Size of Americium in a Smoke Detector

[Garlic]

Hello,
A few days ago I tried to calculate the decay heat of Americium in a Smoke Detector, but as I was trying to find the volume it should have, my calculations showed that its volume should be really small (smaller than a cubic micrometer), but it was obvious that it couldn't be so small (I've seen the Americium piece inside a Smoke detector, and there was at least a cubic milimeter of it).
I thought I probably made a mistake in my calculations, so I did it again.

The amount of Americium (not Americium Oxide) in a smoke detector is 0.28 μg, but I am taking it roundly as 0.3μg.

The isotope Americium 241 has a molar mass of roundly 241 g/moles, and we have 3x10^-7 grams of it. So it must be 1.245x10^-9 moles.

The molecule AmO2 (Americium Oxide) mas a molar mass of 275 g/moles and a density of 11.68 g/cm^3.

275 g/moles x 1.245x10^-9 moles = 3.423x10^-7 grams.

(11.68 g/cm^3) / (3.423x10^-7 g) = 3.4122x10^-7 cm^3 = 3.4122x10^-4 mm^3 = 0.34122 μm^3

(Where) did I make a mistake here?

Thank you

 

[DEvens]

There are how many $\mu$m in a mm? And the cube of that number is what?

The Americium in a smoke detector is probably contained in some kind of material that is just there for handling and mechanical purposes. Just because the amount is so small, you need it fixed to something that is easily visible and handled by ordinary means.

 

[SteamKing]

Hello,
A few days ago I tried to calculate the decay heat of Americium in a Smoke Detector, but as I was trying to find the volume it should have, my calculations showed that its volume should be really small (smaller than a cubic micrometer), but it was obvious that it couldn't be so small (I've seen the Americium piece inside a Smoke detector, and there was at least a cubic milimeter of it).
I thought I probably made a mistake in my calculations, so I did it again.

The amount of Americium (not Americium Oxide) in a smoke detector is 0.28 μg, but I am taking it roundly as 0.3μg.

The isotope Americium 241 has a molar mass of roundly 241 g/moles, and we have 3x10^-7 grams of it. So it must be 1.245x10^-9 moles.

The molecule AmO2 (Americium Oxide) mas a molar mass of 275 g/moles and a density of 11.68 g/cm^3.

275 g/moles x 1.245x10^-9 moles = 3.423x10^-7 grams.

I started here:

(11.68 g/cm^3) / (3.423x10^-7 g) = 3.4122x10^-7 cm^3 = 3.4122x10^-4 mm^3 = 0.34122 μm^3

g / cc divided by g ≠ cc. It equals 1 / cc. I also cannot get 3.4122×10^-7 from the indicated arithmetic.

mass / density = volume, rather than density / mass = volume

 

[Garlic]

Okay, I corrected my mistake, I calculated that it must be 2.93x10^-2 μm^3
(Apparently brackets are necessary in the calculator app)

DEvens, If mixed with something, most of the α rays coming from Americium be blocked, all of the Americium should be placed in the uppermost part of it, otherwise most of the Americium would be unused. (Because even a sheet of paper blocks most of the α rays from Americium). Isn't that true?

 

[SteamKing]

Okay, I corrected my mistake, I calculated that it must be 2.93x10^-2 μm^3
(Apparently brackets are necessary in the calculator app)

DEvens, If mixed with something, most of the α rays coming from Americium be blocked, all of the Americium should be placed in the uppermost part of it, otherwise most of the Americium would be unused. (Because even a sheet of paper blocks most of the α rays from Americium). Isn't that true?

The americium is placed inside a special ionization chamber located inside the smoke detector. The alpha radiation ionizes a small amount of air also inside the chamber, making it more sensitive to the passage of an electrical current. If any smoke particles enter the chamber, these will attach themselves to the ionized air, which will then cause a tiny drop in the amount of current which is able to pass thru the chamber. The alarm sounds when the current passing thru the air in the chamber drops below a certain amount:

https://en.wikipedia.org/wiki/Smoke_detector#Ionization

You don't want to be spraying alpha radiation willy-nilly throughout the household.

 

[Garlic]

The americium is placed inside a special ionization chamber located inside the smoke detector. The alpha radiation ionizes a small amount of air also inside the chamber, making it more sensitive to the passage of an electrical current. If any smoke particles enter the chamber, these will attach themselves to the ionized air, which will then cause a tiny drop in the amount of current which is able to pass thru the chamber. The alarm sounds when the current passing thru the air in the chamber drops below a certain amount:

https://en.wikipedia.org/wiki/Smoke_detector#Ionization

You don't want to be spraying alpha radiation willy-nilly throughout the household.

Thanks, but I already know. But what do you mean by "You don't want to be spraying alpha radiation willy-nilly throughout the household." ?

 

[SteamKing]

Thanks, but I already know. But what do you mean by "You don't want to be spraying alpha radiation willy-nilly throughout the household." ?

Just that. People get very concerned when radiation is around.

