Calculate number of photons absorbed

  • #1
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Homework Statement:
The energy of the x-ray is 50 keV, how many photon has been absorbed?
Relevant Equations:
E=(hc)/lambda
I've tried to solve this by calculating the number of photons. I've done this by calculating the energy of one photon, taking h*c divided by lambda. h*c is 1240 eV*nm and lambda is 10 nm. This gives me 124 eV. I then divide the total energy by the energy of one photon, 50 keV/ 124 = 403 photons. Is this correct? And how do I get the number of absorbed photons?
 

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  • #2
Steve4Physics
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The question is incomplete. May I suggest you post the complete question, word-for-word.
 
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  • #3
kuruman
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The number of photons is correct. You get the number of photons absorbed by first considering the process whereby they are absorbed. What do you know about that?
 
  • #4
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The additional information that I forgot to put in is that a body part with the mass of 1,2 kg is being x-rayed and gets an equivalent dose of 0,40 mSv.
 
  • #5
kuruman
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OK, so you know that 1 Sv = 1 Joule/kilogram. Therefore ...

On edit: Something does not sit well with me. You seem to interpret the 50 keV as the total energy in a burst of 402 photons. What exactly does the problem state about that number?
 
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  • #6
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OK, so you know that 1 Sv = 1 Joule/kilogram. Therefore ...
Therefore the absorbed dose is 0,33 mGy?
 
  • #7
kuruman
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Therefore the absorbed dose is 0,33 mGy?
I think that, for the purposes of answering this question, you can assume a 1:1 conversion factor. However, since you still have not provided the full statement of the problem, I cannot be sure.
 
  • #8
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I think that, for the purposes of answering this question, you can assume a 1:1 conversion factor. However, since you still have not provided the full statement of the problem, I cannot be sure.
There is no more information in the question.... The only things that are stated are the equivalent dose (0,40 mSv), the energy of the x-ray (50 keV) and the mass of the body part (1,2 kg).
 
  • #9
kuruman
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OK, it seems that 50 keV is the energy of a single photon. You don't have 402 photons totaling 50 keV of energy. You need to find how many Joules are absorbed by the body and then figure out how many 50 keV photons must be absorbed to amount to that much energy.
 
  • #10
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OK, it seems that 50 keV is the energy of a single photon. You don't have 402 photons totaling 50 keV of energy. You need to find how many Joules are absorbed by the body and then figure out how many 50 keV photons must be absorbed to amount to that much energy.
I see! Could you please elaborate on how I do this?
 
  • #11
hutchphd
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My understanding here is that there is only one actual unit (the Gray=1 Joule/kg) and then a phalanx of units called Sievert =Q x Gray where Q is a factor defined to make the math "easier". I will assume Q=1 and Sievert=1J/kg.
  1. So if 1.2 kg of flesh is exposed to 0.4 mSievert how many J of energy are deposited?
  2. How many eV?
  3. So how many photons?
 
  • #12
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My understanding here is that there is only one actual unit (the Gray=1 Joule/kg) and then a phalanx of units called Sievert =Q x Gray where Q is a factor defined to make the math "easier". I will assume Q=1 and Sievert=1J/kg.
  1. So if 1.2 kg of flesh is exposed to 0.4 mSievert how many J of energy are deposited?
  2. How many eV?
  3. So how many photons?
Homework Statement:: The energy of the x-ray is 50 keV, how many photon has been absorbed?
Relevant Equations:: E=(hc)/lambda

I've tried to solve this by calculating the number of photons. I've done this by calculating the energy of one photon, taking h*c divided by lambda. h*c is 1240 eV*nm and lambda is 10 nm. This gives me 124 eV. I then divide the total energy by the energy of one photon, 50 keV/ 124 = 403 photons. Is this correct? And how do I get the number of absorbed photons?
Isn't this what I've done here? I'm a bit confused, is the 50 keV the energy released from one photon or all of them? In the question it says "the energy of the radiation is 50 keV".
 
  • #13
Steve4Physics
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Isn't this what I've done here? I'm a bit confused, is the 50 keV the energy released from one photon or all of them? In the question it says "the energy of the radiation is 50 keV".
I'll chip in. 50keV is the energy in ONE photon. The wording in the question may not make this clear, but that is what is meant.

But forget photons for the moment! The first question you have to answer is:

How much energy (in joules) is absorbed when 1.2kg of tissue receives an equivalent dose of 0.40mSv of X-rays?
 
  • #14
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I'll chip in. 50keV is the energy in ONE photon. The wording in the question may not make this clear, but that is what is meant.

But forget photons for the moment! The first question you have to answer is:

How much energy (in joules) is absorbed when 1.2kg of tissue receives an equivalent dose of 0.40mSv of X-rays?
The amount of energy absorbed should be 0,33 Gy (0,4/1,2) right?
 
  • #15
jbriggs444
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The amount of energy absorbed should be 0,33 Gy (0,4/1,2) right?
So if you had twice as much tissue you'd absorb half as much energy? That sounds badly wrong.
 
