Particles And Nuclear Equations

Mr. Shields Regents Chemistry U02 L02

Radioactive Emissions•There are several types of particles that can be expelled from a nucleus undergoing nuclear decay…

Alpha, Beta &Gamma wereFirst identifiedBy Rutherford

(see Table O)

Penetrating abilities

•Particles emitted during radioactive decay have specific penetrating power into various materials.

o This provides an Indication of how much material or “SHIELDING” would be needed to stop the particle

The greater the penetrating power the greater the health risk !

Particle Penetrating Power•Alpha particles are easy to stop, gamma rays are hard to stop.

                                                                                                                                        

N to P Ratio

Remember the Neutron to Proton Ratio?

Natural radioactivity is an attempt to correct a bad Neutron to Proton ratio

It’s an attempt to get back into the “band of stability”

Or to get below Atomic number 83 a nucleus may shednuclear mass

Alpha Emission

Nuclear equation for alpha emission

Parent NuclideDaughter Nuclides

o Notice that this equation is BALANCED. Why?

o Mass and Atomic number on one side of the equation equals Mass and Atomic number on the other side

Alpha Emission

•Why would emitting an alpha particle from U-238 increase nuclear stability?

o Atomic mass is decreased by 4 (shedding mass)o Atomic number is decreased by 2 (decreases Atomic #)

Beta Emission

    

Hey! How do we get an electron out of the nucleus?

Neutrons can decay to a Proton and a beta particle.

The proton remains in the nucleusand the beta particle is expelled.

Notice that theatomic number increases by 1But the mass stays the same

Beta Emission

•Favored by Nuclides above the “Band of Stability”

• N/P ratio is too high• Beta emission Decreases the N/P ratio

• neutron decays into a proton

C: n/p ratio 8/6 = 1.33N: n/p ratio 7/7 = 1.00

Positron Emission

    

Hey! So how do we get a positron out of the nucleus?

Like the Neutron, Protons can decay .However, Proton decay into a positronAnd a neutron.

The neutron remains in the nucleusand the positron is expelled.

The positron is the opposite of an beta particle (electron)

Positron Emission

•Favored by Nuclides below the “Band of Stability”

• N/P ratio is too Low• Positron emission Increases the N/P ratio

• Proton decays into a neutron

O: n/p ratio 7/8 = 0.88N: n/p ratio 8/7 = 1.14

Gamma Rays

The last decay mode doesn’t actually involve a particle.

Gammaradiation isactually apart of thehigh energy end of the EMSpectrum.

Increasing Energy

Gamma RadiationIn nuclear decay processes, gamma radiation is a way For an excited, or metastable, nucleus to get rid of excess energy.

It occurs typically as part of an alpha or beta decayprocess in which the nucleus is left in a high energy state.

Nuclear Equations• Describes the decay process.

• Has a reactant side (left) and a product side (right)– starting isotope – ending isotope and emitted particle.

• An “” separates the two sides of the equation

• Several isotopes and their decay mode can be found in Table N.

Nuclear Equations

• Must Demonstrate the law of conservation of mass

– The SUM of the Atomic numbers and Mass numbers on each side of the arrow must be the same

Determining Unknowns in Nuclear Equations

• Write the equation for the decay of Thorium-232

• Use Table N to find the decay mode:

• Write the initial equation:

232232Th Th 44He + He + XX9090 22

We need to figure out what X isWe need to figure out what X is

Alpha decay, thorium-232

232Th 4He + X

Conservation of Mass NumberConservation of Mass Number

The sum of the mass numbers on the left side The sum of the mass numbers on the left side must equal the sum of the mass numbers on the must equal the sum of the mass numbers on the right side.right side.

9090 22

228228

Decay Equations

232Th 4He + 228X9090 22

Conservation of Atomic NumberConservation of Atomic Number

The sum of the atomic numbers on the left The sum of the atomic numbers on the left side must equal the sum of the atomic side must equal the sum of the atomic numbers on the right side.numbers on the right side.

8888

Decay of Am-241

241Am 4He + X X has to be?9595 22

237237

9393XX

So “X” is Np-237

14C X + 0e6 -1 X has to be? 14N

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Problems:

• Write the nuclear equation for the natural decay of Co-60. (Hint: Table N)