I'll bet 99% of people who own an ionizing smoke detector do not realize that theirs contains a radioactive isotope. Even though alpha radiation can be easily blocked by most materials, regulatory authorities still like to make sure that any exposure of the general population to radiation is minimal

 

[DEvens]

Okay, I corrected my mistake, I calculated that it must be 2.93x10^-2 μm^3
(Apparently brackets are necessary in the calculator app)

DEvens, If mixed with something, most of the α rays coming from Americium be blocked, all of the Americium should be placed in the uppermost part of it, otherwise most of the Americium would be unused. (Because even a sheet of paper blocks most of the α rays from Americium). Isn't that true?

I think it's painted on the surface of this little grain of something-or-other.

3.4E-7 gm
11.68 g/cm^3
2.91E-8 cm^3
2.91E+4 $\mu$m^3

A cube about 30 $\mu$m on a side. This is very roughly twice the thickness of a human hair.

 

[Garlic]

I think it's painted on the surface of this little grain of something-or-other.

3.4E-7 gm
11.68 g/cm^3
2.91E-8 cm^3
2.91E+4 $\mu$m^3

A cube about 30 $\mu$m on a side. This is very roughly twice the thickness of a human hair.

1- Why and how is a cubic micrometer is equal to 10^-12 cubic micrometer?
There are 1000 cubic millimeters in a cubic centimeter. And there are 1000 cubic micrometers in a cubic millimeter. Shouldnt there be a million cubic micrometers in a cubic centimeter?

2- What do you mean by "A cube about 30 μm on a side."? It clearly can't be a cube of americium in the ionizing piece.
See image: http://bit.ly/1EAoYrI

 

[DEvens]

1- Why and how is a cubic micrometer is equal to 10^-12 cubic micrometer?
There are 1000 cubic millimeters in a cubic centimeter. And there are 1000 cubic micrometers in a cubic millimeter. Shouldnt there be a million cubic micrometers in a cubic centimeter?

Heh heh. How many Romans?

https://en.wikipedia.org/wiki/Micrometre

 

[Garlic]

Heh heh. How many Romans?

https://en.wikipedia.org/wiki/Micrometre

Oops... Looks like I am unimaginably stupid..
At least I diddn't make this mistake in one of our physics exams..

-After 10 minutes- OH MY GOD I REMEMBERED

 

[Garlic]

Still, I don't understand how it's volume is 30 μm^3. There is a disc containing Americium. Let's say it has 1.5 mm radius. In order it to have 3x10^-8 cm^3 volume, its thickness must be roughly 4.2 nm.

V=π r^2 h

3x10^-5 mm^3 = 7.1 mm^2 x h
h= 4.2x10^-6 mm= 4.2x10^-3 μm= 4.2 nm

How can it be 4.2 nm thick?

 

[SteamKing]

Still, I don't understand how it's volume is 30 μm^3. There is a disc containing Americium. Let's say it has 1.5 mm radius. In order it to have 3x10^-8 cm^3 volume, its thickness must be roughly 4.2 nm.

V=π r^2 h

3x10^-5 mm^3 = 7.1 mm^2 x h
h= 4.2x10^-6 mm= 4.2x10^-3 μm= 4.2 nm

How can it be 4.2 nm thick?

The volume of a cube which measures 30 μm on each side ≠ 30 μm³.

The volume of this size cube is (30×10^-3 mm)³ = 27,000×10^-9 mm³ = 27×10^-6 mm³

If a cube measures a quarter inch on a side, then the volume of the cube is not 1/4 in³ but 1/4 in × 1/4 in × 1/4 in = 1/64 in³

 

[Garlic]

The volume of a cube which measures 30 μm on each side ≠ 30 μm³.

The volume of this size cube is (30×10^-3 mm)³ = 27,000×10^-9 mm³ = 27×10^-6 mm³

If a cube measures a quarter inch on a side, then the volume of the cube is not 1/4 in³ but 1/4 in × 1/4 in × 1/4 in = 1/64 in³

I don't know how I don't remember this.. I probably hit the tree with my head harder than I thought :/

Are my calculations about the thickness of the Americium disc true?

 

[SteamKing]

Well, let's say there's 30 μg of Americium in the smoke detector, and the density of Am is 11.68 g/cc

The volume of this button would be 30×10^-6 g / 11.68 g/cc = 2.568×10^-6 cc

Let's also say the button is a cylinder with a radius of 1.5 mm. How thick is the button?

V = π r² h = 2.568×10^-6 cm³ and r = 0.15 cm

Then, h = 2.568×10^-6 / (π×0.15²) = 36.34×10^-6 cm = 36.34×10^-8 m = 0.3634 μm
Since 1 μm = 10^-6 m, which is pretty thin any way you slice it.

 

[Garlic]

Well, let's say there's 30 μg of Americium in the smoke detector, and the density of Am is 11.68 g/cc

The volume of this button would be 30×10^-6 g / 11.68 g/cc = 2.568×10^-6 cc

Let's also say the button is a cylinder with a radius of 1.5 mm. How thick is the button?

V = π r² h = 2.568×10^-6 cm³ and r = 0.15 cm

Then, h = 2.568×10^-6 / (π×0.15²) = 36.34×10^-6 cm = 36.34×10^-8 m = 0.3634 μm
Since 1 μm = 10^-6 m, which is pretty thin any way you slice it.

Thank you :)