  • #16
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So if you had twice as much tissue you'd absorb half as much energy? That sounds badly wrong
Agree! That doesn't make sense, should it be 0,4*1,2=0,48 Gy? Since the absorbed energy is in Grey or J/kg I need to scale it to 1,2 kg, right?
 
  • #17
jbriggs444
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Agree! That doesn't make sense, should it be 0,4*1,2=0,48 Gy? Since the absorbed energy is in Grey or J/kg I need to scale it to 1,2 kg, right?
Yes.

Without having paid enough attention to the problem in detail to be sure that no factors of 1000 or whatever have been missed, this sounds right. You multiply the dose per kilogram by the number of kilograms to get the total dose.
 
  • #18
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Yes.

Without having paid enough attention to the problem in detail to be sure that no factors of 1000 or whatever have been missed, this sounds right. You multiply the dose per kilogram by the number of kilograms to get the total dose.
Perfect! From this, how do I get the number of absorbed photons?
 
  • #19
jbriggs444
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Perfect! From this, how do I get the number of absorbed photons?
You could do it with some simple algebra. Though eventually it becomes second nature and you just know whether it is a multiplication or division and whether a unit conversion constant is required.

We know how much energy there is per photon.
We know how much total energy is absorbed.
We want to know how many photons it takes to embody that much total energy.

I think to myself: "Self, the more energy there is per photon, the fewer we need. So we divide by energy per photon".

I think to myself "Self, the more energy there is total, the more photons we need. So we multiply by total energy".

With this in mind, I can immediately write down: NumberOfPhotons = TotalEnergy / EnergyPerPhoton.

Now you just have to worry about whether TotalEnergy and EnergyPerPhoton are using the same units for energy.
 
  • #20
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You could do it with some simple algebra. Though eventually it becomes second nature and you just know whether it is a multiplication or division and whether a unit conversion constant is required.

We know how much energy there is per photon.
We know how much total energy is absorbed.
We want to know how many photons it takes to embody that much total energy.

I think to myself: "Self, the more energy there is per photon, the fewer we need. So we divide by energy per photon".

I think to myself "Self, the more energy there is total, the more photons we need. So we multiply by total energy".

With this in mind, I can immediately write down: NumberOfPhotons = TotalEnergy / EnergyPerPhoton.

Now you just have to worry about whether TotalEnergy and EnergyPerPhoton are using the same units for energy.
I have the total absorbed energy in Gy or J/kg (0,48 mGy) and the energy of one photon in keV (50 keV)
 
  • #21
jbriggs444
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I have the total absorbed energy in Gy or J/kg (0,48 mGy) and the energy of one photon in keV (50 keV)
No. The total absorbed energy is not in units of J/kg. Because J/kg is not a unit of energy.
 
  • #22
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No. The total absorbed energy is not in units of J/kg. Because J/kg is not a unit of energy.
Isn't 1 Gray = 1 J/kg?
 
  • #23
jbriggs444
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Isn't 1 Gray = 1 J/kg?
Yes, one Gray = 1 J/kg.
But it is not a unit of energy. So a claim that total energy is some number of Grays is certainly wrong.

So we need to back up and figure out the units on the number you calculated previously.

1.2kg of tissue receives an equivalent dose of 0.40mSv
You multiplied 0,4 by 1,2 to get 0.48. And asserted that the result is in units of Gray.

What we know for sure is that the unit would properly be kg mSv.

Personally, I do now know milli-Sieverts from Grays from a hole in the ground. But there is Google...

Ahh, so a Gray is the same as a Sievert except for the quality factor, Q. So a mSv is the same as a mGray and a kg mSv is the same as a kg mGray: 1/1000 of a kg Gray.

Do you know what a kg Gray is?
 
  • #24
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Yes, one Gray = 1 J/kg.
But it is not a unit of energy. So a claim that total energy is some number of Grays is certainly wrong.

So we need to back up and figure out the units on the number you calculated previously.


You multiplied 0,4 by 1,2 to get 0.48. And asserted that the result is in units of Gray.

What we know for sure is that the unit would properly be kg mSv.

Personally, I do now know milli-Sieverts from Grays from a hole in the ground. But there is Google...

Ahh, so a Gray is the same as a Sievert except for the quality factor, Q. So a mSv is the same as a mGray and a kg mSv is the same as a kg mGray: 1/1000 of a kg Gray.

Do you know what a kg Gray is?
Now I'm a bit confused, if you are saying that 1 Gy = 1 J/kg and that 1 Sv = 1 Gy (assuming Q=1), isn't then the absorbed energy expressed in J/kg? Since I have calculated it to be 0,48 Gy?
 
  • #25
jbriggs444
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Now I'm a bit confused, if you are saying that 1 Gy = 1 J/kg and that 1 Sv = 1 Gy (assuming Q=1), isn't then the absorbed energy expressed in J/kg? Since I have calculated it to be 0,48 Gy?
No. Total absorbed energy would be reported in units of energy. You have multiplied dose rate in mSv by target mass in kg. The correct unit for that result is not mSv. The correct unit for that result would be kg mSv. Which is the same as kg mGray which is the same as kg mJoule/kg.

What is a better name for the unit kg mJoule/kg?

And what happened to the m in mSv?
 
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