Question:

Half-life of 32P is

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Radioisotope Half-Lives:
$^{32}\text{P} = 14.3\text{ days}$.
$^{35}\text{S} = 87.4\text{ days}$.
$^{131}\text{I} = 8\text{ days}$.
$^{3}\text{H} = 12.3\text{ years}$.
  • 164 days
  • 87 days
  • 14 days
  • 8.1 days
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The Correct Option is C

Solution and Explanation


Step 1: Understanding the Concept:

Phosphorus-32 ($^{32}\text{P}$) is a high-energy $\beta$-emitting radioisotope widely used for nucleic acid radiolabeling in molecular biology.
Key Formula or Approach:
\[ N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}}, \quad t_{1/2}(^{32}\text{P}) \approx 14.29\text{ days} \]

Step 2: Detailed Explanation:

Phosphorus-32 ($^{32}\text{P}$):
- Pure $\beta^-$-emitter ($E_{\max} = 1.71\text{ MeV}$) used to label nucleotides at the $\alpha$- or $\gamma$-phosphate position for Southern blotting, Northern blotting, DNA sequencing, and kinase assays.
- Its radioactive half-life ($t_{1/2}$) is 14.28 days ($\approx 14$ days).
(Compare with $^{35}\text{S} = 87.4\text{ days}$, $^{14}\text{C} = 5730\text{ years}$, $^{131}\text{I} = 8.02\text{ days}$, $^{45}\text{Ca} = 163\text{ days}$).

Step 3: Final Answer:

Thus, the half-life of 32P is 14 days, matching option (C).
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