Radioactiv edecay will occur as follows
`overset(220)86Rnrarroverset(216)84PO + overset(4)2He ` Half life =55s
`overset(216)84Poarroverset(212)82Pb + overset(4)2He ` Half life =0.66s
`overset(812)82Pbrarroverset(212)82Bi + overset(0)-1e ` Half life =10.6 s
If a certain mass of radon (Rn=220) is allowed to decay in a certain container,then after 5 minutes the element with the greater mass will be

Options

A. radon
B. polonium
C. lead
D. bismuth

Explanation

To solve the problem, we need to analyze the decay of the radioactive elements step by step and determine which element will have the greatest mass after 5 minutes (300 seconds).

### Step 1: Determine the number of half-lives for each element
First, we need to calculate how many half-lives each element undergoes in 5 minutes (300 seconds).

1. **Radon (Rn-220)**: Half-life = 55 seconds
\[ \text{Number of half-lives} = \frac{300 \text{ seconds}}{55 \text{ seconds}} \approx 5.45 \text{ half-lives} \]

2. **Polonium (Po-216)**: Half-life = 0.66 seconds
\[ \text{Number of half-lives} = \frac{300 \text{ seconds}}{0.66 \text{ seconds}} \approx 454.55 \text{ half-lives} \]

3. **Lead (Pb-212)**: Half-life = 10.6 seconds
\[ \text{Number of half-lives} = \frac{300 \text{ seconds}}{10.6 \text{ seconds}} \approx 28.30 \text{ half-lives} \]

### Step 2: Calculate the remaining mass for each element
Assuming we start with a mass \( M_0 \) of Radon (Rn-220), we can calculate the remaining mass after the decay for each element.

1. **Radon (Rn-220)**:
\[ M_{\text{Rn}} = M_0 \left(\frac{1}{2}\right)^{5.45} \approx M_0 \cdot 0.028 \]

2. **Polonium (Po-216)**:
Since Polonium is produced from Radon, we need to calculate how much Polonium is produced before it decays. The amount of Polonium produced can be approximated using the decay of Radon:
\[ M_{\text{Po}} \approx M_0 - M_{\text{Rn}} \approx M_0 - M_0 \cdot 0.028 \approx M_0 \cdot 0.972 \]
However, this will also decay into Lead, so we need to account for that as well.

3. **Lead (Pb-212)**:
Lead is produced from the decay of Polonium. The amount of Lead produced can be calculated from the decay of Polonium:
\[ M_{\text{Pb}} \approx M_{\text{Po}} \left(1 - \left(\frac{1}{2}\right)^{454.55}\right) \approx M_{\text{Po}} \approx M_0 \cdot 0.972 \]
Since the decay of Polonium is very rapid, almost all of it will convert to Lead.

### Step 3: Compare the masses
Now we need to compare the remaining masses of Radon, Polonium, and Lead after 5 minutes.

- **Remaining mass of Radon**: \( M_{\text{Rn}} \approx M_0 \cdot 0.028 \)
- **Remaining mass of Polonium**: \( M_{\text{Po}} \) will be very small due to its rapid decay.
- **Mass of Lead**: Since Polonium decays rapidly into Lead, the mass of Lead will be significant.

### Conclusion
After 5 minutes, the element with the greatest mass will be **Lead (Pb-212)**, as it is produced from the decay of Polonium and has a much slower decay rate compared to Polonium.