Transmutation reactions and half-life problems?
240384
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
244558
A lot of radioactive elements have decayed and no longer exist in nature.  Those that remain, have half lives long enough that they have not completely decayed.  238-Uranium, for example, has a half life of 4.468x10^9 years, which is about the age of the earth.  Only one-half of the original uranium has decayed.
1007912
We did a similar lab in my math class. It will be about half whenever you flip 100 pennies, then half again, ect... This is roughly the same thing as the radioactive element ("half-life").\n\nNow, make sure you do the work so you will know the material, and explain the relationship in words your teacher would like to see.\n\nGood luck.
452881
False, because the binding energy increases steeply till iron and again reduces. So I would require enormous accelerators to transmute the element with Z>20.
848476
Measuring the background radiation.\nCalculations involving half life.\nCalculating the energy of fusion and fission reactions.
251198
id say reactions can occur at any temperature above 0k. so 0.00000000000000000000000000000001K, the reaction would happen but really really really really relaly slowly. but thatz in theory. i have no clue bout pracitcally. i think the lowest temp theyve got to so far is either 4K or 0.6K. so ya.
231768
Well, if the radioactivity was 1 to begin with, then after 5700 years it would be 0.5, then after another 5700 years ot would be half of that again, 0.25, so the answer lies between 5700 and 11400 years.\n\nIn equation form you can say \n\nRadioactivity = (0.5 ^ x) * Halflife (lambda symbol.. like the game?!)\n\nSo: 0.33 = 0.5^x  ----- Need to take logs of both sides\nln (0.33) = ln(0.5) * x\nx = ln(0.33)/ln(0.5)\nx = 1.5994\n\n1.5994 * 5700 = 9117 years.
500933
13. in isotopes, only the neutron number changes, not the protons. if the proton number changes, then it becomes another element.
973296
hi Fellow Genius !,\ngood question.\n\nYES. Protons do decay. they have an average lifetime of about 10 to the power 32 years. Now, since this is an average, it should be possible to monitor a large enough collection of protons, and observe a few decays in the space of an year or so. Such experiments, involving 100s of tons of H2O or some other material, are underway in many parts of the world [eg.: a cave 600m below Lake Erie, Ohio]. However, none of them have yet [so far as I know] found any convincing evidence for the "predicted" decays of protons. [will answer part 2&3 of ur Q seperately] contact me at email address: fizamaker@hotmail.com or IM with me on Y!Messenger [ I am usually online from 9 pm to 5 am India time all night ]. my ID there is :  fizanaqsh . Add my ID to ur contacts so we can talk back and forth on physics and metaphysical questions bothering u and/ or me !  I have a Skype ID also and its : homeworknetguru.\ndo take the trouble of replying if my answer satisfies ur curiosity or not. i am v. much interested in these questions. look out for my answer 2 the third part of ur question re the "arrow of time"
840804
Others ways to change an element (nuclear chemistry)\n\nProton emission \nK^40 ------> Ar^40 + e\n\nElectron Capture.\nHg^197 + e-  ------->  Au^197\n\nAlpha Radiation 4/2He+2\nU^238 ----->  (4/2He2+) +  Th^234\n\n\nBeta Radiation which causes a neutron in the nucleus to spontanously decay into a proton pluse an electron which is then ejected\n\nI^131 ------> Xe^131 +  e-
240255
The definition of the half-life is the amount of time for half the substance to react.  \n\nWhen the last molecule (or atom) reacts, the reaction is complete.  One might argue that the half-life ends when the reaction ends.  However, for many reactions, the half-life is independent of the amount of that substance (a good example is  nuclear fission), so you could argue since the half-life doesn't change, it doesn't end.  This is really a matter of semantics.
546696
Ok, So you know the Time, the Starting mass we'll say Ao, and the Half life.  So by the equation:\nA/Ao = (1/2)^n,\n\nwhere A is the amount remaining, Ao is the starting amount and n is an integer, which needs to be determined.\n\nn = T/(t)\n\nWhere t is the half life and T is the time elapsed.  \n\nSo you first solve for n, and then derive the expression for A.  you should get the equation,\n\nA = ((1/2)^n)*(Ao)\n\nThat'll give you the new mass and from the mass you can use Avogadros number to determine the particles emitted.
82368
U get precipitate of Lead Iodide. The reaction is already mentioned!!!!\n\nPb(NO3)2 (aq) + 2KI (aq) --> 2KNO3 (aq) + PbI2 (s)
207847
It's a Chemical/Fusion reaction...
926537
The decay is of the form N(t)/N0 = e^-t/h, where h is a constant.  When t = HL (half life time) N(t)/N0 = .5  therefore N(HL)/N0 = .5 = e^-HL/h.  Take the natural log of both sides to get ln(.5) = -HL/h, so h = HL/ln(2).  Therefore the formula for decay is \n\nN(t)/N0 = e^-t*ln(2)/HL\n\nYou want to find tx such that N(t)/N0 = 100/10^6 = 10^-4.  That means 10^-4 = e^-tx*ln(2)/HL.  Again, taking log (base 10) of both sides gives you -4 = log(e^-tx*ln(2)/HL) or[ -tx*ln(2)/HL]*ln(10).  The time tx is then 4*HL*ln(10)/ln(2).  This comes out 74,411years
669411
Yes! The proton has a half life estimated by the electroweak theory to be about 10^32 years.\n\nThis means that in our lifetime, or probably ANY human being's lifetime, there will be no effect, because the half life is so large.\nYou's have to wait trillions of years to see even a remote difference.\nEventually, though, the universe will end up as a big jumble of fermions. (I'm assuming that bosons are unstable, and fermions (like the electron) are stable. Correct me if i'm wrong, someone!)\n\nParticle physicists are currently looking for experimental evidence of proton decay to validate the elctroweak theory (the unification of the electromagnetic and weak forces).
109774
the simple answer is that it's due to the radioactive decay series. how isotopes decay and how long...