Nuclear physics And you……..

What happens when: The nucleus of an atom gets larger and larger? Is there a finite limit to just how big it can be? Well let’s look at a graph of #n o –vs- # P + and find out !!!!!!!!

#n o –vs- # P + for isotopes #n o #p + at ~ 10 nuclear decay in some isotopes Bi (w/126 n o ) = last stable ele. Every element with more than 83 p + breaks down over time! Stable if #n o > #p + m = 1 Band of stability

Why? It is a battle between 2 opposing forces! F electrostatic repulsion (between P + s) – about 100 x weaker than strong F but acts over longer distances (F  w/  d). F strong limited to ~ m – outside that d, no strong F exists! When F ele. repulsion > F strong we observe….

Nuclear Decay

The energy that holds the nucleotides in the nucleus is called The binding energy Binding energy for a nucleotide = the energy required to remove the nucleotide from the nucleus to infinity Think of the parallel ionization energy for electrons More stable nuclei require more E to break them apart = greater binding energy

A strange observation The Σ masses of the unbound P + and n o is greater than the Σ masses of P + and n o in a stable nucleus. The difference in masses is called the “mass defect” = Δm ΔBinding E = Δm (c 2 )

Example He mass = x kg -vs- 2 P n o mass = x kg Mass defect (Δm) = x kg So what would the binding E be for the He atom in J and in ev ????????????

Did you calculate: 4.53 x J and 2.83 x 10 7 ev ????? Great!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! This is about 2 million times the E required to remove a valence e- from an atom.

In nuclear reactions E = mc 2 … ΔBinding E = E released by the rxn. A fission reaction of U-235 into the products Ba-141 and Kr-92 would result in a mass defect in the daughter nuclei and an increase in the binding E of the products. A fusion reaction of H-2 and H-3 to form He-4 results in a mass defect in the product of the reaction and a higher binding E per nucleon Show: fission and fusion cassiopeia (youtube)

Check this out - Fusion endothermic Fission exothermic Fusion exothermic Fission endothermic Electrostatic repulsion becomes greater than binding E = nuclear decay

Types of Nuclear Decay Alpha (α) U  4 2 α Th Beta (β) Th  Pa β + υ (anti neutrino) Neutron changes into a proton!!!!! Releasing an electron from the nucleus with a lot of energy – when n o /p + ratio is too high

Positron emission Na  Ne β + υ (+ beta particle) (neutrino) positron proton changes into a neutron releasing electron antimatter with a lot of energy – when p+/n o ratio is too high Gamma decay U  U γ (photon of electromagnetic E)

Decay particles have discrete amounts of energy Quantized E ? E levels in nucleus? WHAT?????

How about some fun and exciting questions for you then.

1) The number of neutrons in 17 O is: Mass # - #p + = #n o so 17 – 8 = 9 n o

2) What is produced when 83 Sr emits an alpha particle? Sr α = Kr

3) For a nucleus that undergoes alpha decay, what is the disintegration energy if the mass of the alpha particle, daughter and parent are A, B, and C, respectively. a.(A+B+C)/c 2 b.(A+B+C) * c 2 c.(A+B-C)/c 2 d.(A+C-B)/c 2 e.(C-A-B) * c 2

4) When a gamma ray is emitted from the excited state of 12 C, the resulting nucleus is: 12 C

5) When a neutron strikes an unknown nucleus, the observed products are 28 Si and 2 H. What is the unknown nucleus? Si H = 1 o n o P

t ½ The fun applied math part And you

t ½ Half life Time interval for ½ of the original sample to decay to the product Can be anywhere from a fraction of a second to years depending on the isotope.

Recall The decay curve of a radioactive decay process is exponential. That makes it much more difficult to determine how much of an initial sample will remain after a given time interval (unless you graph out the entire decay process) But we can calculate what we are looking for using applied math --- YES!!!!!!!!!!

The decay curve

If I start with 100 atoms After 1 half life I have 50 orig. atoms left After 2 half lives, I have 25 orig. atoms left Or I can use this relationship: The # of orig. atoms left = orig. # atoms/ 2 n n = the number of ½ lives that have elapsed Or express it like this: N n = N O / 2 n Then rearrange to get:

The Fundamental expression N n = (1/2) n N o

Example problem just for you If N o (orig. # nuclei) = 100 N n = 20 t ½ = 5 days t = ? (how much time has elapsed)

The Solution 20 = (1/2) n 100 Divide out N O and take the log of both sides log 20/100 = - n log = - n so n = 2.33 log 2 Therefore: 2.33 ½ lives have elapsed 2.33 x 5 days = days

Decay activity based on a decay constant (probability) Activity = #disintegrations/time = ΔN/Δt ΔN/Δt = - KN (N = #nuclei not decayed) Where –K is a decay constant based on the probability of decay of a nuclei in that time t N = N o e –k t so N/N o = e -k t so taking natural log Ln (N/N o ) = - K t so when N/N o = ½ = -K t½ so.693/K = t½

I think you get the idea – just a few more then 1) 194 Po has a t ½ of 0.7 seconds. How many grams of a 3 kg sample remain after 2 minutes Answer: 7.6 x g. 2) 210 Po has a t ½ of 138 days. How much of a 2.2 kg sample remains after 1.5 minutes? Answer: 2.199kg (2199 g)

NOW That was fun!!!!!!!!!