Calculate number of photons absorbed

[Luxdot]

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?

 

[Steve4Physics]

The question is incomplete. May I suggest you post the complete question, word-for-word.

 

[kuruman]

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?

 

[Luxdot]

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.

 

[kuruman]

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?

 

[Luxdot]

OK, so you know that 1 Sv = 1 Joule/kilogram. Therefore ...

Therefore the absorbed dose is 0,33 mGy?

 

[kuruman]

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.

 

[Luxdot]

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).

 

[kuruman]

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.

 

[Luxdot]

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?

 

[hutchphd]

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?

 

[Luxdot]

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".

 

[Steve4Physics]

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?

 

[Luxdot]

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?

 

[jbriggs444]

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.

 

[Luxdot]

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?

 

[jbriggs444]

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.

 

[Luxdot]

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?

 

[jbriggs444]

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.

 

[Luxdot]

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)

 

[jbriggs444]

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.

 

[Luxdot]

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?

 

[jbriggs444]

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?

 

[Luxdot]

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?

 

[jbriggs444]

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?

 

[Luxdot]

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?

mJ?
What I have calculated (the 0,48 Gy) is the absorbed dosage. What is the total absorbed energy then?

 

[Luxdot]

Just to clarify, what I'm after is the number of absorbed atoms. I have given the energy of one photon (50 keV), the equivalent dose (0,4 mSv) and the mass (1,2 kg). I have from this (as stated earlier) calculated the absorbed dose (0,48 Gy). With this information, how do I calculate the number of absorbed photons?
My initial though was to divide the total energy (which I don't have?) by the energy of one photon (50 keV).

 

[Steve4Physics]

My initial though was to divide the total energy (which I don't have?) by the energy of one photon (50 keV).

That is the correct approach. You can easily find the total energy, but you haven't understood the underlying concepts yet. Try this short exercise (the values used are unrealistic and for illustration only). Post your answers.

5kg of tissue absorbs 10J of ionising radation energy.
Q1. What is the absorbed dose in Gy?
Q2. What is the absorbed dose in mGy?
Q3. If the type of radiation is X-rays, what is the effective (equivalent) dose in mSv?
Q4. If the type of radiation is alpha radiation, what is the effective dose in mSv?

(Edited - typo's corrected.)

 

[Luxdot]

Q1. 2 Gy (10J/5kg)
Q2. 2000 mGy
Q3. 2 Sv or 2000 mSv(Assuming Q=1)
Q4. 40 Sv (Q=20)

 

[kuruman]

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?

Wrong. This is where your confusion lies and you need to settle this first. Let's put units in this equation. You have $0.4~\dfrac{\text{(mJ)}}{\text{(kg)}}$, and $1.2~\text{(kg)}$. Thus, $0.4~\dfrac{\text{(mJ)}}{\text{(kg)}}\times1.2~\text{(kg)}=0.48~\text{(what units?)}.$ Then you proceed..

 

[Luxdot]

Wrong. This is where your confusion lies and you need to settle this first. Let's put units in this equation. You have $0.4~\dfrac{\text{(mJ)}}{\text{(kg)}}$, and $1.2~\text{(kg)}$. Thus, $0.4~\dfrac{\text{(mJ)}}{\text{(kg)}}\times1.2~\text{(kg)}=0.48~\text{(what units?)}.$ Then you proceed..

mJ! So this is my total energy? If so, what is the absorbed dose?

 

[jbriggs444]

mJ! So this is my total energy? If so, what is the absorbed dose?

Who cares? You wanted energy absorbed. You got energy absorbed. Now compute number of photons from that.

 

[Luxdot]

Who cares? You wanted energy absorbed. You got energy absorbed. Now compute number of photons from that.

The first part of the question was what the absorbed dose was, I though I had done this correctly but obviously not...

 

[kuruman]

And don't forget that the absorbed energy is in mJ.

 

[jbriggs444]

The first part of the question was what the absorbed dose was, I though I had done this correctly but obviously not...

I do not see that you have ever shown us the first part of the question or your attempted solution to it.

 

[kuruman]

The first part of the question was what the absorbed dose was, I though I had done this correctly but obviously not...

I thought the absorbed dose was given as you posted in #4. Is it a calculated number from a part of the problem that you didn't post?

