Half-life period of a radio-isotope is independent of — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - The Nature of Radioactive Decay Radioactive decay is a **purely nuclear** process. It depends only on the instability of the nucleus itself (the neutron-to-proton ratio, binding energy, etc.) and is characterized by the decay constant $\lambda$: $$t_{1/2} = \frac{\ln 2}{\lambda}$$ Since $\lambda$ is an intrinsic nuclear property, $t_{1/2}$ is independent of any external macroscopic conditions. Step 2 - Evaluate Each Factor **(A) Temperature**: Higher temperature increases kinetic energy of atoms but cannot affect nuclear binding forces ($\sim 10^6$ times stronger than chemical bonds). Half-life is **independent** of temperature. ✓ **(B) State of chemical combination**: Whether the element is in elemental form, oxide, sulfide, ionic solution, etc., the nucleus is completely unaffected. Chemical bonding involves only orbital electrons. Half-life is **independent** of chemical state. ✓ **(C) Amount of radioisotope**: The half-life is defined as the time for half of any given number of nuclei to decay. Whether you have 1 mg or 1 kg, $t_{1/2}$ is the same. Half-life is **independent** of the amount. ✓ **(D) Pressure**: Pressure acts on the electron cloud (via mechanical force on atoms/molecules) and cannot penetrate the nucleus. Half-life is **independent** of pressure. ✓ Step 3 - Conclusion The half-life is independent of all four factors listed. $$\boxed{\text{Answer: A, B, C, D — All four options are correct}}$$