There are $$10^{10}$$ radioactive nuclei in a given radioactive element. Its half-life time is 1 min. How many nuclei will remain after 30 s? $$(\sqrt{2} = 1.414)$$

We begin with the radioactive-decay law, which gives the number of undecayed nuclei $$N$$ at any time $$t$$: $$N = N_0 \left(\dfrac{1}{2}\right)^{\dfrac{t}{T_{1/2}}}$$ where $$N_0$$ is the initial number of nuclei and $$T_{1/2}$$ is the half-life.

In the present problem the data are:

Initial count $$N_0 = 10^{10}$$,

Half-life $$T_{1/2} = 1 \text{ min} = 60 \text{ s}$$,

Elapsed time $$t = 30 \text{ s}$$.

Substituting these numerical values into the decay formula, we have

$$ N = 10^{10} \left(\dfrac{1}{2}\right)^{\dfrac{30\;\text{s}}{60\;\text{s}}}. $$

The exponent simplifies first:

$$ \dfrac{30}{60} = 0.5. $$

So the expression becomes

$$ N = 10^{10} \left(\dfrac{1}{2}\right)^{0.5}. $$

Recall that raising $$\dfrac{1}{2}$$ to the power $$0.5$$ is the same as taking its square root, because $$a^{0.5} = \sqrt{a}$$. Hence,

$$ \left(\dfrac{1}{2}\right)^{0.5} = \sqrt{\dfrac{1}{2}} = \dfrac{1}{\sqrt{2}}. $$

Using the given value $$\sqrt{2} = 1.414$$, we find

$$ \dfrac{1}{\sqrt{2}} = \dfrac{1}{1.414} \approx 0.707. $$

Multiplying this numerical factor with the initial number of nuclei, we obtain

$$ N = 10^{10} \times 0.707 = 7.07 \times 10^9. $$

When rounded to one significant figure, the result is simply $$7 \times 10^9$$.

Hence, the correct answer is Option A.