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

Correct answer: C. 92.6 min The average life of a radioactive isotope is given by the reciprocal of its decay constant λ . = (1)/(λ) Given λ = 1.8 × 10^ -4 s^ -1 , substituting this value gives: = 1 1.8 × 10^ -4 s = (10000)/(1.8) s = (50000)/(9) s To convert the average life from seconds to minutes, divide by 60 : = (50000)/(9 × 60) min = (2500)/(27) min 92.59 min 92.6 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}

. Determine the average-life of this isotope.

Options

A $185.2 \text{ min}$
B $64.2 \text{ min}$
C $92.6 \text{ min}$
D $46.3 \text{ min}$

Correct answer

C. $92.6 \text{ min}$

Step-by-step solution

The average life

\tau

of a radioactive isotope is given by the reciprocal of its decay constant

\lambda

\tau = \dfrac{1}{\lambda}

Given

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

, substituting this value gives:

\tau = \dfrac{1}{1.8 \times 10^{-4}} \text{ s} = \dfrac{10000}{1.8} \text{ s} = \dfrac{50000}{9} \text{ s}

To convert the average life from seconds to minutes, divide by

60

\tau = \dfrac{50000}{9 \times 60} \text{ min} = \dfrac{2500}{27} \text{ min}

\tau \approx 92.59 \text{ min} \approx 92.6 \text{ min}

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