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Decay problem

  • #1
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Homework Statement


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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
 

Answers and Replies

  • #2
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You're given two different decay mechanisms/modes from which to choose. What are the characteristics of the two?
 
  • #3
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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
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The water line is 0.4 m underground. Which is mode are you going to be able to detect?
 
  • #5
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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
BvU
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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 :)
 
  • #7
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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
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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?

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
DEvens
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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
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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
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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
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Have "alpha, beta, and gamma" decays been defined/explained to you at all?
 
  • #13
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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
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How much shielding is required to "protect" you from each of the types of radiation?
 
  • #15
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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
DEvens
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Dude! Why are you carefully ignoring my post?
 
  • #17
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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.
Gamma rays have much more penetrating power
 
  • #18
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Dude! Why are you carefully ignoring my post?
i read it, but through a pipe, won't both radiation be equally inappropriate?
 
  • #19
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so, gamma is even worst, right?
 
  • #20
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should we stop the radiations here or should they actaully reach a detector? is beta stopped in the ground?
 
  • #22
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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.
 
  • #23
DEvens
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  • #24
BvU
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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.
 

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