Calculate number of photons absorbed

[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?

 

[kuruman]

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?

Yes.

 

[kuruman]

This is a new question. Please post on separate thread with separate title.

 

[jbriggs444]

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?

Ummm. You need to track your units more carefully.

If the energy absorbed per kg is $4 \times 10^{-4}$ J then we multiply by the number of kilograms to get the total energy absorbed, yes. Don't just multiply the numbers. Multiply the units as well.

The unit for total energy absorbed is the Joule, not a Joule per kilogram. If you find yourself expressing energy in milliGrays then you know that you have bungled something.

 

[Steve4Physics]

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?

One mistake (but a very serious one). Others have pointed it out, but I will point it out again!

You cannot say total absorbed energy is [some number] of Gy. It is [some number] of J.

For example if the object's mass is 2kg and the aborbed dose is 3Gy, then the total energy absorbed by the object is 2*3 = 6J. It is not 6Gy.

 

[kuruman]

When I posted #51, I thought, without really thinking, that one cannot get the correct number of photons from the wrong premises. I suspect OP has the same conceptual difficulty as people who, when told that a power plant generates 200 MW, ask "200 MW per what?"
To @Luxdot: Until you get used to the idea of Grays as a unit, should carry the units when multiplying numbers and not put them in only some times. Here is what I mean using the soda can question as an example.

Hourly wage = $W = 15.00~\frac{\text{dollars}}{\text{hour}}$
Hours worked = $T=4~\text{hour}$
Can price = $P=1.20~\frac{\text{dollars}}{\text{can}}$
Formula for solution
$$\text{Number of cans}=\frac{\text{Hours worked}\times \text{Hourly wage}}{\text{Can price}}=\frac{T\times W}{P}.$$ Now $$T\times W \times \frac{1}{P}=4~\cancel{\text{hour}}\times 15.00~\frac{\cancel{\text{dollars}}}{\cancel{\text{hour}}}\times \frac{1}{1.20}~\frac{\text{can}}{\cancel {\text{dollars}}}=50~\text{can(s)}.$$
Let's see you do this using your numerical variables.

 

[Luxdot]

One mistake (but a very serious one). Others have pointed it out, but I will point it out again!

You cannot say total absorbed energy is [some number] of Gy. It is [some number] of J.

For example if the object's mass is 2kg and the aborbed dose is 3Gy, then the total energy absorbed by the object is 2*3 = 6J. It is not 6Gy.

So, since the equivalent dose is 0,40 mSv = 0,0004 Sv and the mass 1,2 kg, the absorbed dose is 0,00048 Gy (or J/kg). To get the total absorbed energy, wouldn't I have to divide the absorbed dose with the mass again to get rid of the kg? I.e. 0,00048 [J/kg] / 1,2 [kg] = 0,00040 [J]? This would mean that the number of photons are 0,00040[J]/8*10^-15[J] = 5*10¹⁰ photons. Given that the energy of one photon is 50 keV or 8*10^-15 J.

 

[Steve4Physics]

So, since the equivalent dose is 0,40 mSv = 0,0004 Sv and the mass 1,2 kg, the absorbed dose is 0,00048 Gy (or J/kg). To get the total absorbed energy, wouldn't I have to divide the absorbed dose with the mass again to get rid of the kg? I.e. 0,00048 [J/kg] / 1,2 [kg] = 0,00040 [J]? This would mean that the number of photons are 0,00040[J]/8*10^-15[J] = 5*10¹⁰ photons. Given that the energy of one photon is 50 keV or 8*10^-15 J.

No. Go through each of the following steps very carefully.

1. The equivalent dose is 0.40mSv = 4x10⁻⁴ Sv.

2. Because we are dealing with X-rays, the 'Q-factor' =1, so

$absorbed \space dose = \frac {equivalent \space dose}{Q-factor} = \frac {4 \times10⁻⁴}{1} = 4\times10⁻⁴ Gy$

Converting from Sv to Gy has nothing to do with the mass.

Note that 4x10⁻⁴ Gy means the same as 4x10⁻⁴ J/kg

3. Then, to get the total energy in joules, you multiply absorbed dose by mass:

E = (4x10⁻⁴ J/kg) x (1.2kg) = 4.8x10⁻⁴J (the ‘kg’ unit cancels giving the answer in joules)

4. Finally, to get the number of photons, you divide E (from step 3) by the energy (in J) of one photon.

(Note. If we were told that we had alpha-radiation rather than X-rays, then the Q-factor would be 20 and the absorbed dose would be 0.40mSv/20 = 0.020mGy. If you then wanted the total energy absorbed (in joules) you would multiply 0.020 by the mass.)

[Edit - typo' corrected.]