A sample contains 1010 atoms of a radioactive isotope with a half-life of 30 days. Calculate (i) the initial activity in disintegrations/sec and (ii) the time for the number of atoms to reduce to 104.

The initial number of atoms is $N_0 = 10^{10}$ atoms.
The half-life is $T_{1/2} = 30$ days.

To work with seconds for activity calculations, convert the half-life to seconds:

$$T_{1/2} = 30 \text{ days} \times 24 \frac{\text{hours}}{\text{day}} \times 60 \frac{\text{minutes}}{\text{hour}} \times 60 \frac{\text{seconds}}{\text{minute}}$$ $$T_{1/2} = 2,592,000 \text{ s}$$

The decay constant is related to the half-life by the formula:

$$\lambda = \frac{\ln(2)}{T_{1/2}}$$ $$\lambda = \frac{0.693147}{2,592,000 \text{ s}}$$ $$\lambda \approx 2.67495 \times 10^{-7} \text{ s}^{-1}$$

i) Calculate the number of disintegrations in the first second.
The number of disintegrations per second is the activity $A$. At the initial instant, this is $A_0$.

$$A_0 = \lambda N_0$$ $$A_0 = (2.67495 \times 10^{-7} \text{ s}^{-1}) \times (10^{10} \text{ atoms})$$ $$A_0 = 2674.95 \text{ disintegrations/second}$$

Rounding to four significant figures:

$$\boxed{\text{2.675} \times 10^3 \text{ disintegrations/second}}$$

ii) What time will elapse before $10^4$ atoms remain?
We use the radioactive decay formula:

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

Given $N(t) = 10^4$ atoms, $N_0 = 10^{10}$ atoms, and $T_{1/2} = 30$ days.

$$10^4 = 10^{10} \left(\frac{1}{2}\right)^{t/30}$$

Divide both sides by $10^{10}$:

$$\frac{10^4}{10^{10}} = \left(\frac{1}{2}\right)^{t/30}$$ $$10^{-6} = \left(\frac{1}{2}\right)^{t/30}$$

Take the natural logarithm of both sides:

$$\ln(10^{-6}) = \ln\left(\left(\frac{1}{2}\right)^{t/30}\right)$$ $$-6 \ln(10) = \frac{t}{30} \ln\left(\frac{1}{2}\right)$$

Since $\ln\left(\frac{1}{2}\right) = -\ln(2)$:

$$-6 \ln(10) = -\frac{t}{30} \ln(2)$$ $$6 \ln(10) = \frac{t}{30} \ln(2)$$

Solve for $t$:

$$t = 30 \text{ days} \times \frac{6 \ln(10)}{\ln(2)}$$ $$t = 30 \text{ days} \times \frac{6 \times 2.302585}{0.693147}$$ $$t = 30 \text{ days} \times \frac{13.81551}{0.693147}$$ $$t = 30 \text{ days} \times 19.93157$$ $$t = 597.947 \text{ days}$$

Rounding to one decimal place:

$$\boxed{\text{597.9 days}}$$

iii) What is the Count rate at this time?
"This time" refers to when $N(t) = 10^4$ atoms. The count rate is the activity $A(t)$ at this specific time.

$$A(t) = \lambda N(t)$$

Using the decay constant $\lambda \approx 2.67495 \times 10^{-7} \text{ s}^{-1}$ and $N(t) = 10^4$ atoms:

$$A(t) = (2.67495 \times 10^{-7} \text{ s}^{-1}) \times (10^4 \text{ atoms})$$ $$A(t) = 2.67495 \times 10^{-3} \text{ disintegrations/second}$$

Rounding to four significant figures:

$$\boxed{\text{2.675} \times 10^{-3} \text{ disintegrations/second}}$$