Chemistry · Chemical Kinetics · NEET
Radioactive decay is always FIRST ORDER. The rate of decay depends only on the number of undecayed atoms present: rate = k[N]. The unit of the decay constant k is s⁻¹ (or year⁻¹), which is the tell-tale sign of a first order process. Because it is first order, you can use every first order formula (t = 2.303/k · log([N]₀/[N]) and t½ = 0.693/k) directly for decay problems.
For any first order reaction, t½ = 0.693/k. Look carefully: this formula has NO initial concentration in it. So the time to fall to half is fixed no matter whether you start with 1 gram or 1 kilogram. That is why a radioactive sample keeps the same half-life from start to end. This is a favourite NEET point, so remember it.
A living thing keeps a fixed ratio of carbon-14. When it dies, C-14 stops coming in and starts to decay with half-life 5730 years. To find age, use the first order formula: t = (2.303/k) log([N]₀/[N]), where [N]₀ is the C-14 in a fresh living sample and [N] is the C-14 left now, and k = 0.693/5730 year⁻¹. If the amount has simply halved a whole number of times, just count half-lives: each halving = 5730 years.
No. For first order (and so for radioactive decay), t½ = 0.693/k depends only on k, not on the starting amount. This is different from a ZERO order reaction, where t½ = [A]₀/2k does depend on the initial amount. NEET often tests this exact contrast, so keep the two separate.
First check if the drop is a clean number of halvings. 20 g → 10 g → 5 g is two halvings, so time = 2 × t½. This shortcut is fast and safe. If the drop is not a clean halving (like 7.2 → 0.9, which is a factor of 8 = 2³, i.e. 3 half-lives), still count halvings, or use t = (2.303/k) log([N]₀/[N]).
A first order reaction has a specific reaction rate of 10⁻² s⁻¹. How much time will it take for 20 g of the reactant to reduce to 5 g?
A first order reaction has a rate constant of 2.303×10⁻³ s⁻¹. The time required for 40 g of this reactant to reduce to 10 g will be (Given log₁₀2 = 0.3010):
If the half-life t½ for a first order reaction is 1 minute, then the time required for 99.9% completion of the reaction is closest to:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Because each atom decays on its own, independent of the others. So the overall rate is proportional to the number of atoms present: rate = k[N]. That single dependence on the first power of amount makes it first order, with k in units of time⁻¹.
Carbon-14 has a half-life of 5730 years, so k = 0.693/5730 ≈ 1.21×10⁻⁴ year⁻¹. Use this k in t = (2.303/k) log([N]₀/[N]) to date old objects.
No. Carbon dating works only up to about 50,000 years, because after too many half-lives almost no C-14 is left to measure. Very old rocks are dated using isotopes with much longer half-lives, such as uranium-lead.
They describe the same decay but in opposite ways. The decay constant k tells how fast decay happens (bigger k = faster). The half-life t½ is the time to fall to half. They are linked by t½ = 0.693/k, so a large k gives a short half-life.
Yes. If the rate constant has units of s⁻¹, min⁻¹ or year⁻¹ (just inverse time), the reaction is first order. Radioactive decay always has k in inverse-time units, confirming it is first order.
