# B How are activity and decay constant different?

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1. May 4, 2017

What's the difference between activity of a radioactive sample and the decay constant. Both are measured in second inverse.
Please distinguish among the two?

2. May 4, 2017

### BvU

Activity is the number of decays / unit of time:

$$dN\over dt$$

When a sample decays, the number of candidates that have not decayed yet $\ \$ decreases, so the activity decreases with time. The time to decrease the activity to 1/e from an initial activity is the decay time $\tau$

$${ dN\over dt} \left ( t_0+\tau\right ) = {1\over e} \ {dN\over dt}\left(t_0 \right )$$

$\tau$ clearly has the dimension of time. Its inverse is the decay constant.

Check here

3. May 4, 2017

### sophiecentaur

One is the first time derivative and the other is a second time derivative OR a first derivative of a first derivative.

4. May 4, 2017

### mjc123

I don't think so; they wouldn't have the same units, for a start.
Activity is the number of decays per unit time. If N = N0exp(-kt), where N0 is the number of radioactive atoms present at t = 0, the activity = -dN/dt = kN0exp(-kt). k is the decay constant (reciprocal of the decay time described above).
Important distinction: Activity varies with time and amount of material. Decay constant does not; it is characteristic of a particular nucleus.

5. May 4, 2017

### sophiecentaur

Rate of decay is rate of change of rate of events.
Dimensional analysis of the situation is perfectly consistent. No 'paradox'.

6. May 4, 2017

### BvU

Well observed. Some confusion arose from the term 'second time derivative' in post #3.

The idea is that activity is $dN\over dt$, or rather $-{dN\over dt}\$, and that the activity is proportional to the amount of decay candidates present, with a proportionality constant that we call the decay constant: $$-{dN\over dt}\ = \lambda N$$

7. May 4, 2017

### sophiecentaur

I can't see where the idea of "the same units" came from. How can they be the same units? The only similarity is the 'per second'.

8. May 4, 2017

### BvU

I agree. But rate of decay is not decay constant.

9. May 4, 2017

### mjc123

Exactly, the units for both are s-1. A second time derivative would have units s-2.
This makes no sense to me. Rate of decay is the rate of occurrence of decay events, equal to the rate of change of the number of undecayed nuclei. This is a first time derivative. "Rate of change of rate of events" is how the activity changes with time. Because of the exponential form of the equation, this is proportional to the activity, and has the same decay constant. Is this what you meant (interpreting "decay" as "decrease in activity" rather than "nuclear decay events")?

10. May 4, 2017

### Staff: Mentor

As an example, You can express torque in J. That doesn't make torque an energy. Typically it is written as Nm to make it look different, but it has the same units as energy.
Length and height of an object have the same units, that does not mean they are the same.
Two different things can have the same units, in this case dimensional analysis alone doesn't work.

The activity is the (negative) first time-derivative of the number of radioactive atoms in the sample.
The decay constant can be calculated as activity divided by the number of atoms in the sample. Note that the number of atoms is dimensionless.

11. May 4, 2017

### sophiecentaur

No. It's there rate of decay of the rate of events. The term 'decay' doesn't refer to the radioactive fission event of an individual nucleus.
A property of the exponential function eaxis that the derivative is the same - apart from the constant.
so d(eax)/dx = aeax
The count at any time depends on the probability of an event, which is proportional to the number of unstable nuclei. The decay constant is the same whatever the number of nuclei are present.
Edit: The exponential decay of a discharging capacitor is not treated as a probabilistic process, so that's why I made the above comment.
Right. The dimensions are not the relevant argument here then.

12. May 4, 2017