The expression for calculating half-life for a radioactive decay is:

The correct answer is Option (3) → t_1/2 = 0.693/K

All natural and artificial radioactive decay follows first order kinetics. For a first order reaction

    \( \displaystyle k = \frac{2.303}{t}\log\frac{[R]_0}{[R]} \)

At the half-life, [R] = [R]₀/2:

    \( \displaystyle t_{1/2} = \frac{2.303}{k}\log 2 = \frac{2.303 \times 0.3010}{k} = \frac{0.693}{k} \)

The half-life of a radioactive isotope is therefore constant and does not depend on the amount of the isotope present.

Why the other options are incorrect

• Option (1) t_1/2 = (R₀ − R)/K — Incorrect. This is the zero order expression for the time to reach [R] (from [R] = [R]₀ − kt); the zero order half-life is [R]₀/2k, and radioactive decay is not zero order.

• Option (2) t_1/2 = K/0.693 — Incorrect. It is the inverse of the correct relation; a larger rate constant must give a shorter half-life, not a longer one.

• Option (4) t_1/2 = K/(R₀ − R) — Incorrect. This has the wrong form and wrong units; it is not obtained from any rate law.

NCERT (2026-27 edition):
Half-Life of a Reaction – “… \( t_{1/2} = \displaystyle\frac{0.693}{k} \)   (3.17)
It can be seen that for a first order reaction, half-life period is constant, i.e., it is independent of initial concentration of the reacting species. The half-life of a first order equation is readily calculated from the rate constant and vice versa.”