Given that the decay constant of ^ 238 U is 4.9 × 10^ -18 s^ -1 , find the half-life of ^ 238 U .

Correct answer: D. 4.5 × 10^ 9 y The half-life of a radioactive substance is given by: T_ 1/2 = ( 2)/(λ) = (0.693)/(λ) Substituting λ = 4.9 × 10^ -18 s^ -1 : T_ 1/2 = 0.693 4.9 × 10^ -18 s = 1.414 × 10^ 17 s Converting the half-life from seconds to years ( 1 year 3.154 × 10^7 s ): T_ 1/2 = 1.414 × 10^ 17 3.154 × 10^7 y T_ 1/2 4.48 × 10^9 y 4.5 × 10^9 y

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

Given that the decay constant of ^ 238 U is 4.9 × 10^ -18 s^ -1 ,...

Given that the decay constant of ^{238}\text{U}

is

4.9 \times 10^{-18} \text{ s}^{-1}

, find the half-life of ^{238}\text{U}

Options

A $2.25 \times 10^{9} \text{ y}$
B $6.5 \times 10^{9} \text{ y}$
C $9.0 \times 10^{9} \text{ y}$
D $4.5 \times 10^{9} \text{ y}$

Correct answer

D. $4.5 \times 10^{9} \text{ y}$

Step-by-step solution

The half-life of a radioactive substance is given by:

T_{1/2} = \dfrac{\ln 2}{\lambda} = \dfrac{0.693}{\lambda}

Substituting

\lambda = 4.9 \times 10^{-18} \text{ s}^{-1}

T_{1/2} = \dfrac{0.693}{4.9 \times 10^{-18}} \text{ s} = 1.414 \times 10^{17} \text{ s}

Converting the half-life from seconds to years (

1 \text{ year} \approx 3.154 \times 10^7 \text{ s}

T_{1/2} = \dfrac{1.414 \times 10^{17}}{3.154 \times 10^7} \text{ y}

T_{1/2} \approx 4.48 \times 10^9 \text{ y} \approx 4.5 \times 10^9 \text{ y}

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