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Radioactive gold-198 is used in the diagnosis of liver problems. The half-life of this isotope is 2.7 days. If you begin with a 5.6-mg sample of the isotope, how much of this sample remains after 1.0 day?

Short Answer

Expert verified
After 1.0 day, approximately 4.32 mg of the gold-198 sample remains.

Step by step solution

01

Understand the Half-Life Concept

The half-life of a radioactive substance is the time it takes for half of the substance to decay. For gold-198, this half-life is 2.7 days.
02

Identify the Decay Formula

The decay formula for radioactive substances is given by: \[ A(t) = A_0 \times \left(\frac{1}{2}\right)^{t / T_{1/2}} \]where \( A(t) \) is the amount of substance left after time \( t \), \( A_0 \) is the initial amount, and \( T_{1/2} \) is the half-life.
03

Plugging Values into the Formula

Using the decay formula, substitute the given values:- \( A_0 = 5.6 \) mg (initial amount),- \( t = 1.0 \) day (time elapsed),- \( T_{1/2} = 2.7 \) days.The formula becomes: \[ A(1) = 5.6 \times \left(\frac{1}{2}\right)^{1.0 / 2.7} \]
04

Calculate the Exponent

First, compute the exponent: \[ \frac{1.0}{2.7} \approx 0.3704 \]
05

Compute the Remaining Amount

With the exponent calculated, compute the remaining amount: \[ A(1) = 5.6 \times \left(0.5\right)^{0.3704} \]\( \left(0.5^{0.3704} \approx 0.7711 \right) \)Then, compute:\[ A(1) \approx 5.6 \times 0.7711 \approx 4.32 \text{ mg} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Half-Life
The concept of half-life is integral to understanding how radioactive materials decay over time. The half-life of a substance is the length of time required for half of the radioactive atoms present to decay. In our example with gold-198, the half-life is precisely 2.7 days. This means that every 2.7 days, half of the gold-198 atoms will have decayed into another element or isotope.
When you start with a certain amount of radioactive material, you know that after one half-life, you'll only have half of that material remaining. This is why half-life is a useful measurement: it shows how quickly or slowly a substance loses its radioactivity. It's important to note that although half of the material decays after one half-life, what remains continues to decay at the same rate. This constant rate is key to the predictable nature of radioactive decay.
Understanding the half-life helps in planning how long a radioactive sample will last and how it can be used, such as in medical diagnostics like the example with gold-198.
Decay Formula
The decay formula is a crucial mathematical expression used in calculating the amount of radioactive substance left after any given time. Its importance cannot be overstated when dealing with radioactive decay because it precisely determines how much of a radioactive sample remains. The decay formula is:
\[ A(t) = A_0 \times \left(\frac{1}{2}\right)^{t / T_{1/2}} \]
- \( A(t) \) represents the amount of substance left after time \( t \).
- \( A_0 \) is the initial amount of the radioactive material.
- \( T_{1/2} \) refers to the half-life of the substance.
With this formula, we can see that the decay is an exponential process. As time \( t \) increases, \( A(t) \) decreases by half every period, corresponding to the substance's half-life. By plugging in values, we can predict the remaining quantity of the substance after a certain period.
For instance, in the example of gold-198, the initial amount \( A_0 \) is 5.6 mg, and we solve to find \( A(t) \) after 1 day, using the half-life of 2.7 days.
Exponential Decay
Exponential decay is a specific type of decay where the quantity of a substance decreases at a rate proportional to its current value. This is the nature of radioactive decay, where the process is not linear but exponential.
In our example, the term \( 0.5^{t/T_{1/2}} \) in the decay formula represents this exponential nature. The fraction \( \frac{1}{2} \) shows that with each passing half-life, any remaining amount of the substance is halved. This is contrary to linear decay, where the substance would lose the same absolute amount over equal time spans, exponential decay signifies that the rate of decay slows down as less of the material remains.
When calculating the remaining amount of gold-198 after 1 day, we apply this concept of exponential decay. We see that after applying the decay formula, the exponent \( 0.3704 \) reveals the proportion of the half-life that has elapsed. This demonstrates how the exponential function \( 0.5^{0.3704} \approx 0.7711 \) leads to a slower decay as more time passes. Importantly, understanding exponential decay allows you to predict just how long a radioactive material will remain useful or hazardous.

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Most popular questions from this chapter

Experimental data are listed here for the reaction \(A \rightarrow 2 B.\) $$\begin{array}{cc}\text { Time (s) } & {[\mathrm{B}](\mathrm{mol} / \mathrm{L})} \\\\\hline 0.00 & 0.000 \\\10.0 & 0.326 \\\20.0 & 0.572 \\\30.0 & 0.750 \\\40.0 & 0.890 \\\\\hline\end{array}$$ (a) Prepare a graph from these data; connect the points with a smooth line; and calculate the rate of change of [B] for each 10 -second interval from 0.0 to 40.0 seconds. Does the rate of change decrease from one time interval to the next? Suggest a reason for this result. (b) How is the rate of change of [A] related to the rate of change of \([\mathrm{B}]\) in each time interval? Calculate the rate of change of \([\mathrm{A}]\) for the time interval from 10.0 to 20.0 seconds. (c) What is the instantaneous rate, \(\Delta[\mathrm{B}] / \Delta \mathrm{t},\) when \([\mathrm{B}]=0.750 \mathrm{mol} / \mathrm{L} ?\)

Describe each of the following statements as true or false. If false, rewrite the sentence to make it correct. (a) The rate-determining elementary step in a reaction is the slowest step in a mechanism. (b) It is possible to change the rate constant by changing the temperature. (c) As a reaction proceeds at constant temperature, the rate remains constant. (d) A reaction that is third order overall must involve more than one step.

Give the relative rates of disappearance of reactants and formation of products for each of the following reactions. (a) \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NOBr}(\mathrm{g})\) (b) \(\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{g})\)

At \(573 \mathrm{K},\) gaseous \(\mathrm{NO}_{2}(\mathrm{g})\) decomposes, forming \(\mathrm{NO}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g}) .\) If a vessel containing \(\mathrm{NO}_{2}(\mathrm{g})\) has an initial concentration of \(2.8 \times 10^{-2} \mathrm{mol} / \mathrm{L},\) how long will it take for \(75 \%\) of the \(\mathrm{NO}_{2}(\mathrm{g})\) to decompose? The decomposition of \(\mathrm{NO}_{2}(\mathrm{g})\) is second order in the reactant and the rate constant for this reaction, at \(573 \mathrm{K},\) is \(1.1 \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}.\)

The reaction cyclopropane \(\rightarrow\) propene occurs on a platinum metal surface at \(200^{\circ} \mathrm{C}\). (The platinum is a catalyst.) The reaction is first order in cyclopropane. Indicate how the following quantities change (increase, decrease, or no change) as this reaction progresses, assuming constant temperature. (a) [cyclopropane] (b) [propene] (c) [catalyst] (d) the rate constant, \(k\) (e) the order of the reaction (f) the half-life of cyclopropane

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