 

[Luxdot]

I thought the absorbed dose was given as you posted in #4. Is it a calculated number from a part of the problem that you didn't post?

That's the equivalent dose... To clarify, the first part is to find the absorbed dose. I did this by multiplying the equivalent dose (0,4 mSv) by the mass (1,2 kg) and got 0,48 mJ. Since the absorbed dose should be in Gray I'm a bit confused now since I made the error of thinking that I got the answer in mSv... I don't think I understand how to find the absorbed dose since I have mixed this up with absorbed energy...
The second part of the question is to find the number of absorbed photons. This I was trying to do by dividing the total energy (0,48 mJ) by the energy of one photon (50 keV), but I get the answer 0,0096 which seems wring? The total energy should be larger than the energy of one photon, right?
Hope this clears up my issues a bit!

 

[jbriggs444]

The total energy should be larger than the energy of one photon, right?

What are the units on the photon energy?
What are the units on the total energy?

This problem is not about radiology. It is about units.

 

[Luxdot]

What are the units on the photon energy?
What are the units on the total energy?

This problem is not about radiology. It is about units.

One is Joule and one is eV... My bad! So the total energy should be 0,48 mJ = about 3*10¹⁵ eV and the energy of one photon was 50 keV, so it makes more sense

 

[kuruman]

One is Joule and one is eV... My bad! So the total energy should be 0,48 mJ = about 3*10¹⁵ eV and the energy of one photon was 50 keV, so it makes more sense

Looks like you are on the right track now.

 

[Luxdot]

Yes I think so! But I'm still a bit confused on how I find the absorbed dose?

 

[kuruman]

Yes I think so! But I'm still a bit confused on how I find the absorbed dose?

See your answers to Q1 and Q2 to the hypothetical example in #29. Why do you need it anyway to calculate the number of photons if you have the total absorbed energy and the energy of one photon?

 

[Luxdot]

See your answers to Q1 and Q2 to the hypothetical example in #29. Why do you need it anyway to calculate the number of photons if you have the total absorbed energy and the energy of one photon?

So let´s say, hypothetically that I have a radiation energy (energy of one photon) of 10 keV (which is equal to 1,6*10^-15 J) and an equivalent dose of 1 Sv and a mass of 1 kg. The absorbed dose is then 1 J/kg or 1 Gy (assuming Q=1). This is also the total energy when multiplied by the mass (in this case 1 so no difference). The number of absorbed photons is therefore 1 [J/kg]/1,6*10^-15 [J] = 6,25*10^-15 photons. Is this correct?

 

[Luxdot]

What's confusing me is that for example if I have an equivalent dose of 1 Sv and a mass of 1,5 kg. Then the absorbed dose should be 1,5 Gy? But then I have multiplied J/kg with kg which should remove kg? Am I wrong when I'm scaling the absorbed dose to the mass? Or should the absorbed dose be 1 Gy even though the mass is 1,5 kg? Do you see my concern?

 

[Steve4Physics]

Q1. 2 Gy (10J/5kg)
Q2. 2000 mGy
Q3. 2 Sv or 2000 mSv(Assuming Q=1)
Q4. 40 Sv (Q=20)

Well done @Luxdot (though the answer to Q4 should have been in mSv).

It is worth noting that the answer to Q1 can also be be written as 2 J/kg.

Maybe part of your confusion is that 'absorbed dose' does not apply to an entire object - it is the aborbed energy per kg of the object.$$absorbed \, dose\, (in \, Gy) = \frac {total \, energy \, absorbed \, (in\,J)}{mass\, of\, object\,(in\, kg)}$$If you want the total energy absorbed, you rearrange the above formula.

 

[kuruman]

So let´s say, hypothetically that I have a radiation energy (energy of one photon) of 10 keV (which is equal to 1,6*10^-15 J) and an equivalent dose of 1 Sv and a mass of 1 kg. The absorbed dose is then 1 J/kg or 1 Gy (assuming Q=1). This is also the total energy when multiplied by the mass (in this case 1 so no difference). The number of absorbed photons is therefore 1 [J/kg]/1,6*10^-15 [J] = 6,25*10^-15 photons. Is this correct?

It is not correct on many grounds. Maybe in this example the equivalent dose is 1 J/kg and the mass also 1 kg, but you do not divide dose by photon energy to get number of photons. Also, 1/1.6 must be a number less than 1. Also, the number of photons is nonsense. You cannot have such a number 15 orders of magnitude less than 1. Do you understand the algebra of powers of 10?

