When $^{210}_{82}Pb$ decays to $^{206}_{80}Pb$, it emits

1. Analyze the question

We are given a radioactive decay process where a parent nucleus, Lead-210 ($^{210}_{82}Pb$), transforms into a daughter nucleus, Lead-206 ($^{206}_{80}Pb$). We need to determine the type and number of particles emitted during this process.

2. Explain the concept

In nuclear physics, radioactive decay follows the Law of Conservation of Mass Number and the Law of Conservation of Atomic Number:

• Mass Number ($A$): The total number of protons and neutrons must remain the same on both sides of the equation.
• Atomic Number ($Z$): The total charge (number of protons) must remain the same on both sides of the equation.

Particle characteristics:

• Alpha particle ($\alpha$): Represented as $^{4}_{2}He$. It has a mass number of 4 and an atomic number of 2.
• Beta particle ($\beta$): Represented as $^{0}_{-1}e$. It has a mass number of 0 and an atomic number of -1.

3. Show the solution

Let the decay equation be: $^{210}_{82}Pb \rightarrow ^{206}_{80}Pb + x(^4_2He) + y(^0_{-1}e)$

Step 1: Calculate the change in Mass Number ($\Delta A$) $\Delta A = 210 - 206 = 4$ Since an alpha particle has a mass of 4, we divide the change by 4: $4 / 4 = 1$ alpha particle.

Step 2: Calculate the change in Atomic Number ($\Delta Z$) $\Delta Z = 82 - 80 = 2$ If we emit 1 alpha particle, the atomic number decreases by 2 ($82 - 2 = 80$). Since the atomic number of the daughter nucleus is exactly 80, we do not need to emit any beta particles to balance the charge.

Verification: $^{210}_{82}Pb \rightarrow ^{206}_{80}Pb + ^4_2He$

• Mass: $206 + 4 = 210$ (Balanced)
• Atomic: $80 + 2 = 82$ (Balanced)

The decay process involves the emission of exactly one alpha particle.

4. Identify the answer

Comparing our result to the options provided, the correct choice is B.

5. Tip

When solving decay problems, always check the Mass Number first. If the mass number decreases by 4, it is a strong indicator of an alpha decay. If the mass number remains the same but the atomic number increases by 1, it is a beta decay.