What is the exponential decay function of an isotope with a half life of 3 minutes?
209362
The exponential growth/decay formula is N=No*e^(kt) where No is the initial amount, k is some constant, and t is time.  So if you started with No=1, then after 3 minutes N = .5\n\n.5 = e^(3k) take ln of both sides\n\nln(.5) = 3k\nk = ln(.5)/3\nk = -.231\n\nSo the equation would be N = e^(-.231t)
461568
The its age is really close to 5730 years since almost 1/2 of the Carbon 14 remains.  The percentage left P = (1/2)^(T/H) where T is the time that has passed, and H is the half-life in years.  We know H is 5730 years, the current percentage left is 51% (0.51), and we want to find T.\n0.51 = 0.5^(T/5730)\nlogbase0.5 0.51 = T/5730\nT = 5730 logbase0.5(0.51)\nlog base 0.5 of 0.51 = ln(0.51) / ln(0.5) or 0.971431\nSo T = 0.971431 x 5730 or 5,566 years ago
374975
Well, mv = h/λ\nm = h/(λv)\n= 6.626E-34 / (3.31E-15 x 0.01 x 3E8)\n= 6.673E-26\nConvert this to Da\n= 6.673E-26 / 1.66E-27\n= 40.2\nYou know, this is Ca. (look for it in the periodic table)
8383
In 1.6 s, 1.2 g decays to 0.6 g\nIn 1.6+1.6 s, 0.6 g decays to 0.3 g\nIn 1.6+1.6+1.6s, 0.3 g decays to 0.15 g\nin 1.6+1.6+1.6+1.6 s, 0.15 g decays to 0.0750g\n\nSo after 6.4 secs, 0.0750 g remains unchanged
351505
4Al + 3O2--->2Al2O3\n4Al -4(3e-)--->4Al+3\n6O+6(2e-) --->6O-2\n\nYou can take out the coefficients for single e- reactions, but the ones given are for the whole reaction
560152
I am not sure of the final answer you are looking for.  If you are looking for the number of molecules transferred after a certain time t, the answer is simply n=-DAt(dc/dz).  This of course assumes that dc/dz remains relatively constant.  If not you will need a dc/dz as a function of t.
557682
(a) There have been four half lives, so (1/2)^4 *2 =\n\n.125g\n\n(b) 2*[(1/2)^(t/15)]\n\n(c) 2*[(1/2)^(96/15)] = .0237g
53387
Here's an example of how e ties in with real life.  Put water in a can with a hole at the bottom.  Measure the rate of water flow with the can full, and calculate how long it would take to empty the can at that rate.  That's the "time constant".  Say it's a 32 ounce can and the initial flow is 1 oz/sec, then the TC is 32 sec.  But what really happens after 32 sec?  As the water goes down, so does the flow rate which is proportional to the amount of water remaining above the hole.  At 32 sec, the can is actually 1/e full.  At 64 sec the can is 1/e^2 full, and so on.  Any time the rate of approach to final value of some quantity is proportional to the value of the quantity relative to the final value, you get similar behavior.  This is called exponential decay, which is an asymptotic approach to the final value (in this case 0).  Acoustic echoes, radioactive decay, temperature change of an object subjected to heat or cooling, the amplitude of a damped oscillation and the charge or discharge of a capacitor through a resistance are other examples, and the principal remaining in an account that pays out a fixed percentage rate approximates this behavior (with TC ~ 1/payout rate).  At any time t, what remains from an initial X value of X0 that approaches a final value Xf is X = X0 + (Xf-X0)/e^(t/TC).\nAnother kind of e-dominated behavior is unbounded proportional change, called exponential growth.  There is no final value, and rate of change only depends on the size of X.  Thus at time t, X = X0 * (1 + e^(k*t)), where k is the growth rate constant.  Examples of this growth (which in real life, generally hit a limit caused by finite resources) are an account that accumulates a fixed interest, chain reactions, population growth, etc.
112911
The exponential law of decay, the one with the lambda constant you speak of, is a statistical one. The lambda constant we use is taken from observing large quantities of atoms and seeing how may decayed after some time. For any one particular atom we can't be sure when it's going to decay, we can only speak of probabilities. Tfis is the meaning of the lambda - the probability of decay in a unit of time.\nThink of rolling six sided dice a couple million times. You can assume you'll get a '4' ON AVERAGE in 1 of 6 rolls, but there is no telling, in which rolls.
463071
3 years, 72 days, 3 hours, 15 minutes, 7 seconds\n                                      OR\n1167 days, 3 hours, 15 minutes, 7 seconds\n                                      OR\n28011 hours, 15minutes, 7 seconds\n                                       OR\n1680675 minutes and 7 seconds\n                                        OR\n100840507 seconds\n\nWant anymore?
379546
15 hours\n2^-(T/H) = left/original = 20 mg/160mg = 1/8\nln 2^-(T/H) = ln 0.125 = -(T/H) * ln 2 = ln 0.125\n\nT = 45, H = half-life.\n\nH = -T * ln 2/ln 0.125 = 15
77999
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.
97088
"The half-life of a quantity subject to exponential decay is the time required for the quantity to fall to half of its initial value. The concept originated in the study of radioactive decay, but it now also occurs in many other fields."\n\nIn laymans terms I would guess it would be easiest to say its a term for a rate of decay.\n\nAn analogy.. hmm..  a bottle of milk has a half life of about 1 day before it starts going off.
308383
Radioactive decay is the set of various processes by which unstable atomic nuclei (nuclides) emit subatomic particles (radiation). Decay is said to occur in the parent nucleus and produces a daughter nucleus. This is a random process, i.e. it is impossible to predict the decay of individual atoms.\n\n \nThe trefoil symbol is used to indicate radioactive material. The Unicode encoding of this symbol is U+2622 (☢).The SI unit for measuring radioactive decay is the becquerel (Bq). If a quantity of radioactive material produces one decay event per second, it has an activity of one Bq. Since any reasonably-sized sample of radioactive material contains very many atoms, one becquerel is a tiny level of activity; numbers on the order of gigabecquerels are seen more commonly.
524706
An exponential function diverges from a base value by a factor of e (or converges to a base or asymptote value, reducing the difference by a factor of 1/e) for each time interval known as a "time constant". In the equation you provided, the time constant is T, the base value is 'a', b is a scale factor for the rate of increase, and the time variable is t. This equation represents an asymptotic or converging function since the power of e is negative. In each successive interval of T, the function's difference from 'a' decreases by a factor of 1/e. For an example of a real-world application of this equation, consider a capacitor of C farads being charged from 0 volts by a 5-volt source through a resistor of R ohms. The time constant of charging is R*C seconds. The equation for the voltage V on the capacitor is V = 5 - 5 * e^(-t/(R*C)). When T = 0, e^(-t/(R*C)) = 1 and V = 0; when T is large, e^(-t/(R*C)) approaches 0 and V approaches 5.
545437
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"
406273
You do not need the equations because all fits well between 2 days and 12 hours, that will be 4 decay cycles.\n\nSo after 12 hours you have 9/2 = 4.5\nafter 12 hours you have 4.5/2 = 2.25\nafte 12 hours you have 2.25 / 2 = 1.125\nafter 12 hours (2 days) you have 1.125 / 2 = 0.5625