The activity of a radioactive sample falls from 700 s$$^{-1}$$ to 500 s$$^{-1}$$ in 30 minutes. Its half life is close to:

We know that the rate of radioactive disintegration, called activity, follows the exponential decay law

$$A = A_0 e^{-\lambda t}$$

where $$A_0$$ is the initial activity, $$A$$ is the activity after time $$t$$, and $$\lambda$$ is the decay constant.

In the present problem the activity falls from $$700\;\text{s}^{-1}$$ to $$500\;\text{s}^{-1}$$ in $$30\;\text{min}$$. So we put

$$A_0 = 700\;\text{s}^{-1}, \qquad A = 500\;\text{s}^{-1}, \qquad t = 30\;\text{min}.$$

Substituting these values in the decay equation, we get

$$500 = 700\, e^{-\lambda(30)}.$$

First we isolate the exponential term by dividing both sides by $$700$$:

$$\frac{500}{700} = e^{-\lambda(30)}.$$

Simplifying the fraction gives

$$\frac{5}{7} = e^{-30\lambda}.$$

Now we take the natural logarithm (base $$e$$) of both sides. Using the property $$\ln(e^{x}) = x$$, we obtain

$$\ln\!\left(\frac{5}{7}\right) = -30\lambda.$$

To find $$\lambda$$, we divide by $$-30$$:

$$\lambda = -\frac{1}{30}\,\ln\!\left(\frac{5}{7}\right).$$

Because $$\displaystyle\ln\!\left(\frac{5}{7}\right)$$ is negative, the minus sign makes $$\lambda$$ positive, as expected. Let us evaluate the logarithm step by step:

$$\ln\!\left(\frac{5}{7}\right) = \ln 5 - \ln 7.$$

Using the common values $$\ln 5 \approx 1.6094$$ and $$\ln 7 \approx 1.9459$$, we get

$$\ln\!\left(\frac{5}{7}\right) \approx 1.6094 - 1.9459 = -0.3365.$$

Substituting this in the expression for $$\lambda$$, we have

$$\lambda = -\frac{1}{30}\,(-0.3365) = \frac{0.3365}{30}\;\text{min}^{-1}.$$

So

$$\lambda \approx 0.01122\;\text{min}^{-1}.$$

The half-life $$T_{1/2}$$ is related to $$\lambda$$ by the well-known formula

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

We substitute $$\ln 2 \approx 0.6931$$ and $$\lambda \approx 0.01122\;\text{min}^{-1}$$:

$$T_{1/2} = \frac{0.6931}{0.01122}\;\text{min}.$$

Carrying out the division,

$$T_{1/2} \approx 61.8\;\text{min}.$$

This value is very close to $$62\;\text{min}$$.

Hence, the correct answer is Option B.