A radioactive nucleus decays by two different processes. The half-life for the first process is $$10\,\text{s}$$ and that for the second is $$100\,\text{s}$$. The effective half-life of the nucleus is close to:

First recall the fundamental relation between the decay constant $$\lambda$$ and the half-life $$T_{1/2}$$ of any radioactive process:

$$T_{1/2}=\frac{\ln 2}{\lambda}.$$

In the present problem the nucleus can disintegrate independently by two different modes. Let the decay constants be $$\lambda_1$$ and $$\lambda_2$$ for these modes. Then:

$$T_{1/2}^{(1)}=10\;\text{s}\quad\Longrightarrow\quad \lambda_1=\frac{\ln 2}{T_{1/2}^{(1)}}=\frac{\ln 2}{10}.$$

Similarly,

$$T_{1/2}^{(2)}=100\;\text{s}\quad\Longrightarrow\quad \lambda_2=\frac{\ln 2}{T_{1/2}^{(2)}}=\frac{\ln 2}{100}.$$

When two independent decay routes are available simultaneously, their probabilities add. Therefore the total or effective decay constant $$\lambda_{\text{eff}}$$ is given by

$$\lambda_{\text{eff}}=\lambda_1+\lambda_2.$$

Substituting the values just obtained, we have

$$\lambda_{\text{eff}}=\frac{\ln 2}{10}+\frac{\ln 2}{100}=\ln 2\left(\frac{1}{10}+\frac{1}{100}\right)=\ln 2\left(\frac{10}{100}+\frac{1}{100}\right)=\ln 2\left(\frac{11}{100}\right)=0.11\,\ln 2.$$

Now we wish to find the corresponding effective half-life $$T_{1/2}^{\text{(eff)}}$$. Using the same basic formula, but with the effective decay constant, we write

$$T_{1/2}^{\text{(eff)}}=\frac{\ln 2}{\lambda_{\text{eff}}}.$$

Substituting $$\lambda_{\text{eff}}=0.11\,\ln 2$$:

$$T_{1/2}^{\text{(eff)}}=\frac{\ln 2}{0.11\,\ln 2}=\frac{1}{0.11}\,\text{s}\approx 9.09\,\text{s}.$$

Rounding to the nearest whole second gives approximately $$9\,\text{s}$$.

Hence, the correct answer is Option A.