On edit: Here is an equivalent problem that I am sure you can do. It has the same solving logic as the one you posted.
You work at a grocery store with pay $15.00 per hour for 4 hours. How many cans of soda can you buy if one can costs $1.20?

Once you do this problem, consider the following equivalence:
Dollars → Energy (Joules)
Your hourly pay → Dose (Joules/kilogram)
Hours → Mass of object (Kilograms)
Soda cans → Photons (just a number).

Do you see how it works?

 

[Luxdot]

It is not correct on many grounds. Maybe in this example the equivalent dose is 1 J/kg and the mass also 1 kg, but you do not divide dose by photon energy to get number of photons. Also, 1/1.6 must be a number less than 1. Also, the number of photons is nonsense. You cannot have such a number 15 orders of magnitude less than 1. Do you understand the algebra of powers of 10?

On edit: Here is an equivalent problem that I am sure you can do. It has the same solving logic as the one you posted.
You work at a grocery store with pay $15.00 per hour for 4 hours. How many cans of soda can you buy if one can costs $1.20?

Once you do this problem, consider the following equivalence:
Dollars → Energy (Joules)
Your hourly pay → Dose (Joules/kilogram)
Hours → Mass of object (Kilograms)
Soda cans → Photons (just a number).

Do you see how it works?

I understand that logic fine, I would be able to buy 50 cans... But there is something else that doesn't add up for me. 50 keV is 8*10^-15 joules, right?
If the energy of one photon is 50 keV (8*10^-15 J), the dose would be the energy (Dollars) times by mass (Hours), i.e. 8*10^-15*1,2 = 9,6 *10^-15 J/kg which I do not think is right... And what number would I divide this by to get the number of photon (Soda cans), i.e. what would the $1.20 represent in my case? Isn't is as simple as dividing the total energy by the energy of one photon to get the number of absorbed photons?

 

[jbriggs444]

Isn't is as simple as dividing the total energy by the energy of one photon to get the number of absorbed photons?

Yes.

 

[Steve4Physics]

I understand that logic fine, I would be able to buy 50 cans... But there is something else that doesn't add up for me. 50 keV is 8*10^-15 joules, right?
If the energy of one photon is 50 keV (8*10^-15 J), the dose would be the energy (Dollars) times by mass (Hours), i.e. 8*10^-15*1,2 = 9,6 *10^-15 J/kg which I do not think is right... And what number would I divide this by to get the number of photon (Soda cans), i.e. what would the $1.20 represent in my case? Isn't is as simple as dividing the total energy by the energy of one photon to get the number of absorbed photons?

You have hit a ‘blind spot’ with regards to finding the total energy absorbed. See if this helps.

For X-rays, 1Sv corresponds to 1Gy, so the absorbed dose is 0.40mGy.
This is the same as 4.0x10⁻⁴ Gy.

1Gy is the same as 1J/kg. So we can also express the absorbed dose as 4.0x10⁻⁴ J/kg.

This means each kg absorbed 4.0x10⁻ ⁴ J.

You had 1.2kg, so how much energy was absorbed in total?

Once you have the total energy, then as you say, it ”is as simple as dividing the total energy by the energy of one photon to get the number of absorbed photons”.

 

[Luxdot]

You have hit a ‘blind spot’ with regards to finding the total energy absorbed. See if this helps.

For X-rays, 1Sv corresponds to 1Gy, so the absorbed dose is 0.40mGy.
This is the same as 4.0x10⁻⁴ Gy.

1Gy is the same as 1J/kg. So we can also express the absorbed dose as 4.0x10⁻⁴ J/kg.

This means each kg absorbed 4.0x10⁻ ⁴ J.

You had 1.2kg, so how much energy was absorbed in total?

Once you have the total energy, then as you say, it ”s as simple as dividing the total energy by the energy of one photon to get the number of absorbed photons”.

I see! Since the energy absorbed per kg is 40x1^-4 J, the total absorbed energy is 4x10^-4x1,2 = 4,8x10^-4 Gy = 0,48 mGy?
Would then the number of photons be 4,8x10^-4 [J/kg]/8*10^-15 = 6x10¹⁰ photons?