Solving a Radioactive Decay Problem: C is the Answer

In summary: You got the half life part, but just to expand on it. If you are doing a test that lasts maybe an hour, you don't want to leave radiation around any longer than you need to. So the short half life material will disappear much faster. After 20 half lives it has decreased by a factor of over a million. So if it had a half life of 3 hours, 20 half lives is 60 hours, two and a half days.
  • #1
PhysicStud01
174
0

Homework Statement


pic.png


Homework Equations

The Attempt at a Solution


i believe that a half-life of several years is too long to gather the data

but i can't figure which of the 2 emitter and why?
the answer is given to be C
 
Physics news on Phys.org
  • #2
You're given two different decay mechanisms/modes from which to choose. What are the characteristics of the two?
 
  • #3
Bystander said:
You're given two different decay mechanisms/modes from which to choose. What are the characteristics of the two?
i've taken a look at the characteristic of both, but can't figure which one of them to use here.

is penetrating power, ionization power, ... poisoning of the water, ...??
 
  • #4
The water line is 0.4 m underground. Which is mode are you going to be able to detect?
 
  • #5
Bystander said:
The water line is 0.4 m underground. Which is mode are you going to be able to detect?
it was not said in the question. should this be known?
 
  • #6
Yes, it should be known somewhat. ##\beta## is charged, so at low-energies it interacts a lot heftier than ##\gamma##. And: Roentgen rays are ##\gamma## rays.

Don't drink the water :)
 
  • Like
Likes PhysicStud01
  • #7
PhysicStud01 said:
it was not said in the question. should this be known?
Step 1: read the question. Step 2: when step 1 fails, re-read the question. Step 3: repeat step 2 until you are familiar with the information given in the question.
 
  • #8
BvU said:
Yes, it should be known somewhat. ##\beta## is charged, so at low-energies it interacts a lot heftier than ##\gamma##. And: Roentgen rays are ##\gamma## rays.

Don't drink the water :)
i don't understand. the answer is gamma rays. if the water can't be drunk, it should not be the emitter?

Bystander said:
Step 1: read the question. Step 2: when step 1 fails, re-read the question. Step 3: repeat step 2 until you are familiar with the information given in the question.
I read again, but it's really not said in the question?
 
  • #9
You got the half life part, but just to expand on it. If you are doing a test that lasts maybe an hour, you don't want to leave radiation around any longer than you need to. So the short half life material will disappear much faster. After 20 half lives it has decreased by a factor of over a million. So if it had a half life of 3 hours, 20 half lives is 60 hours, two and a half days.

You won't want to drink the water with either beta or gamma emitters in it. Your class should have taught you something about protection from these radiation sources. If it didn't, you need to do some reading on beta and gamma radiation protection. Google for it, and take a look on Wikipedia.

Beta particles, at least for common energies from radioactive decay, are stopped by very thin shielding. Most of the time they will be stopped by your clothes or even by the outer dead layer of your skin. So as long as you don't ingest it (eat, drink, breathe) a beta emitter is unlikely to cause harm. If you ingest it then the emitter is going to be mixed into your cells and so can cause harm.

Gamma rays have much more penetrating power. To shield from gamma you want mass, in particular you want lots of protons. Often lead is used because it is very dense and, compared to other materials of similar density, fairly cheap and easy to obtain. But any kind of matter will provide some shielding. You just need thicker shielding for less dense material.

So now think about detecting these particles on the other side of some ground, and through a pipe.
 
  • #10
You included it with the problem statement in the template; "In order to trace the line of a water pipe buried 0.4 m beneath the surface of a field ..."
 
  • #11
Bystander said:
You included it with the problem statement in the template; "In order to trace the line of a water pipe buried 0.4 m beneath the surface of a field ..."
so, it is detected through a pipe. but how does the 2 differ?
 
  • #12
Have "alpha, beta, and gamma" decays been defined/explained to you at all?
 
  • #13
Bystander said:
Have "alpha, beta, and gamma" decays been defined/explained to you at all?
basically they are all radiations, with different properties and characteristics.

but which characteristics are we considering and how does it affect the situation here. Why is it not beta, if both will contaminate the water?
 
  • #14
How much shielding is required to "protect" you from each of the types of radiation?
 
  • #15
Bystander said:
How much shielding is required to "protect" you from each of the types of radiation?
it depends.
for beta: few mm of aluminium
for gamma: few cm of lead
 
  • #16
Dude! Why are you carefully ignoring my post?
 
  • #17
DEvens said:
Beta particles, at least for common energies from radioactive decay, are stopped by very thin shielding. Most of the time they will be stopped by your clothes or even by the outer dead layer of your skin.

DEvens said:
Gamma rays have much more penetrating power
 
  • #18
DEvens said:
Dude! Why are you carefully ignoring my post?
i read it, but through a pipe, won't both radiation be equally inappropriate?
 
  • #19
so, gamma is even worst, right?
 
  • #20
should we stop the radiations here or should they actaully reach a detector? is beta stopped in the ground?
 
  • #22
Dear Phys,

Looks as if By and DE can't really imagine your difficulties. But groping through terra incognita is always a bit awkward. Good thing it's just an exercise and not a lab experiment!

The contamination side issue may be solved by assuming this is about a drain pipe, or else by supposing the water isn't supplied to households. It's unhealthy. On the other hand, for medical purposes they even inject people with technetium-99 (##\tau_{1\over 2}## 6 h).
And 0.4 m of soil is a lot of shielding for the ##\gamma## too, so the engineer will be fine.

Key in the exercise is that the ##\beta##'s don't get through at all.

This sheet shows some of the ranges, half lives, etc.

And: this is an exercise. I would be surprised if the water authorities used radioactivity to locate pipes.
 
  • #24
Making life difficult, eh ? Cs with a half-life of 37 years. yuck. Fortunately, the exercise wants the engineer to locate a water pipe. But I confess to lack of domain knowledge: the "I would be surprised" clause is from a physicists point of view.
 

1. How do I solve a radioactive decay problem using C as the answer?

To solve a radioactive decay problem using C as the answer, you will need to use the following equation: N(t) = N₀e^(-λt). N(t) represents the amount of the substance remaining after a certain amount of time, N₀ is the initial amount of the substance, λ is the decay constant, and t is the time elapsed. You will also need to know the half-life of the substance in question.

2. What is the decay constant?

The decay constant, represented by the Greek letter λ (lambda), is a value that indicates the rate at which a substance decays. It is a characteristic property of each radioactive isotope and is used to calculate the amount of a substance that remains after a certain amount of time has passed.

3. What is the half-life of a substance?

The half-life of a substance is the amount of time it takes for half of the initial amount of the substance to decay. This is a constant value for a specific radioactive isotope and is used in the calculation of the amount of a substance remaining after a certain amount of time has passed.

4. How do I determine the initial amount of a substance in a radioactive decay problem?

The initial amount of a substance, represented by N₀ in the equation N(t) = N₀e^(-λt), can be determined by using the information given in the problem. This could be the mass of the substance, the number of atoms, or the activity of the substance. Once you have the initial amount, you can use the decay constant and the half-life to calculate the amount remaining after a certain amount of time has passed.

5. Can I use the same equation for all radioactive decay problems?

Yes, the equation N(t) = N₀e^(-λt) can be used for all radioactive decay problems as long as you have the necessary information, such as the initial amount, decay constant, and half-life. However, keep in mind that different isotopes have different decay constants and half-lives, so the values used in the equation will vary depending on the substance in question.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
852
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
4K
Back
Top