The rate constant of a nuclear decay is k = 6.5 × 10−14 s−1. Which of…

The correct answer is Option (3) → t_1/2 = 1.07 × 10¹³ s and it will go to completion in infinite time.

Every radioactive (nuclear) decay follows first order kinetics. This is clear from the unit of the rate constant here, s⁻¹. For a first order reaction the integrated rate equation and the half-life are:

    \( \displaystyle [R]=[R]_0\,e^{-kt} \)

    \( \displaystyle t_{1/2}=\frac{0.693}{k} \)

Two facts follow:

• The half-life is independent of the starting amount; it depends only on k.

• After every half-life only half of what is left decays, so [R] becomes 1/2, 1/4, 1/8 ... of [R]₀ but never becomes zero. A first order reaction therefore reaches completion only in infinite time. (In practice about 99.9% is over after 10 half-lives.)

Here k = 6.5 × 10⁻¹⁴ s⁻¹:

    \( \displaystyle t_{1/2}=\frac{0.693}{6.5\times10^{-14}\ \mathrm{s^{-1}}} \)

    \( \displaystyle t_{1/2}=0.1066\times10^{14}\ \mathrm{s}\approx1.07\times10^{13}\ \mathrm{s} \)

So t_1/2 = 1.07 × 10¹³ s, and the decay goes to completion only in infinite time.

Why the other options are incorrect

• Option (1) t_1/2 = 1.26 × 10¹³ s and it will go to completion in time equal to 5 half lives — Incorrect. The half-life value is wrong, and after 5 half-lives 1/32 (about 3%) of the substance is still left.

• Option (2) t_1/2 = 1.26 × 10¹³ s and it will go to completion in time equal to 50 half lives — Incorrect. The half-life value is wrong, and even after 50 half-lives a fraction (1/2)⁵⁰ remains; a first order reaction is never fully complete.

• Option (4) t_1/2 = 1.07 × 10¹³ s and it will go to completion in time equal to 14 half lives — Incorrect. The half-life is correct, but after 14 half-lives (1/2)¹⁴ of the substance is still left, so the decay is not complete.

Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.

NCERT (2026-27 edition):
Half-Life of a Reaction – …