Questions: The radioactive decay of T1-206 to Pb-206 has a half-life of 4.2 s. Starting with 6.4 x 10^12 atoms, how many will remain after 12.6 s? 1.6 x 10^12 8.0 x 10^11 4.0 x 10^11 3.2 x 10^12

Solution Steps

The half-life of T1-206 is 4.2 seconds. We need to determine how many half-lives have passed in 12.6 seconds.

\[ \text{Number of half-lives} = \frac{12.6 \, \text{s}}{4.2 \, \text{s/half-life}} = 3 \]

Starting with \(6.4 \times 10^{12}\) atoms, we will calculate the remaining atoms after each half-life.

After the first half-life (4.2 s): \[ \frac{6.4 \times 10^{12}}{2} = 3.2 \times 10^{12} \]

After the second half-life (8.4 s): \[ \frac{3.2 \times 10^{12}}{2} = 1.6 \times 10^{12} \]

After the third half-life (12.6 s): \[ \frac{1.6 \times 10^{12}}{2} = 8.0 \times 10^{11} \]

Final Answer