Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At $$t = 0$$ it was 1600 counts per second and $$t = 8$$ seconds it was 100 counts per second. The count rate observed, as counts per second, at $$t = 6$$ seconds is close to:

For any radioactive sample the count rate (or the number of undecayed nuclei) varies with time according to the exponential decay law

$$N(t)=N_0\,e^{-\lambda t},$$

where $$N_0$$ is the initial count rate at $$t=0$$, $$\lambda$$ is the decay constant, and $$N(t)$$ is the count rate at a later time $$t$$.

We are told that at $$t=0$$ the counter registers $$N_0 = 1600\ \text{counts s}^{-1}$$. At $$t = 8\ \text{s}$$ the rate has fallen to $$N(8)=100\ \text{counts s}^{-1}$$. Substituting these data into the decay law gives

$$100 = 1600\,e^{-\lambda\,(8)}.$$

Dividing both sides by $$1600$$ we obtain

$$\frac{100}{1600}=e^{-8\lambda}.$$

The left-hand side simplifies to

$$\frac{100}{1600}=\frac{1}{16},$$

so we have

$$\frac{1}{16}=e^{-8\lambda}.$$

To extract $$\lambda$$ we take the natural logarithm on both sides. Using the identity $$\ln(e^{x})=x$$, we write

$$\ln\!\left(\frac{1}{16}\right)=\ln\!\left(e^{-8\lambda}\right)=-8\lambda.$$

The natural logarithm of the reciprocal turns a division into a negative sign:

$$\ln\!\left(\frac{1}{16}\right)=-\ln 16.$$

Hence,

$$-\ln 16=-8\lambda \quad\Longrightarrow\quad \lambda=\frac{\ln 16}{8}.$$

Because $$16=2^4$$, we have $$\ln16=4\ln2$$. Using $$\ln2\approx0.693$$ we find

$$\ln16=4\times0.693\approx2.772,$$

and therefore

$$\lambda=\frac{2.772}{8}\approx0.3465\ \text{s}^{-1}.$$

We now need the count rate at $$t = 6\ \text{s}$$. Inserting $$t=6\ \text{s}$$ and the just-found $$\lambda$$ into the decay formula gives

$$N(6)=1600\,e^{-\lambda\,(6)}=1600\,e^{-0.3465\times6}.$$

The product in the exponent is

$$0.3465\times6\approx2.079,$$

so

$$N(6)=1600\,e^{-2.079}.$$

Evaluating the exponential: $$e^{2.079}\approx7.999$$ (very close to 8), hence

$$e^{-2.079}\approx\frac{1}{7.999}\approx\frac{1}{8}.$$

Substituting this approximation we get

$$N(6)\approx1600\times\frac{1}{8}=200\ \text{counts s}^{-1}.$$

So the observed count rate after six seconds is about $$200\ \text{counts s}^{-1}$$, which matches option B.

Hence, the correct answer is Option B.