How many Bq equals to 1 Ci?

[Drizzy]

Homework Statement

1 Curie corresponds to the number of decay in 1 gram radium-226 in 1 second. How many Bequerel is 1 Curie?

Homework Equations

The Attempt at a Solution

Half life: 1600 years

1 gram = 0,001 kg = (0,001/1,660539*10^-27) u
one Radium atom = 226,0260974 u

Number of decays: (0,001/1,660539*10^-27)/226,0260974

Then I thought that since the half-life is so high that the activity won't change much in only 1 second so I though I could pretend that the activity is constant. So I just divided the number of decays with one second but my answer is wrong. In the key that I got from my teacher I am supposed to use this formula: A = lambda * N.

Is it wrong to assume that the activity doesn't change so much in 1 second?

 

[SteamKing]

Homework Statement

1 Curie corresponds to the number of decay in 1 gram radium-226 in 1 second. How many Bequerel is 1 Curie?

Homework Equations

The Attempt at a Solution

Half life: 1600 years

1 gram = 0,001 kg = (0,001/1,660539*10^-27) u
one Radium atom = 226,0260974 u

Number of decays: (0,001/1,660539*10^-27)/226,0260974

Then I thought that since the half-life is so high that the activity won't change much in only 1 second so I though I could pretend that the activity is constant. So I just divided the number of decays with one second but my answer is wrong. In the key that I got from my teacher I am supposed to use this formula: A = lambda * N.

Is it wrong to assume that the activity doesn't change so much in 1 second?

Possibly.

But I can't figure out your calculations. 1 g of radium 226 contains a certain number of atoms initially. Even figuring a constant decay rate over the HL of this isotope gives you a much closer figure than the one you didn't finish calculating, BTW.

Remember Avogadro!

 

[Drizzy]

okay I will try to explain it. 1 g is 0,001 kg. Then I divided it by1,66 * 10^-27 because I want to convert it from kg to u. And then when I have it in u I took it and divided it by how much one atom weighs so that I know how many atoms there are.

My book uses this method at times.. they say that we pretend that the activity is constant when the time is small compared to the half-life so I thought I could apply that to this question too. I don't know why they didn't do it on this question.. is it maybe because we know that the atom is known, that we now tht it is radium and we can calculate A through the formula?

 

[SteamKing]

okay I will try to explain it. 1 g is 0,001 kg. Then I divided it by1,66 * 10^-27 because I want to convert it from kg to u. And then when I have it in u I took it and divided it by how much one atom weighs so that I know how many atoms there are.

You don't care how much an atom of radium weighs, only how many atoms of radium you start with.

You know that 1 mole of Ra-226 will contain a fixed number of atoms, namely Avogadro's number, which is 6.022 × 10²³ atoms. You figure out how many moles 1 g of radium is by dividing this amount by the molecular weight of radium, which in this case is 226. The corollary to this is that 226 g of Ra-226 = 1 mole of atoms, by definition.

https://en.wikipedia.org/wiki/Mole_(unit)

My book uses this method at times.. they say that we pretend that the activity is constant when the time is small compared to the half-life so I thought I could apply that to this question too. I don't know why they didn't do it on this question.. is it maybe because we know that the atom is known, that we now tht it is radium and we can calculate A through the formula?

I just don't think you have calculated the initial number of atoms of radium correctly, regardless of what method your book seems to use.

 

[Drizzy]

my book got 2,66 * 10^21 atoms. if I calculate this:

(0,001/1,660539*10^-27)/226,0260974 I et exactly 2,66 * 10^21 So it seems to be right

 

[SteamKing]

my book got 2,66 * 10^21 atoms. if I calculate this:

(0,001/1,660539*10^-27)/226,0260974 I et exactly 2,66 * 10^21 So it seems to be right

That's the same number as dividing Avogadro by the MW of radium (226). Less fuss using moles.

So, this means that your decay rate is calculated incorrectly. The number which you claim is the number of decays in the OP doesn't seem to use the HL of radium.

 

[Drizzy]

I don't know how to use moles.

the key= 1 mole Radiums is 226 grams. How am I supposed to know this?

 

[Drizzy]

never mind i figured it out :) Thanx for helping out :P

 

[SteamKing]

I don't know how to use moles.

the key= 1 mole Radiums is 226 grams. How am I supposed to know this?

By knowing the definition of what a mole is, just like you know what a kilogram is or what a meter is. I even provided a link.

The mole is not some obscure concept. It's what most of chemistry is based on as well as working with the ideal gas laws in physics.

PV = nRT, the n represents the number of moles of gas.