Need help understanding Half life, decay constant & activity equations

Click For Summary
SUMMARY

The discussion focuses on the relationships between half-life, decay constant, and activity in nuclear physics. The half-life (T1/2) is defined as the time required for half of the nuclides to decay, while the decay constant (λ) is calculated using the formula λ = 0.693 / T1/2. The activity (A) is expressed as A = -dN/dt = λN, where N represents the number of undecayed nuclides. The conversation highlights the importance of calculus in understanding these equations, particularly the derivative dN/dt, which represents the rate of decay.

PREREQUISITES
  • Understanding of half-life in nuclear decay
  • Familiarity with decay constant calculations
  • Basic knowledge of differential equations
  • Concept of radioactivity and its measurement
NEXT STEPS
  • Study the derivation of the decay constant using λ = 0.693 / T1/2
  • Learn about the mathematical principles of differential equations
  • Explore the concept of activity in radioisotopes and its measurement
  • Review calculus fundamentals, particularly derivatives and their applications in physics
USEFUL FOR

Students of nuclear physics, educators teaching radioactivity concepts, and anyone seeking to understand the mathematical foundations of decay processes.

immuno
Messages
4
Reaction score
0
Hi guys

i'm having much difficulties in understanding the relations between half life, decay constant and activity.

I'll explain what i known here.

T^{\frac{1}{2}} is the half life which is the time to reduce the nuclides by half.

\lambda is the decay constant which can be solved from

\lambda = 0.693/ T^{\frac{1}{2}}. Am i correct? Where do 0.693 come from?

The decay/sec can be find out using A = -\frac{dN}{dt} = \lambda N if the half life and N is known.

I want to clarify N is in grams? So like 10g of C14 would be the No?

I don't really understand this equation:

<br /> \frac{dN}{dt} = -\lambda N<br />

i know why it is negative \lambda N as it's decreasing exponentially but what do the d and t stands for? Does the t stands for the half life? What is the use of the equation?

Thanks a bunch!
 
Physics news on Phys.org
Have you had calculus? d/dt is the time derivative.

Zz.
 
No.. I've not done calculus.

so how do i solve for an equation involving \frac{dN}{dt} = -\lambda N? What context do i use this equation?
 
immuno said:
No.. I've not done calculus.

so how do i solve for an equation involving \frac{dN}{dt} = -\lambda N? What context do i use this equation?

Just so you're aware of this (in case you aren't) that you're having a problem with the mathematics, not the physics. I'm pointing this out because, not just in your case, but as often the case, students just learning physics often think that physics is "difficult", when they are actually having issues with the mathematics.

It's hard to teach you how to solve this when you haven't had calculus, because we will end up tell you how to solve it completely. I'm surprised that you can take this class without having had calculus as a prerequisite.

Zz.
 
dN/dt is the activity that is the number of decays in one second.Since radioactivity is a random process it may not be possible to get an exact equation for activity but let's make some assumptions:
1.Assume that the activity is proportional to N,the number of atoms?For example two moles of U 235 as twice as active as one mol of U 235.
2.Assume the activity depends on what the radioisotope is for example it seems likely that one mol of U 235 has a different activity to one mol of U 238.
If the assumptions are reasonable we can write dN/dt=-lambda *N.When the equation is then tested experimentally it is shown that it works very well if N is large.(lambda=decay constant,a property of the isotope).When you think about it it makes sense,the activity depends on what the atoms are and how many there are.The rest is maths.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K