Download To balance a nuclear equation, the mass number and

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Calutron wikipedia , lookup

Compact Muon Solenoid wikipedia , lookup

Lepton wikipedia , lookup

Electron scattering wikipedia , lookup

Identical particles wikipedia , lookup

Standard Model wikipedia , lookup

Dirac equation wikipedia , lookup

Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup

Atomic nucleus wikipedia , lookup

Elementary particle wikipedia , lookup

Relativistic quantum mechanics wikipedia , lookup

Transcript
To balance a nuclear equation, the mass number and atomic numbers
of all particles on either side of the arrow must be equal.
LEARNING OBJECTIVE [ edit ] [ edit ] [ edit ] [ edit ]
Produce a balanced nuclear equation
KEY POINTS [ edit ] [ edit ] [ edit ] [ edit ]
A balanced nuclear equation is one where the sum of the mass numbers (the top number in
notation) and the sum of theatomic numbers balance on either side of an equation.
Nuclear equation problems will often be given such that one particle is missing.
Instead of using the full equations, in many situations a compact notation is used to describe
nuclear reactions.
TERM [ edit ] [ edit ] [ edit ] [ edit ]
baryon
A heavy subatomic particle created by the binding of quarks by gluons; a hadron containing three
quarks. They have half-odd integral spin and are thus fermions.
Give us feedback on this content: FULL TEXT [edit ] [ edit ] [edit ] [ edit ]
Nuclear reactions may be shown in a form similar to chemical equations, for which invariant
mass, which is the mass not considering the mass defect, must balance for each side of the
equation. The transformations of particles must follow certain conservation laws, such as
conservation of charge and baryon number, which is the total atomic mass number. An
example of this notation follows:
6
3
2
Li + H
1
→
4
H e + ?
2
To balance the equation above for mass,
charge, and mass number, the
second nucleus on the right side must
have atomic number 2 and mass number
4; it is therefore also helium-4. The
complete equation therefore reads:
6
3
2
Li + H
1
→
4
4
2
2
H e + H e
Or, more simply:
6
3
2
Li + H
1
→
4
2 H e
2
Register for FREE to stop seeing ads
Lithium-6 plus deuterium gives two helium-4s.
The visual representation of the equation we used as an example.
Compact Notation of Radioactive Decay
Instead of using the full equations in the style above, in many situations a compact notation
is used to describe nuclear reactions. This style is of the form A(b,c)D, which is equivalent to
A + b gives c + D. Common light particles are often abbreviated in this shorthand,
typically p for proton, n forneutron, d for deuteron, α representing an alpha particle or
helium-4, β for beta particle or electron, γ for gamma photon, etc. The reaction in our
example above would be written as Li-6(d,α)α.
Balancing a Radioactive Decay Equation
In balancing a nuclear equation, it is important to remember that the sum of all the mass
numbers and atomic numbers, given on the upper left and lower left side of
the elementsymbol, respectively, must be equal for both sides of the equation. In addition,
problems will also often be given as word problems, so it is useful to know the various names
of radioactively emitted particles.
Example
235
92
U
→
231
90
T h + ?
This could be written out as uranium-235 gives thorium-231 plus what? In order to solve, we
find the difference between the atomic masses and atomic numbers in
the reactant and product. The result is an atomic mass difference of 4 and an atomic number
difference of 2. This fits the description of an alpha particle. Thus, we arrive at our answer:
235
92
U
→
231
90
4
T h + H e
2
Example
214
84
4
0
Po + 2 H e + 2−1 e
2
→
?
This could also be written out as polonium-214, plus two alpha particles, plus two electrons,
give what? In order to solve this equation, we simply add the mass numbers, 214 for
polonium, plus 8 (two times four) for helium (two alpha particles), plus zero for the electrons,
to give a mass number of 222. For the atomic number, we take 84 for polonium, add 4 (two
times two) for helium, then subtract two (two times -1) for two electrons lost through
beta emission, to give 86; this is the atomic number for radon (Rn). Therefore, the equation
should read:
214
84
4
0
Po + 22 H e + 2−1 e
→
222
86
Rn
Writing nuclear equations
Describes how to write the nuclear equations for alpha and beta decay.