91Ó°ÊÓ

The half-life of a radioactive substance is the time \(H\) that it takes for half of the substance to change form through radioactive decay. This number does not depend on the amount with which you start. For example, carbon 14 is known to have a half-life of \(H=5770\) years. Thus if you begin with 1 gram of carbon 14, then 5770 years later you will have \(\frac{1}{2}\) gram of carbon 14. And if you begin with 30 grams of carbon 14 , then after 5770 years there will be 15 grams left. In general, radioactive substances decay according to the formula $$ A=A_{0} \times 0.5^{t / H} $$ where \(H\) is the half-life, \(t\) is the elapsed time, \(A_{0}\) is the amount you start with (the amount when \(t=0\) ), and \(A\) is the amount left at time \(t\). a. Uranium 228 has a half-life \(H\) of \(9.3\) minutes. Thus the decay function for this isotope of uranium is $$ A=A_{0} \times 0.5^{t / 9.3} $$ where \(t\) is measured in minutes. Suppose we start with 8 grams of uranium 228 . i. How much uranium 228 is left after \(2 \mathrm{~min}\) utes? ii. How long will you have to wait until there are only 3 grams left? b. Uranium 235 is the isotope of uranium that can be used to make nuclear bombs. It has a halflife of 713 million years. Suppose we start with 5 grams of uranium 235 . i. How much uranium 235 is left after 200 million years? ii. How long will you have to wait until there are only 3 grams left?

Step by step solution

01

Decay Calculation for Uranium 228 (2 minutes)

Starting with 8 grams of uranium 228 and using the given decay function, we calculate the amount left after 2 minutes: \[ A = 8 \times 0.5^{2/9.3} \] First, calculate the exponent: \[ \frac{2}{9.3} \approx 0.215 \] Then, calculate the decay factor: \[ 0.5^{0.215} \approx 0.871 \] Finally, find the amount left: \[ A \approx 8 \times 0.871 = 6.968 \, \text{grams} \]

02

Time Calculation for Uranium 228 (3 grams remaining)

To find the time for uranium 228 to decay to 3 grams, use the decay formula:\[ 3 = 8 \times 0.5^{t/9.3} \]Solve for \( t \):\[ 0.375 = 0.5^{t/9.3} \]Take the logarithm of both sides:\[ \log_{10}(0.375) = \log_{10}(0.5^{t/9.3}) \]This becomes:\[ \log_{10}(0.375) = \frac{t}{9.3} \times \log_{10}(0.5) \]Solve for \( t \):\[ t \approx 9.3 \times \frac{\log_{10}(0.375)}{\log_{10}(0.5)} \approx 15.12 \, \text{minutes} \]

03

Decay Calculation for Uranium 235 (200 million years)

Using the decay formula for uranium 235 starting with 5 grams, calculate the amount left after 200 million years:\[ A = 5 \times 0.5^{200/713} \]Calculate the exponent:\[ \frac{200}{713} \approx 0.28 \]Calculate the decay factor:\[ 0.5^{0.28} \approx 0.832 \]Then, find the remaining amount:\[ A \approx 5 \times 0.832 = 4.16 \, \text{grams} \]

04

Time Calculation for Uranium 235 (3 grams remaining)

To find the time for uranium 235 to decay to 3 grams, use the decay formula:\[ 3 = 5 \times 0.5^{t/713} \]Solve for \( t \):\[ 0.6 = 0.5^{t/713} \]Take the logarithm of both sides:\[ \log_{10}(0.6) = \log_{10}(0.5^{t/713}) \]This becomes:\[ \log_{10}(0.6) = \frac{t}{713} \times \log_{10}(0.5) \]Solve for \( t \):\[ t \approx 713 \times \frac{\log_{10}(0.6)}{\log_{10}(0.5)} \approx 384.8 \, \text{million years} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Radioactive Decay

Radioactive decay is a natural and spontaneous process where unstable atomic nuclei lose energy by emitting radiation. Atoms of certain elements are inherently unstable, and to reach stability, they transform to other atoms by releasing particles and energy. This process reduces the number of initially unstable nuclei over time.

- **Half-life**: A crucial concept in radioactive decay, half-life refers to the time necessary for half of the original nuclei in a sample to transform into different atomic forms. It remains constant regardless of the size of the sample and is an intrinsic property of the radioactive material. - **Importance**: Knowing the half-life helps in determining how long a radioactive substance will remain active and is crucial in fields such as archaeology, medicine, and nuclear power.
For example, in the exercise, Carbon 14 has a half-life of 5770 years. This means if you start with 1 gram, after 5770 years, only half would remain. Simple, right?

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. They frequently appear in many areas of science and mathematics to model growth or decay processes.

- **General Form**: The general form is given as \( A=A_0 \times b^{t/H} \) where: - \( A \) is the final amount. - \( A_0 \) is the initial amount. - \( b \) is the base of the exponential function. - \( t \) is the time elapsed. - \( H \) is a constant time period (like the half-life).
In radioactive decay, the base \( b \) is 0.5. Each increment in time by one half-life reduces the substance to half its current amount. This results in a continual halving pattern, accurately captured by exponential decay functions.

Logarithms

Logarithms are inherently tied to exponential functions and are used to find the power to which a number, known as the base, is raised to produce another number. They help solve equations involving exponentials by providing a mechanism to "undo" the exponentiation.

- **Definition**: If you know that \( b^x = a \), then \( \log_b(a) = x \). In simpler terms, if \( 2^3 = 8 \), then \( \log_2(8) = 3 \).- **Application in Decay**: When solving how long it takes for a radioactive substance to decrease to a certain amount, logarithms break down the equation where the exponential form \( A = A_0 \times b^{t/H} \) needs to be rearranged to solve for the time \( t \).
For instance, in the exercise, to find how long it will take for Uranium 228 to reduce from 8 grams to 3 grams, we rearrange and take logarithms of the equation. This makes it easy to isolate and solve for \( t \). Understanding logarithms empowers us to work backwards from known quantities to unknown time variables.