Calculating Am-241 Content from Radioactivity Measurements

[triac]

Homework Statement

A piece of Am-241 has a radioactivity of 10kBq. Determine how much Am-241 it contains.

Homework Equations

N(t)=N_0(\frac{1}{2})^{t/T_{1/2}}

The Attempt at a Solution

Let A be the activity
Let N be the number of atoms
We know that A(t)=A_0(\frac{1}{2})^{t/T_{1/2}} We can set our initial time to zero, which gives us A(t)=10kBq=A_0.
Furthermore, we know that A(t)=-N(t). We also know that N(t)=N_0(\frac{1}{2})^{t/T_{1/2}} => N'(t)=-\frac{N_0ln2}{T_{1/2}}(\frac{1}{2})^{t/T_{1/2}}=>N'(0)=-\frac{N_0ln2}{T_{1/2}}=10kBq => N_0=\frac{10kBqT_{1/2}}{ln2}. Now we use that one atom weights 241,0568229u and that the half-life is 432,2 y (I converted it to seconds). Then we get that the mass of our "piece" is approximately 78,7 micrograms. However, in the key it says 1,83 ng.
What am I doing wrong here?

 

[mgb_phys]

It's much simpler than that
The number of decays in a second is just the number of atoms * the chance of a decay/second
Which is just 1/mean lifetime - which you can easily get form the half life

 

[Rajini]

Use this equation.
<br /> A=A_0\exp\left(\frac{-t\ln 2}{T_{1/2}}\right),<br />
A_0=10000 Bq,
T_{1/2} is halflife
PS: What is 't' (times passed from initial activity)?