Half life: solving for mineral age, the half life, and the p/d ratio.?
954316
parent : daughter ratio is 1 to 15 so 1/16 of the original atoms are present . After 500 megayears it would be about 1/2, 1000Ma about 1/4, 1500Ma about 1/8 and after about 2000Ma it would be 1/16. \n\nto get a parent-daughter ratio of 1 to 7 then 1/8 of the original atoms are still present so 3000Ma is 3 half-lives. \n\nBest of luck - Mike
177541
lambda = 38 s\nInitial Mass (m1) = 12 g\nFinal Mass (m2) = 6 g\n\nEasy - after 1 half life what remains is 6 grams.\nAfter 2 half lives (t=76 s) only 6/2 = 3 grams remain.\n\nThis is the equation you should use for this type of problem.\n\nmf = mi exp(-kt)\n\nwhere mf is final mass\nmi is initial mass\nk is what you are looking for\nt for this should be t=lambda=38 sec\n\n6=12exp(-38k) -> 6/12=.5 = exp(-38k) -> multiply both sides\nby ln and you should get\nln(.5) = -38k -> -ln(.5)/38 = k = .018241\nnow you could solve for 114s.\n\nm2 = 12exp(-.018241*114) = 1.5 grams\n\nSo, after 114 seconds, only 1.5 grams of the material should remain.  That is 3 halflives.\n\n1 halflife - > 12/2 = 6 grams remains\n2 halflives - > 6/2 = 3 grams remains\n3 halflives -> 3/2 = 1.5 grams remain\n\nBut sometimes, if your teacher wants to screw you, use that equation.
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.
54817
Are you referring to radioactive decay rates of isotopes of the three elements (copper, aluminum and iron) that you listed? If so, the first answer gave you some info, but you should check it (e.g., because copper has more than one isotope and they all have different half-lives). One can't specify a half-life for stainless steel because it's a mixture of metal elements whose ratio isn't specified by the generic name.\nIf instead you're referring to some measure of chemical activity, such as corrosion life, you haven't given enough details. (Well actually you haven't given enough details either way.) You could google corrosion rate, reaction rate and similar terms if you want to go further with that.
777912
Multiplu the time it takes to achieve half-life by the days required for a given amount of decay...\n\nSuch as 18.72 * .75\n\nI know you said not to give the answer, but you must understand that the question asks, how many days are required for 3/4 of the decay to decompose... \n\n The answer is by the days it takes for its half-life, multiplied by the desired ammount of decay.
929019
With such a large quantity of ore you will have a problem using the two-beaker method. But the first two suggestions contain half the answer each. Just combine what they are saying.
1122499
pH = pKa - log (.25/.4) ~= 4.3
745292
the definition of 1/2 life is that 1/2 of the material will be left after that amount of time.\n100 g to start, 50 g after 1 half life.\n25 g after 2 half lives.\n\nb) 60 yrs is 2 half lives.  see ans to part A)\nc) 25 g will be left after 60 yrs.  See answer to part A)\n-->
194428
I assume you are using a half-life equation of the form:\nN(t) = N_0*e(-kt) where N(t) is the amount of radium remaining at time t and N_0 is the initial amount of radium. You should really provide this sort of information as the constant k is in no way universal (some people use r or lambda). Decay constant would be a more appropriate term.\n\nSo if the initial amount of radium is N_0 = 1 gram, then\nN(1622) = 0.5 grams(ie. there will only be a half a gram left after 1622 years). Take a natural log and solve for k.\n\nOnce you have k, for the second part let N_0 = 2.3 grams and solve for t where N(t) = 1.7 grams. You're going to need a natural log again for this guy, I trust you know how they work.\n\nAlso I'm pretty sure the previous answer is dead wrong.
165799
You want us to calculate acitivites?\n\nYou need to determine the ionic strength and atomic radius in order to do that.
951626
It is very simple.  This method will help you when the fraction is not a nice one like 1/16.\n\n\nAt time zero you have, let's say N0 = 1600 atoms of protactinium-234.  After 26.76 hours, 1/16 remain:\n\nNt = 1600 / 16 = 100.\n\nNow in radioactive decay:\n\nNt = N0 * exp (-Lamda * t).\n\nYou have t, Nt, N0.  Calculate lambda\n\n100 = 1600 exp (-Lambda * 26.76)\n0.0625 = exp (-Lambda * 26.76)\nTaking Ln on both sides\nLn 0.0625 = - Lambda * 26.76\nLambda = - (Ln 0.0625) / 26.76 = 0.10361 h^(-1)\n\nHalf life = (Ln 2) / Lambda = 6.69 hours
416430
There are various means of determining this, and without any specific data on what elements are involved in the answer you are looking for, the link below is the best reference I can find that can answer this question in a mostly generalized way.
221300
Measured by the decay of long-lived radioactive isotopes of elements that occur naturally in rocks and minerals and that decay with half lives of 700 million to more than 100 billion years to stable isotopes of other elements, scientists believe that the age of the earth is approximately 4.4 to 4.6 billion years old.
