Paper: CSIR-NET June 2013 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Chemical Kinetics – General · Marks: 2 · Difficulty: Medium
A radioisotope ${ }^{41} \mathrm{Ar}$ initially decays at the rate of 34, 500 disintegrations/minute, but decay rate falls to 21, 500 disintegrations/minute after 75 minutes. The $\mathrm{t}_{1 / 2}$ for ${ }^{41} \mathrm{Ar}$ is:
(a)90 minutes
(b)110 minutes
(c)180 minutes
(d)220 minutes
Answer
Answer: B ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
k = (1/75) ln(34500/21500) = 0.4729/75 = 6.3×10⁻³ min⁻¹; t₁/₂ = 0.693/6.3×10⁻³ = 110 min.