A radioactive isotope has a decay constant of 10⁻⁸ s⁻¹. Calculate its half-life.

$$T_{1/2} = 0.693 \times 10^8 \text{ s}$$ $$T_{1/2} = 6.93 \times 10^7 \text{ s}$$

Comparing this to the options:
A. $5.93 \times 10^6 \text{ s}$
B. $6.93 \times 10^7 \text{ s}$
C. $6.93 \times 10^9 \text{ s}$
D. $5.93 \times 10^7 \text{ s}$

The calculated half-life matches option B.

The final answer is $\boxed{\text{B}}$

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Let $x$ be the number of alpha particles ($^{4}_{2}\text{He}$) and $y$ be the number of beta particles ($^{0}_{-1}\text{e}$).

$$^{210}_{82}\text{Pb} \to ^{206}_{80}\text{Pb} + x \cdot ^{4}_{2}\text{He} + y \cdot ^{0}_{-1}\text{e}$$

$$210 = 206 + 4x + 0y$$ $$210 - 206 = 4x$$ $$4 = 4x$$ $$x = 1$$

This means one alpha particle is emitted.

$$82 = 80 + 2x + (-1)y$$

Substitute $x=1$:

$$82 = 80 + 2(1) - y$$ $$82 = 82 - y$$ $$y = 0$$

This means no beta particles are emitted.

Therefore, the decay process emits one alpha particle.

Comparing this to the options:
A. two alpha and two beta particles.
B. a beta particle
C. one alpha and (assuming this means "one alpha particle")

Based on the calculation, one alpha particle is emitted.

The final answer is $\boxed{\text{C}}$