**Radioactive decay is the process where unstable nuclei break down and emit radiation**. The nuclide before decay is the parent nuclide and that after decay is daughter nuclide.

1)

**α-decay**:^{A}X →^{A-4}Y +^{4}He (α), a**helium nuclei**is emitted, atomic number - 2 while mass number -4, e.g.^{238}U →^{234}Th + α2)

**β-decay**:^{A}X →^{A}Y + e (β), the atomic number of daughter nuclei +1 while mass number unchanged, a**neutron decays into a electron and a proton**. (a antineutrino as well, but this is out of the discussion). Electron was written as atomic number -1 to balance the total number of protons. e.g.,^{14}C →^{14}N + β3)

**γ-decay**:^{A}X* →^{A}X + γ, an**excited nuclei emit an energetic photon and become stable**. e.g.,^{60}Ni* →^{60}Ni + γSometimes we use A-Z graph or N(number of neutrons)-Z graph to show the decay series.

**Radioactive decay – a random process**

Every radioactive nucleus has a certain probability to decay within a certain period of time but it’s unknown to know the time to decay. In macroscopic view the decay activity is more stable and can be calculated.

Number of disintegration of a source per second is called the activity (A) of the source, measured in (Becquerel) Bq or s

^{-1}. It’s proportional to undecayed nuclei N, i.e.,**A = kN**where**k is the decay constant in s**.^{-1}**Half life: it’s the time taken for half of the parent nuclei in any given sample to decay**. i.e., N at a certain time is equal to

**N**where N

_{0}(0.5)^{n}_{0}is the original number of parent nuclei and n is the number of period of half life passed by. Similarly we have

**A = A**.

_{0}(0.5)^{n}Now consider

**A = -dN/dt = kN**, dN/N = -k dt, integrate both sides gives ln N = -kt+C, when t = 0, N = N_{0}, so we have**N = N**. Similarly_{o}e^{-kt}**A = A**._{0}e^{-kt}Assume t be the half life. Then N

_{0}/2 = N_{0}e^{-kt}, then half life**t**. So by plotting the graph_{1/2}= ln 2/k__ln A-t or ln N-t, the y-intercept is ln A__. Then we can find the half life of the sample._{0}(N_{0}) while the slope is k**Nuclear energy**

*Binding energy of an atom is the energy to separate those neutrons and protons to infinite far*. They are originally bonded by

**strong nuclear force**(which is stronger than electrostatic force), so energy is released during the combination of nucleus. Under nuclear reaction, binding energy is provided to nucleus to separate the nucleus, and another binding energy is released through combining into a new nucleus.

__Higher binding energy means more energy is released in combination__, hence it’s

**more stable**. The maximum of binding energy occurs around Fe to Co, to lighter elements undergoes fusion to release energy to achieve stableness, while heavier elements undergo fission to release energy.

1)

**Nuclear fission**: a**large nucleus splits into 2 or more nuclei of comparable masses**. Neutrons are used to bombard the nucleus and to supply energy to it. Proton won’t be used to prevent electrostatic repulsion between proton and the nuclei. For example consider a piece of uranium-235:^{235}U + n →^{236}U (very unstable) →^{92}Kr +^{141}Ba + 3n. Note that more energetic neutrons are produced, they can also used to bombard other uranium atoms and cause a**chain reaction**.**Uncontrolled chain reaction**were used in nuclear weapons like**atomic bomb**.2)

**Nuclear fusion**:**two lighter nuclei unions into a heavier nucleus**. For example consider a fusion occurring in stars:^{2}D +^{3}T →^{4}He + n, where D and T are deuterium and tritium which are isotopes of hydrogen. Unfortunately,**extremely high temperature**is needed (10^{7}~ 10^{8}K) to**overcome the electrostatic repulsion**(in order for strong nuclear force to take place). Therefore uncontrolled fusion is invented (**hydrogen bomb**) but the technological level is**not enough to generate electricity by fusion**, through the advantages include plentiful source of hydrogen and non-radioactive waste.Unit of smaller mass

We

**define the weight of**or simply u.^{12}C as 12a.m.u.**1u = 1.66*10**. However by calculation scientists find that m^{-27}kg_{p}= 1.0073u, m_{n}= 1.0087u, m_{e}=0.00055u, considering^{12}C contains 6 protons, 6 neutrons and 6 electrons, total mass = 12.099u, however the mass of that atom is only 12u, then we say that it has a**mass defect**of 0.099u. In fact, the energy is released in terms of binding energy.**Mass-energy equivalence**

When

**mass are released in forms of energy**, we found that there’s a certain relationship between mass and energy.**Einstein**states the**E = mc**(ΔE = Δmc^{2}^{2}), where c is the speed of light in vacuum.For example, consider n → p

^{+}+ e^{-}, mass defect = (1.0087-1.0073-0.00055)u = 0.00085 u, then energy released = (0.00085)(1.66*10^{-27})(3*10^{8})^{2}= 1.27 * 10^{-13}J. When one mole of neutrons decays, (1.27*10^{-13})(6.02*10^{23}) = 7.65*10^{10}J of energy, so we can see nuclear reaction can give out a large amount of energy.
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