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:

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


  • $\lambda$ is the radioactive decay constant

  • $N_0$ is initial number of nuclides present

  • $N_t$ is the co. of parent nuclides present at time $t$

  • $\begingroup$ Clarify, what part of the exercise you have problems with $\endgroup$ – permeakra Jun 28 '14 at 15:57
  • $\begingroup$ I had already mentioned that what does question mean by $\beta$ decay. Means I has $N_0$, $t_{\frac{1}{2}}$ and $N_t$. With all this I can even find $\lambda$. But does $\beta$ or any other kind of particle gonna make difference by emission or they are having no use in the question. $\endgroup$ – Saharsh Jun 28 '14 at 16:03
  • $\begingroup$ Absolutely no. This is decay, not chain reaction or breeder reactor. For the purpose of the question you can ignore it completely. $\endgroup$ – permeakra Jun 28 '14 at 16:35

$$\ce{^198Au -> ^198Hg + e^{-} ~~~~~~~~ \beta~ decay}$$

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 $$\ce{\frac{N_t}{N_o}= e^{-\frac{ln(2)}{t_{1/2}}*4}=1/16}$$

  • $\begingroup$ This is what I also got but in my book the answer is given as $\frac{15}{16}$. This is quite confusing. $\endgroup$ – Saharsh Jun 28 '14 at 17:23
  • $\begingroup$ The question asks how much Hg will be formed and the correct answer is 15/16. The "equation" shows how much Au remains $\ce{(N_t/N_o)}$ which is 1/16. $\endgroup$ – ron Jun 28 '14 at 17:29
  • $\begingroup$ Oh gosh! I didn't paid attention on that. Ok alright. Thank You. $\endgroup$ – Saharsh Jun 28 '14 at 17:34
  • $\begingroup$ Your welcome, glad to help. $\endgroup$ – ron Jun 28 '14 at 17:34

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