The table shows how the count rate from a sample of a radioactive nuclide varies with time \(t\). The nuclide has a half-life of \(50\,\mathrm{s}\). The average background count rate is constant.
| t / s | Count rate / \(\mathrm{s^{-1}}\) |
|---|---|
| 0 | 72 |
| 50 | 40 |
What count rate will the detector measure when \(t=150\,\mathrm{s}\)?
Correct Answer: \( \boxed{\mathrm{B}} \)
The measured count rate contains both the count rate from the radioactive sample and the constant background count rate.
Let the background count rate be \(B\).
At \(t=0\), the measured count rate is \(72\,\mathrm{s^{-1}}\).
At \(t=50\,\mathrm{s}\), one half-life has passed, so the activity of the sample has halved. Therefore,
\(40-B=\frac{1}{2}(72-B)\)
\(80-2B=72-B\)
\(B=8\,\mathrm{s^{-1}}\)
Therefore, the initial count rate due to the radioactive sample alone is
\(72-8=64\,\mathrm{s^{-1}}\)
At \(t=150\,\mathrm{s}\), three half-lives have passed:
\(150=3(50)\)
The sample count rate is therefore
\(64\left(\frac{1}{2}\right)^3=8\,\mathrm{s^{-1}}\)
Adding the background count rate gives the detector reading:
\(\text{count rate}=8+8=16\,\mathrm{s^{-1}}\)
Thus,
\( \boxed{16\,\mathrm{s^{-1}}} \)
Hence, the correct answer is \( \boxed{\mathrm{B}} \).
Deflection in electric/magnetic fields requires charged particles:
• Alpha particles: positively charged (deflected)
• Beta particles: negatively charged (deflected)
• Gamma photons: neutral (not deflected)
✅ Answer: (A) I and II only
| Count rate / counts s−1 | Time / s |
|---|---|
| 150 | 0 |
| 90 | 20 |
