Question

The disintegration rate of a certain radioactive sample at any instant is 4250 disintegrations per minute. 10 minutes later, the rate becomes 2250 disintegrations per minute. The approximate decay constant is
(Take log₁₀1.88 = 0.274)

• A. 0.02 min^–1
• B. 2.7 min^–1
• C. 0.063 min^–1
• D. 6.3 min^–1

Answer 1

The Correct Option is C

To find the decay constant (\(\lambda\)) of a radioactive sample, we can use the relation between the decay rates at different times. The formula relating the decay rates at two different times is given by:

\text{R}_2 = \text{R}_1 \cdot e^{-\lambda t}

Where:

• R_1 is the initial disintegration rate.
• R_2 is the disintegration rate after time t.
• \lambda is the decay constant.
• t is the elapsed time.

From the problem, we know:

• R_1 = 4250 disintegrations per minute.
• R_2 = 2250 disintegrations per minute.
• t = 10 minutes.

Substitute the values into the formula:

2250 = 4250 \cdot e^{-\lambda \cdot 10}

Dividing both sides by 4250 gives:

\frac{2250}{4250} = e^{-\lambda \cdot 10}

Simplify the fraction:

\frac{2250}{4250} = \frac{9}{17}

Thus:

e^{-\lambda \cdot 10} = \frac{9}{17}

Taking the natural logarithm on both sides:

-\lambda \cdot 10 = \ln\left(\frac{9}{17}\right)

Convert the natural log to base 10 using the provided value \log_{10}(1.88) = 0.274:

First, recognize that:

\ln(x) = \frac{\log_{10}(x)}{\log_{10}(e)}

Calculating:

\log_{10}\left(\frac{9}{17}\right) \approx \log_{10}\left(\frac{1}{1.88}\right) = -0.274

Then, find:

-\lambda \cdot 10 = \frac{-0.274}{0.4343} \approx -0.631

Solving for \lambda:

\lambda = \frac{0.631}{10} = 0.0631 \text{ min}^{-1}

Therefore, the approximate decay constant is 0.063 min^-1, which matches the correct answer option.