At a certain time, a specific sample of a radioactive material decays at the rate of 500 per second. After 50 minutes , the count rate falls to 200 per second. Determine the decay constant of the sample.

Correct answer: B. 3.05 × 10^ -4 s^ -1 Given initial activity A_0 = 500 s^ -1 Activity after time t = 50 minutes = 3000 s is A = 200 s^ -1 Using the radioactive decay law for activity, A = A_0 e^ -λ t Substituting the given values: 200 = 500 e^ -λ × 3000 e^ 3000 λ = (500)/(200) = 2.5 Taking natural logarithm on both sides: 3000 λ = (2.5) λ = (0.916)/(3000) λ = 3.05 × 10^ -4 s^ -1

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

At a certain time, a specific sample of a radioactive material...

At a certain time, a specific sample of a radioactive material decays at the rate of

500

per second. After

50 \text{ minutes}

, the count rate falls to

200

per second. Determine the decay constant of the sample.

Options

A $1.83 \times 10^{-2} \text{ s}^{-1}$
B $3.05 \times 10^{-4} \text{ s}^{-1}$
C $1.52 \times 10^{-4} \text{ s}^{-1}$
D $6.10 \times 10^{-4} \text{ s}^{-1}$

Correct answer

B. $3.05 \times 10^{-4} \text{ s}^{-1}$

Step-by-step solution

Given initial activity

A_0 = 500 \text{ s}^{-1}

Activity after time

t = 50 \text{ minutes} = 3000 \text{ s}

A = 200 \text{ s}^{-1}

Using the radioactive decay law for activity,

A = A_0 e^{-\lambda t}

Substituting the given values:

200 = 500 e^{-\lambda \times 3000}

e^{3000 \lambda} = \dfrac{500}{200} = 2.5

Taking natural logarithm on both sides:

3000 \lambda = \ln(2.5)

\lambda = \dfrac{0.916}{3000}

\lambda = 3.05 \times 10^{-4} \text{ s}^{-1}

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