
#1
Mar2712, 08:14 AM

P: 50

Hi,
My textbook states a decay constant of an isotope as 3.84 x 10 to the minus 12  per second. Surely decay constant can't be the number of decays per second because that wouldn't stay constant. As the isotope decays there are less atoms to decay and therefore the rate reduces. So what does decay constant mean? and why are the units per second? Any help would be appreciated! Thanks 



#2
Mar2712, 08:37 AM

Mentor
P: 14,427

One way to express this is in terms of the half life T_{1/2}: [tex]A(t) = A_0 \, 2^{\,\frac{tt_0}{T_{1/2}}}[/tex] Note that the half life T_{1/2} has units of time. An equivalent form is to use the exponential function [itex]\exp(x) = e^x[/itex] instead of [itex]2^x[/itex]. [tex]A(t) = A_0 \, 2^{\,\frac{tt_0}{T_{1/2}}} = A_0 \, e^{\,\frac{\ln 2}{T_{1/2}}(tt_0)} = A_0 \, e^{\lambda (tt_0)}[/tex] Here, the decay constant λ (which has units of 1/time) is related to the half life via [tex]\lambda \equiv \frac{\ln 2}{T_{1/2}}[/tex] 



#3
Mar2712, 09:32 AM

P: 9

Basically it means that it is decaying at a constant rate, thus allowing its decay to be defined by an exponential function. Think of it like this you start out with a large quantity and then divide it in half. Each time you divide it it is getting smaller and smaller at the same constant rate. This will keep happening following the graph of an exponential function.




#4
Mar2712, 09:38 AM

P: 1,506

What is meant by 'decay constant'?
The decay constant is the fraction of the number of atoms that decay in 1 second.
It is the constant λ in the decay equation: dN/dt = λN The  sign indicates decay, dN/dt is the number of decays per second (also known as 'Activity') and N is the number of atoms present. Eg if λ= 0.1 and we start with 1000 atoms then after 1 second 100 will have decayed and 900 will be left undecayed. After another second 90 will decay leaving 910. After another second 9.1 will have decayed (but you cant have 9.1 atoms!!!) leaving 900.9 (you cant have 900.9 !!!) this shows the pattern of the numbers in decay and plotting these numbers against time will give an exponential decay curve. The decay constant is also known as the probability of decay, I hope you can see that in my example with λ = 0.1 the probability of an atom decaying is 1 in 10....0.1 In your question the decay canstant of 3.84x10^12 is a very small probability but the principle is the same 



#5
Mar2712, 09:55 AM

P: 832

Technician is mostly correct, but it should be noted that the probability of decay is compounded continuously, to use a banking term, rather than compounded once a second. That's where the e comes in.




#6
Mar2712, 11:18 AM

Sci Advisor
P: 2,470

I think it's important to point out that stating dN(t)/dt = λN(t) is the same thing as stating N(t) = N_{0} e^{λt}, because for someone not familiar with diff eqs, that might not be obvious.




#7
Mar2712, 11:27 AM

P: 3,015

Radioactive isotopes are characterized by the following rule. The probability that a radioactive decays in unit time is a constant that depends on the nature of the isotope, but not on external conditions (pressure, temperature, external fields, etc). It is called a decay constant because the expected number of remaining isotopes is an exponentially decaying function:
[tex] N(t) = N_0 \, e^{\lambda \, t} [/tex] 



#8
Mar2712, 04:23 PM

P: 50

Thanks for all the responses. I think i've got it... If I was to define Decay Constant using words only would it be acceptable to say the 'fraction of the number of atoms remaining to decay in one second'? Does decay constant have to be between 0 and 1? Thanks again.




#9
Mar2712, 04:43 PM

P: 1,506

I would say it is the "fraction of the atoms that do decay in 1 second "rather than the "fraction that remain to decay"
It is a fraction so a number between 0 and 1 or a percentage 



#10
Mar2712, 04:43 PM

Sci Advisor
P: 2,470

So you don't look at what fraction would actually decay in a second, because the amount that decays depends on amount that is present. You can only define instantaneous rate. So yes, if the same number of atoms per second was decaying for a full second, that's the fraction that would have decayed in a second. But that's precisely why the decay constant can be higher than one. It just tells you that at a given rate, all of the material would decay before the second is out. Of course, that doesn't happen, because less material means fewer decay events. But the rate can still be greater than 1. Does any of that make sense? Some calculus really does help with this one. 



#11
Mar2712, 04:46 PM

P: 1,506

the decay constant is also a probability of decay....
CORRECTION MADE 



#12
Mar2712, 04:53 PM

Sci Advisor
P: 2,470





#13
Mar2712, 05:03 PM

P: 1,506

Correction




#14
Mar2712, 05:07 PM

Sci Advisor
P: 2,470

A decay constant is always inverse time units. If you look at OP, he gives his in inverse seconds.
Edit: A good way to tell is that it's in the exponent of the formula. You cannot take an exponent of a unit. So arguments to exponents, logarithms, and trigonometric functions must be dimensionless. In this case, you multiply decay constant by time t. Time has units of time, and so decay constant must have units of inverse time for their product to be dimensionless. 



#15
Mar2712, 06:42 PM

Emeritus
Sci Advisor
PF Gold
P: 5,198

I'm not arguing with you, I just want to understand. 



#16
Mar2712, 06:52 PM

Sci Advisor
P: 2,470

Edit: I added an attachment to demonstrate this. Blue curve is exponential decay with rate of 2/second. The red and yellow lines are tangent at 0 and 1 seconds respectively. Notice that they cross the time axis at 0.5 and 1.5 seconds respectively. 



#17
Mar2812, 12:42 AM

P: 1,506

K^2....You are quite correct....I have made correction....decay constant can be greater than 1.
Mental block cleared..cheers 



#18
Mar2812, 07:59 AM

P: 3,015

A decay constant (probability of a decay per unit time) is a physical quantity with dimension {time^{1}}. Therefore it has a difgerent numerical value in different systems of units. The probability, however, that a nucleus would decay in a finite time interval is:
[tex] P_d \left\lbrace [0, t] \right\rbrace = 1  e^{\lambda \, t} [/tex] This is indeed a number between 0 and 1. 


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