A sample containing radioactive ^ 131 I with a half-life of 8.0 days has an initial activity of 20 at t = 0 . Determine the activity of the sample at t = 4.0 days.

Correct answer: C. 14 The activity of a radioactive sample at time t is given by the formula: A(t) = A_0 ((1)/(2))^ t T_ 1/2 Given A_0 = 20 , T_ 1/2 = 8.0 days , and t = 4.0 days . Substituting the values: A(4.0) = 20 ((1)/(2))^ (4.0)/(8.0) A(4.0) = 20 ((1)/(2))^ (1)/(2) A(4.0) = 20 2 A(4.0) = 10 2 10 × 1.414 = 14.14 The closest value is 14 .

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

A sample containing radioactive ^ 131 I with a half-life of 8.0...

A sample containing radioactive ^{131}\text{I}

with a half-life of

8.0 \text{ days}

has an initial activity of

20 \mu\text{Ci}

at

t = 0

. Determine the activity of the sample at

t = 4.0 \text{ days}.

Options

A $28 \mu\text{Ci}$
B $10 \mu\text{Ci}$
C $14 \mu\text{Ci}$
D $7.1 \mu\text{Ci}$

Correct answer

C. $14 \mu\text{Ci}$

Step-by-step solution

The activity of a radioactive sample at time

t

is given by the formula:

A(t) = A_0 \left(\dfrac{1}{2}\right)^{\dfrac{t}{T_{1/2}}}

Given

A_0 = 20 \mu\text{Ci}

T_{1/2} = 8.0 \text{ days}

, and

t = 4.0 \text{ days}

.

Substituting the values:

A(4.0) = 20 \left(\dfrac{1}{2}\right)^{\dfrac{4.0}{8.0}}

A(4.0) = 20 \left(\dfrac{1}{2}\right)^{\dfrac{1}{2}}

A(4.0) = \dfrac{20}{\sqrt{2}}

A(4.0) = 10\sqrt{2} \approx 10 \times 1.414 = 14.14 \mu\text{Ci}

The closest value is

14 \mu\text{Ci}

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