During World War I radium-226 was used in the manufacturing of luminous paint. If it takes 2.12 x 10 4 days for its degradation to be 2.49% complete, what is its decay constant?

Here we're told that during World War 1, radium 226 was used in the manufacturing of luminous paint. If it takes 2.12 times 10 to the 4 days for its degradation to be 2.49% complete, what is its decay constant? Right. So here we need to determine what our k value will be. Our integrated rate law is ln of our final concentration equals negative kt plus ln of our initial concentration. Here they're saying 2.49% complete. What does that mean? Well that means that 2.49% of your beginning material has decomposed. It's gone. So here, since we're dealing with percentages, we can assume that we're dealing with a 100% initially. So initial is a 100. We do a 100% minus 2.49% to determine what's left. That would come out to 97.51%. So that would be my final amount. We don't know what k is, that's what we need to find, and we have 2.12 times 10 to the 4 days. Here we're going to subtract ln100 from both sides. When we do that, we're going to see that our equation was going to be negative 0.025212, 212-252.491 equals negative k times 2.12x10 to the 4 days. Divide both sides by that to isolate your k. So when we do that we're going to see that k equals 1.19 times 10 to the negative 6 days inverse. So this would be our final answer.

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21. Nuclear Chemistry

Rate of Radioactive Decay

General Chemistry

21. Nuclear Chemistry

Rate of Radioactive Decay

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Multiple Choice

During World War I radium-226 was used in the manufacturing of luminous paint. If it takes 2.12 x 10⁴ days for its degradation to be 2.49% complete, what is its decay constant?

1.74 x 10^–4 days^–1

1.08 x 10^–5 days^–1

1.19 x 10^–6 days^–1

2.14 x 10^–5 days^–1

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Verified step by step guidance

Identify the type of decay process: Radium-226 undergoes first-order radioactive decay, which is characterized by a constant half-life and a decay rate that is proportional to the amount of substance present.

Use the first-order decay formula: The formula for first-order decay is \( N(t) = N_0 e^{-kt} \), where \( N(t) \) is the remaining quantity of the substance, \( N_0 \) is the initial quantity, \( k \) is the decay constant, and \( t \) is the time elapsed.

Determine the fraction of radium-226 remaining: Since the degradation is 2.49% complete, 97.51% of the radium-226 remains. Therefore, \( \frac{N(t)}{N_0} = 0.9751 \).

Rearrange the decay formula to solve for the decay constant \( k \): Take the natural logarithm of both sides of the equation \( 0.9751 = e^{-kt} \) to get \( \ln(0.9751) = -kt \).

Calculate the decay constant \( k \): Substitute \( t = 2.12 \times 10^4 \) days into the equation \( \ln(0.9751) = -k(2.12 \times 10^4) \) and solve for \( k \) to find the decay constant.