The decay constant of ^ 197 _ 80 Hg (which decays by electron capture to ^ 197 _ 79 Au ) is 1.8 × 10^ -4 s^ -1 . What is the half-life of this isotope?

Correct answer: A. 64.2 min The half-life T_ 1/2 of a radioactive isotope is related to its decay constant λ by the formula: T_ 1/2 = ( 2)/(λ) Given λ = 1.8 × 10^ -4 s^ -1 and using 2 0.693 : T_ 1/2 = 0.693 1.8 × 10^ -4 s T_ 1/2 = 3850 s Converting the half-life from seconds to minutes: T_ 1/2 = (3850)/(60) min T_ 1/2 64.16 min 64.2 min

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: The Nucleus

Concepts Of Physics MCQ Edition [Volume 2] · Physics · The Nucleus

The decay constant of ^ 197 _ 80 Hg (which decays by electron...

The decay constant of ^{197}_{80}\text{Hg}

(which decays by electron capture to ^{197}_{79}\text{Au}

) is

1.8 \times 10^{-4} \text{ s}^{-1}

. What is the half-life of this isotope?

Options

A $64.2 \text{ min}$
B $128.4 \text{ min}$
C $32.1 \text{ min}$
D $92.6 \text{ min}$

Correct answer

A. $64.2 \text{ min}$

Step-by-step solution

The half-life

T_{1/2}

of a radioactive isotope is related to its decay constant

\lambda

by the formula:

T_{1/2} = \dfrac{\ln 2}{\lambda}

Given

\lambda = 1.8 \times 10^{-4} \text{ s}^{-1}

and using

\ln 2 \approx 0.693

T_{1/2} = \dfrac{0.693}{1.8 \times 10^{-4}} \text{ s}

T_{1/2} = 3850 \text{ s}

Converting the half-life from seconds to minutes:

T_{1/2} = \dfrac{3850}{60} \text{ min}

T_{1/2} \approx 64.16 \text{ min} \approx 64.2 \text{ min}

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