucleus, and a proton
decays into a neutron and a neutrino.
Let us rewrite the above equation
20181Tl+
0−1e → 00γ +?
The atom that balances the
above equation is will have an atomic number of 80 and a mass number of 201.
The above with atomic number of 80 is mercury
20181Tl +
0−1e → 00γ + 20180Hg
LET
us balance the above equation to see if we are right or wrong
For
the superscript
201+0→ 0+201
For
the subscript
81+
(-1) → 80+0
Since
both superscript and subscript are equal for both side, the equation is
therefore balanced.
Solve ?→84be+01e
When
beryllium undergo a beta decay, the atomic number increases by one while the
mass number remain unchanged, during a beta decay, a neutron decays into a proton, an electron
and an electron antineutrino.
For the above decay, let us rewrite the
equation
?→84Be+0+1e
For a positron decay of beryllium-8, the atomic
number of the beryllium decreases by one while the mass number will remain
unchanged. The nuclide formed will have atomic number of 5 and a mass number of
8, the nuclide that fit the above condition is boron-8.
85B→84Be+01e
Let
u balance, the above equation to see if we are right and wrong.
For
the superscript, we have
8 →8+0
For the subscript, we have
5 →4+1
Since both superscript and subscript on both
side for the two side, the equation is balanced and the nuclide formed is
boron-8.
Solve ?→5927co+ 0−1e
When
cobalt-59 undergo a beta decay, the atomic number increases by one while the
mass number remain unchanged, during a beta decay, a neutron decays into a proton, an electron
and an electron antineutrino.
For the above decay, let us rewrite the
equation
?→5927Co+0-1e
For a positron decay of cobalt-5, the atomic number
of the beryllium decreases by one while the mass number will remain unchanged.
The nuclide formed will have atomic number of 5 and a mass number of 8, the
nuclide that fit the above condition is boron-8.
59 26Fe →59
27Co+0-1e
Let
us balance, the above equation to see if we are right and wrong.
For
the superscript, we have
59 →59+0
For the subscript, we have
26→27+(-1)
Since both superscript and subscript on both
side for the two side, the equation is balanced and the nuclide formed is iron-59.
what isotope of what element is produced if krypton-81 undergoes beta decay?
8136Kr
→8137Rb +0-1e
When krypton undergo a beta decay the atomic number
of the daughter nucleus formed increases by one while the mass number remain
unchanged, the new atom formed will have an atomic number of 37 and a mass
number of 81, the nuclide that fit the above description is rubidium-81
Let us
balance the above equation to see if we are right or wrong.
To balance a nuclear equation the atomic number of
the and mass number on both side must be equal, in other word the total sum of the super script on the
reactant side must be equal to the sum of the superscript on the product side
and the total sum of the subs scripts on the reactant side must be equal to
the total sum of the subscript on the
product side.
For the superscript, we have
81→
81+0
And for the
subscript we have
36 →37+(-1)
Since both
side are equal , the equation is balanced, the isotope formed is therefore
rubidium-81
write the identity of the missing nucleus for the following nuclear decay reaction: ?→5927co+ 0−1e
let us write the nuclear equation first
?→5927Co+ 0−1e
We
can find the above equation by using the
law of conservation of matter, the law of conservation of matter requires that
the total sum of the mass numbers in both side of nuclear equation must be
equal and likewise the total atomic number on both side must also be equal.
The mass number is represented in the
superscript, while he atomic number is represented as subscript.
Therefore
Let
us rewrite the above equation
azX →5927Co+
0−1e
assuming
the unknown atom is represented as X
For
the mass number.
a
=59+0
a =
59
and
for the subscript
z=27+(-1)
z=26
therefore the new nuclide formed has an atomic
number of 26 and a mass number of 59 the atom that has the above
characteristics is iron-59
we can rewrite the above equation
5926Fe→5927Co+
0−1e
write the identity of the missing nucleus for the following nuclear decay reaction: 224 88ra→42he+?
LET US START BY
REWRITING THE ABOVE EQUATION
22488Ra→42He+?
let
us balance the above equation to find the missing particle.
let
us rewrite the equation
22488Ra→42He+
azX
To find the value of X, we first start by finding
the value of a and z.
224
=4+a
A=224-4
A=220
And
to find the value of z
We
say
88=2+z
Z=88-2
Z
=86,
The nuclide with atomic number of 86 and a
mass number of 220
The nuclide that has atomic number of 86 is
radon-220
The above equation can be written as
22488Ra → 42He
+ 2202Ra
When
an atom undergo an alpha decay, the atomic number decreases by 2 while the mass
number decreases by 4 the nuclide formed will be in two places above the
original atom on the periodic table of element.
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