Radium-226 decays to radon-222 with the emission of an alpha particle, \( \ce{^{226}_{88}Ra -> ^{4}_{2}He + ^{222}_{86}Rn} \). If the amount of radium-226 in a sample is halved, how does the rate of decay change?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → It becomes half
Radioactive decay of unstable nuclei follows first-order kinetics. The rate is proportional to the first power of the amount of radium present:
\( \displaystyle \text{Rate} = k[\ce{Ra}] \)
\( \displaystyle \text{New rate} = k\left(\frac{1}{2}[\ce{Ra}]\right) = \frac{1}{2} \times \text{Rate} \)
Why the other options are incorrect
Option (1) It remains the same — Incorrect. The rate would stay the same only for a zero-order process.
Option (2) It becomes one-fourth — Incorrect. The rate would become one-fourth only for a second-order process.
Option (4) It becomes double — Incorrect. A smaller amount can never give a faster decay.
NCERT (2026-27 edition):
3.3.2 First Order Reactions – “All natural and artificial radioactive decay of unstable nuclei take place by first order kinetics.”
\( \displaystyle \ce{^{226}_{88}Ra -> ^{4}_{2}He + ^{222}_{86}Rn} \qquad \text{Rate} = k[\ce{Ra}] \)