Question

The half-life of a radioactive isotope (\(^{210}\)Bi) is 5 days. The fraction of the nuclei undecayed at the end of 20 days will be

• A. \(\frac{1}{2}\)
• B. \(\frac{1}{4}\)
• C. \(\frac{1}{5}\)
• D. \(\frac{1}{16}\)

Hint:

You can also think about this step-by-step: - After 5 days (1st half-life): 1/2 remains. - After 10 days (2nd half-life): 1/4 remains. - After 15 days (3rd half-life): 1/8 remains. - After 20 days (4th half-life): 1/16 remains.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The half-life (\(T_{1/2}\)) of a radioactive isotope is the time it takes for half of the initial number of nuclei to decay. After each half-life, the amount of the remaining substance is halved.
Step 2: Key Formula or Approach:
The fraction of nuclei remaining undecayed (\(\frac{N}{N_0}\)) after a certain time \(t\) can be calculated using the number of half-lives, \(n\). \[ n = \frac{\text{Total time}}{\text{Half-life}} = \frac{t}{T_{1/2}} \] The fraction remaining is then given by: \[ \frac{N}{N_0} = \left(\frac{1}{2}\right)^n \] Step 3: Detailed Explanation:
Given values:
Half-life, \(T_{1/2} = 5\) days.
Total time, \(t = 20\) days.
First, calculate the number of half-lives (\(n\)) that have passed in 20 days: \[ n = \frac{20 \text{ days}}{5 \text{ days}} = 4 \] So, 4 half-lives have occurred.
Now, calculate the fraction of undecayed nuclei: \[ \frac{N}{N_0} = \left(\frac{1}{2}\right)^4 = \frac{1}{2 \times 2 \times 2 \times 2} = \frac{1}{16} \] Step 4: Final Answer:
After 20 days (4 half-lives), the fraction of the nuclei remaining undecayed will be \(\frac{1}{16}\). Option (D) is correct.