Final amount of Hg present after $\beta$ decay I have been trying to solve an exercise on nuclear chemistry, but I was not sure whether I should ask this on physics or chemistry but I think that if chemistry had an exercise based on this then chemistry experts might be having the way to overcome this problem. So here is what my nuclear chemistry question says:


  One gram atom of $ _{79}^{198}\mathrm {Au} $ (having a half life of 65
  hours) decays by $\beta$-emission to produce stable nuclide of Hg.
  How much Hg will be present after 260 hours?


The thing which is making me confused is $\beta$ decay. I think I have a formula for this question but that $\beta$ thing is creating a problem. I would like to make that equation public. Here it is:

$$N_t=N_0\cdot\exp(-\lambda\cdot t)$$

where,


$\lambda$ is the radioactive decay constant
$N_0$ is initial number of nuclides present
$N_t$ is the co. of parent nuclides present at time $t$
$$\ce{^198Au -&gt; ^198Hg + e^{-}  ~~~~~~~~ \beta~ decay}$$

260 hours corresponds to 4 half-lifes (4 x 65 = 260).  After each half-life, one-half of the $\ce{^198Au}$ will have been transformed into $\ce{^198Hg}$.  After 4 half-lifes ((1/2)^4 = 1/16) 1/16 of the gold will remain and 15/16 of the gold will have been transformed into mercury. Therefor, after 260 hours 15/16 gram atom of mercury will be present.  15/16 gram atom of mercury corresponds to (15/16 x 198 = 185.63 gm) 185.63 gm of mercury.

Using your equation we (of course) get the same result. 
$$\ce{N_t=N_o * e^{-kt}~~(I)}$$ where k is the rate constant and t the elapsed time.  For radioactive decay (a first order reaction) 
$$\ce{k=\frac{ln(2)}{t_{1/2}}~~(2)}$$ where $\ce{t_{1/2}}$ is the decay half-life.  Substituting into equation (I) we find
$$\ce{\frac{N_t}{N_o}= e^{-\frac{ln(2)}{t_{1/2}}*4}=1/16}$$