Radioactive decay. Given Ax, Ay and t½ for Y, get t½ for X. Not possible?

In summary, the question asks for the half-life of element X given its activity is four times that of element Y, which has a half-life of 20,000 years. The approach involves using the equation A = 0.69N/t_{0.5} and considering the activities of 10 g of both elements. However, without knowing the atomic masses of X and Y, the problem may not be solvable.
  • #1
catkin
218
0
[SOLVED] Radioactive decay. Given Ax, Ay and t½ for Y, get t½ for X. Not possible?

Homework Statement


This is from Advanced Physics by Adams & Allday, spread 8.13 Question 1.

The activity of 20 g of element X is four times the activity of 10 g of element Y. Element Y has a half-life of 20,000 y. What is the half-life of X?

Homework Equations


[tex]A = \lambda N[/tex]
[tex]\lambda t_{0.5} = 0.69[/tex]

The Attempt at a Solution


Rewriting the first relevant equation in [itex]t_{0.5}[/itex], rather than λ, using the proportionality from the second relevant equation
[tex]A = 0.69 N / t_{0.5}[/tex]

Considering 10g of both elements
[tex]A_{X} = 2A_{Y}[/tex]

Expressing these activities in terms of the number of atoms in 10 g and half life
[tex]0.69 N_{X} / t_{0.5X} = 2 \times 0.69 N_{Y} / t_{0.5Y}[/tex]
[tex]N_{X} / t_{0.5X} = 2N_{Y} / t_{0.5Y}[/tex]
[tex]t_{0.5X} = (N_{X} / 2 N_{Y}) t_{0.5Y}[/tex]

Substituting, using years as time units
[tex]t_{0.5X} = (N_{X} / 2 N_{Y}) {20000}[/tex]

If the number of atoms in 10 g of element X were the same as the number of atoms in 10 g of element Y (there is no reason why it should be) then [itex]t_{0.5X}[/itex] would be 10,000 years (the answer the book gives).

4. Question Am I right in thinking there is not enough information in the question to answer it?
 
Physics news on Phys.org
  • #2
10 g of x and y cannot have same number of atoms. Nx and Ny depend on their molicular weights.
 
  • #3
Thanks rl.bhat :smile:

It doesn't answer my question though; is the problem soluble?

[Separate issue: what if the elements had the same atomic number? Say Th-234 and Pa-234? Wouldn't the number of atoms in 10 g be the same, at least to the number of significant figures the question implies?]
 
  • #4
In case of Th- 234 and Pa- 234 the problem is soluble.
 
  • #5
Thanks again. That makes sense. Is the problem soluable as it is posed in the original question?
 
  • #6
Certainly if X -> Y (or Y -> X) by beta decay, then the same mass would have approximately the same number of atoms within 1% or less.

The question becomes - "does X -> Y, or vice versa, i.e. do they represent sequential steps in a decay chain?"

The approach seems correct. The problem hinges on the assumption of Nx and Ny, which would be determined by the atomic masses.
 
  • #7
Astronuc said:
Certainly if X -> Y (or Y -> X) by beta decay, then the same mass would have approximately the same number of atoms within 1% or less.

The question becomes - "does X -> Y, or vice versa, i.e. do they represent sequential steps in a decay chain?"

The approach seems correct. The problem hinges on the assumption of Nx and Ny, which would be determined by the atomic masses.
Thanks, astronuc :smile:

That helps understanding.

There's nothing in the question to indicate either any decay relationship or the atomic mass relationship between X and Y, though. Decay chains are introduced in a later "spread" in the textbook so should not be necessary for the solution.

I'm increasingly coming to think that the question as set is not soluable.
 
  • #8
Raise this concern with the professor.

With mass and activity, one can get the specific activity, but one needs to know the atomic mass to obtain the number of atoms.

Since you obtained the answer given in the book with the assumption that the atomic mass of X and Y are roughly equal, that would seem to indicate an implicit assumption on the part of the author. If beta decay was involved (e.g. X -> Y), then that is a reasonable assumption.
 
  • #9
Thanks Astronuc :smile:

That's enough to consider this one SOLVED
 

1. What is radioactive decay?

Radioactive decay is the process by which an unstable atomic nucleus loses energy by emitting radiation in the form of particles or electromagnetic waves. This process results in the transformation of one element into another.

2. How is radioactive decay measured?

The rate of radioactive decay is measured using the half-life (t½) of a radioactive element. This is the amount of time it takes for half of the initial amount of the element to decay into a more stable form.

3. Can the half-life of one element be used to determine the half-life of another element?

No, the half-life of one element cannot be used to determine the half-life of another element. Each radioactive element has its own unique rate of decay, so the half-life cannot be predicted based on another element's half-life.

4. How is the half-life of an element calculated?

The half-life of an element can be calculated using the equation t½ = ln(2)/λ, where t½ is the half-life, ln is the natural logarithm, and λ is the decay constant of the element.

5. Is it possible to determine the half-life of X if we know the half-life of Y and the decay constant of both elements?

Yes, it is possible to determine the half-life of X if we know the half-life of Y and the decay constant of both elements. This can be done using the equation t½X = (t½Y * ln(2))/ln(Ax/Ay), where t½X is the half-life of X, t½Y is the half-life of Y, and Ax and Ay are the activities of X and Y, respectively.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
4K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
16
Views
2K
Back
Top