570720
The age of a fossil may be specified in both relative terms (how old it is in relation to other fossils or rock units) and in absolute terms (approximately how many years old it is). One principle of relative dating is called superposition, which holds that in any one place, the lower rock layers (and fossils in them) are older than higher ones, unless there is evidence that the layers have been overturned. This is logical, since the lower sediments would have been deposited first. A related concept is correlation, whereby rock layers (strata) are compared with others in another location on the basis of their mineral composition, fossil content, and other features. When a rock unit or series closely matches those of another area, they are said to be correlated and considered to be of similar age. Correlations are especially reliable when they include index fossils, which are fossils with a limited age range. Good index fossils are easily recognized and widely distributed around the world, so that they aid both local and international correlations.\n\nAbsolute dating complements relative dating by providing a specific (not necessarily precise) chronological age for a given specimen, such as "50 million years before present." In recent years reliable forms of absolute dating became available through the development of radiometric dating methods. These methods are based on the known, regular decay of certain radioactive elements (isotopes) into other isotopes or "daughter products." By measuring the amount of "parent" and "daughter" products in a rock sample, its approximate age may be calculated.\n\nMany people have heard of the Carbon 14 method and assume that fossils are dated by it. Actually, in most cases the C-14 method is useful only on organic material less than about 50,000 years old, which includes many human remains and artifacts, but excludes most fossils. Absolute dating of fossils requires other dating methods such as the potassium-argon or rubidium-strontium methods, which involve isotopes with slower decay rates (longer "half-lives"). Such isotopes are rare in fossils themselves, but may occur in surrounding or adjoining rock layers, yielding an approximate age for the fossil-bearing unit. Using the principles discussed above, scientists also can deduce the approximate age of other layers correlated with the same formation, as well as an approximate age for layers above and below the formation (which would be somewhat younger and older respectively).\n\nThe terms "above" and "below" in this context are used in a relative or stratigraphic sense; that is, they are based on the positions of the units relative to each other (or correlated units), rather than their height above sea level. The branch of geology that deals with the correlation of rock layers is known as stratigraphy (or biostratigraphy when focusing on the fossils within the layers), and is an important tool of paleontologists.
496873
Sorry this is so long, but I have to first fix some misinformation given in other people's answers.  Specifically, carbon-14 (a.k.a. C14, carbon dating, or radiocarbon dating) is useful for dating a lot of things, but *not* fossils!  Instead, other forms of radiometric dating are used for fossils.  \n\nThe 'half-life' is a measure of how fast the process of radioactive decay causes one type of atom to decay into another type.  For example carbon-14 (C14) decays into nitrogen-14 (N14) with a half-life of 5,730 years.  This means that if you have a bone with 1 gram of C14, then after 5,730 years, 0.5 grams will have decayed to N14, leaving 0.5 grams of C14.  After another 5,730 years only 0.25 grams would remain.  And so on.\n\nNow, 5,730 years is not very long in geologic terms.  This means that after 50,000 years or so, the amount of C14 left is so small that it is practically unmeasurable.  It's for this reason that C14 dating is not much good for fossil dating, as most fossils are *much* older than 50,000 years.\n\n[Aside:  There's another reason carbon-14 dating is not useful for fossils.  It's only useful for dating once-living matter (bones, wood, teeth, shells, cloth, paper, etc.) that was incorporating carbon into its body while it lived ... either a plant that was once breathing CO2, or an animal that was eating things that contained carbon (like plants or other animals).  A fossil is not once-living matter.  It is rock that has filled in the impression left by a dead organism.  So fossils themselves may contain very little carbon of any type.]\n\nBut there are other atoms with a *much* longer half-life.  For example potassium-40 (K40) decays into argon-40 (Ar40) with a half-life of 1.26 *billion* years.  So when a rock is created by a volcano, it contains only K40 (as argon is a gas, so any Ar40 present at the time the rock forms escapes from the molten rock before it cools).  But over millions of years this K40 decays to Ar40, which remains trapped in the rock.  So a geologist can dig up a rock from a layer of strata, melt it down, and measure the amount of molten K40 and gaseous Ar40 that is released.  By the ratio of K40 to Ar40 in the rock, the geologist can calculate the age of the rock, in millions of years.  The geologist then also knows the age of anything else that was encased in the same layer of strata as the rock ... such as a fossil.\n\nCarbon-14 dating and potassium-argon dating are two forms of what are called 'radiometric dating'.  Other examples are uranium-lead dating (half-life of 700 million years), and rubidium-strontium dating (half-life of 50 billion years ... which is *very* long, considering that the earth is only 4.5 billion years old).
610525
B\n\n1 day = 24 hours\n4 half life periods\n\n10\n5\n2.5\n1.25\n.625
781203
2H3PO4(aq) + 3Mn(OH)2(s)  --> Mn3(PO4)2(s) + 6H2O(l)