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