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 half-life of the sample.

Correct answer: B. 38 min The radioactive decay law is given by A = A_0 e^ -λ t . Given A_0 = 500 s^ -1 and A = 200 s^ -1 at t = 50 min . Substituting the values: 200 = 500 e^ -50λ e^ 50λ = (500)/(200) = 2.5 Taking the natural logarithm on both sides: 50λ = 2.5 λ = ( 2.5)/(50) The half-life T_ 1/2 is given by: T_ 1/2 = ( 2)/(λ) = (50 2)/( 2.5) Using _ 10 2 0.301 and _ 10 2.5 = _ 10 ((10)/(4)) = 1 - 2 _ 10 2 0.398 : T_ 1/2 = 50 × (0.301)/(0.398) 37.8 min Rounding to the nearest integer, we get 38 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

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 half-life of the sample.

Options

A $25 \text{ min}$
B $38 \text{ min}$
C $19 \text{ min}$
D $76 \text{ min}$

Correct answer

B. $38 \text{ min}$

Step-by-step solution

The radioactive decay law is given by

A = A_0 e^{-\lambda t}

.

Given

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

and

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

t = 50 \text{ min}

.

Substituting the values:

200 = 500 e^{-50\lambda}

e^{50\lambda} = \dfrac{500}{200} = 2.5

Taking the natural logarithm on both sides:

50\lambda = \ln 2.5

\lambda = \dfrac{\ln 2.5}{50}

The half-life

T_{1/2}

is given by:

T_{1/2} = \dfrac{\ln 2}{\lambda} = \dfrac{50 \ln 2}{\ln 2.5}

Using

\log_{10} 2 \approx 0.301

and

\log_{10} 2.5 = \log_{10}\left(\dfrac{10}{4}\right) = 1 - 2\log_{10} 2 \approx 0.398

T_{1/2} = 50 \times \dfrac{0.301}{0.398} \approx 37.8 \text{ min}

Rounding to the nearest integer, we get

38 \text{ min}

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