Q.23
Nucleus of a radioactive material can undergo beta decay with half life of 4 minutes. Suppose beta decay starts with 4096 nuclei at \( t = 0 \), the number of nuclei left after 20 minutes would be
The nucleus undergoes beta decay with a 4-minute half-life, starting from 4096 nuclei at t=0. After 20 minutes, exactly 5 half-lives pass, halving the nuclei each time to leave 128 undecayed.
Calculation Steps
Radioactive decay follows N = N₀ (1/2)^(t/T), where N₀ = 4096 is initial nuclei, t = 20 min is elapsed time, and T = 4 min is half-life.
Number of half-lives n = t/T = 20/4 = 5. Thus, N = 4096 × (1/2)^5 = 4096 × 1/32 = 128.
Option Analysis
| Option | Nuclei Count | Half-Lives | Time (min) | Status |
|---|---|---|---|---|
| (A) 1024 | 4096/4 = 1024 | 2 | 8 | Too early |
| (B) 128 | 4096/32 = 128 | 5 | 20 | Correct |
| (C) 512 | 4096/8 = 512 | 3 | 12 | Too early |
| (D) 256 | 4096/16 = 256 | 4 | 16 | Too early |
CSIR NET Exam Insight
In radioactive decay problems like this CSIR NET-style question, beta decay half life calculation determines nuclei remaining over time. Starting with 4096 nuclei and a 4-minute half-life, the beta decay process halves the count every 4 minutes. After 20 minutes (5 half-lives), precise application of the formula yields 128 nuclei left.
This matches option (B), as detailed above. Such beta decay half life calculations test exponential decay mastery for competitive exams.
