PHYS 1300 Lecture Notes - Lecture 29: Exponential Decay, Beta Decay, Beta Particle

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salmonbuffalo710

1 Feb 2017

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University of Guelph

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Physics

Course

PHYS 1300

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298 92 u = 4 a he + b a = 2, b = 294 90 th, 230 90 th alpha decay; 4 2 he. Amount of energy released? delta e = (-8. 5036e-30)(3e8)2. = -7. 7e-13 j, the energy released per decay. 230 90 th = 4 2 he + 226 88 ra. E = mc2 delta e = (delta m) c2 (masses are given) mass before = 230th = 230. 033 u mass after = 226ra + 4he = 226. 0254+ 4. 0026 delta m = (mass after - mass before) = -0. 005121 u, negative means energy is released. Beta (-) decay, one neutron is replaced by a proton and an electron/beta. The proton remains in the nucleus, beta particle is ejected. Z increases by one and a stays the same. 14 6 c = 14 7 n + 0 -1 e + v(line above it) Daughter nucleus: z is one more than the parents nucleus; a is the same.

Related Questions

A nuclear reaction is represented by an equation similar to theequation for a chemical reaction. An example of a nuclear reactionis the decay of radium into radon by the emission of an alphaparticle (a helium nucleus). The equation for this reactionis

^{226}_{\ 88}{\rm Ra}\; \rightarrow \, ^{222}_{\ 86}{\rmRn}+\;^4_2{\rm He}.

The superscript preceding each element indicates the nucleon numberof that particular nucleus (total number of protons and neutrons).The subscript indicates the charge of the nucleus, which is equalto the atomic number for nuclei. Nuclear reactions frequentlyinvolve electrons (sometimes called beta rays) and photons(sometimes called gamma rays). Electrons, which have nucleon number0 and charge -1, are represented in equations as either ^{\ \;0}_{-1}\beta or ^{\ \; 0}_{-1}{\rm e}. Photons are represented as^{0}_{0}\gamma to emphasize that they have no nucleon number andcarry no charge.

In nuclear reactions, several conservation laws are usuallysufficient to determine the products or reactants. One of these isconservation of nucleon number. The total number of neutrons andprotons in all of the reactants must be the same as the totalnumber of protons and neutrons in the products. Although protonsand neutrons sometimes change into one another, neither can simplydisappear. Another important conservation law is conservation ofcharge. The net charge of the reactants must be the same as that ofthe products.

Notice that, in our equation for the decay of radium, the sum ofthe nucleon numbers on each side is 226 and the sum of the chargeson each side is 88. If you knew everything in this equation exceptthe isotope of radon produced, you could simply subtract 4 from 226to find the nucleon number required to make the nucleon number ofthe products equal the nucleon number of the reactants. Similarly,if you knew everything in this equation except for the identity ofthe product element, you could simply subtract 2 from 88 and thenlook in a periodic table to find that the element with atomicnumber 86 is radon.

In parts A and B consider the decay of radon by alpha particleemission:

^{222}_{\ 86}{\rm Rn}\;\rightarrow\;X +\,^{4}_{2}{\rm He}.

a. What is the nucleon number of the element X?
Express your answer as an integer.

b. What is the atomic number (nuclear charge) of the elementX?
Express your answer as an integer.

n parts C and D consider the decay of Potassium-43 into calcium-43via the following reaction:

^{43}_{19}{\rm K} \rightarrow ^{43}_{20}{\rm Ca}+\,X.

c. What is the charge of the emitted particle X?
d. is not visible yet

E. An isotope of potassium with a half-life of roughly one billionyears is often used in radioactive dating. This isotope decaysthrough electron capture, the process of drawing an electron intothe nucleus. What is the chemical symbol for the element thatresults from this process? The equation for the reaction is

^{40}_{19}{\rm K}+\,^{\ \;0}_{-1}{\rm e}\rightarrow \,?.
Express your answer as a chemical symbol (e.g., Na forsodium).

f. The reaction energy of a reaction is the amount of energyreleased by the reaction. It is found by determining the differencein mass between the reactants and products and then using E=mc^2 toconvert the lost mass into the released energy.

Look again at the equation for the decay of radium (Ra), given inthe introduction. Use the following table of masses to determinethe reaction energy of this reaction.
Nucleus or particle Mass in atomic mass units
radium-226 226.025402
radon-222 222.017571
alpha particle 4.002602
Express your answer in joules to three significant figures.

please answer fully... thank you

PHYS 1300 Lecture Notes - Lecture 29: Exponential Decay, Beta Decay, Beta Particle

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