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Assuming the age of the earth to be \(10^{10}\) years, what fraction of the original amount of \(_{92} \mathrm{U}^{238}\) is still in existence on earth \(\left(\mathrm{t}_{1 / 2}\right.\) of \(_{92} \mathrm{U}^{238}=4.51 \times 10^{9}\) years \() ?\) (a) \(10 \%\) (b) \(20 \%\) (c) \(30 \%\) (d) \(40 \%\)

Short Answer

Expert verified
The fraction still existing is about 20% (option b).

Step by step solution

01

Identify the Known Values

We are given that the age of the Earth is \(10^{10}\) years and the half-life of \(_{92} \mathrm{U}^{238}\) is \(4.51 \times 10^{9}\) years.
02

Use the Half-Life Formula

The formula to calculate the remaining amount of a substance based on its half-life is \( N = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}} \), where \( N \) is the remaining amount, \( N_0 \) is the initial amount, \( t \) is the time elapsed, and \( t_{1/2} \) is the half-life.
03

Substitute the Values

Substitute \( t = 10^{10} \) years and \( t_{1/2} = 4.51 \times 10^9 \) years into the formula. Calculate the exponent first: \( \frac{10^{10}}{4.51 \times 10^9} \approx 2.2173 \).
04

Calculate the Remaining Fraction

Use the exponent to find the remaining fraction: \( N = N_0 \times \left(\frac{1}{2}\right)^{2.2173} \). Calculate this to find \( \left(\frac{1}{2}\right)^{2.2173} \approx 0.204 \).
05

Select the Closest Answer

The remaining fraction of \(_{92} \mathrm{U}^{238}\) is approximately 20% of the original amount. Therefore, the correct answer is (b) 20%.

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Key Concepts

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

Half-life Calculation
Understanding the concept of half-life is essential in solving problems related to radioactive decay. The half-life of a radioactive substance is the amount of time it takes for half of the original amount of the substance to decay. This is a constant rate, unique to each radioactive isotope. It means that after each half-life period, the remaining quantity of the substance will always be half of what it was at the start of that period.

To calculate the remaining amount of a substance using its half-life, you use the formula:
  • \[ N = N_0 \times \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \]
Here:
  • \( N \) is the remaining quantity.
  • \( N_0 \) is the initial quantity.
  • \( t \) is the time that has elapsed.
  • \( t_{1/2} \) is the substance's half-life.
By substituting the known values into this formula, we can determine how much of the substance remains after a certain period of time. It illustrates exponential decay, where decay happens rapidly at first and gradually slows down over time.
Uranium-238 Decay
Uranium-238 is a radioactive isotope with a very long half-life of 4.51 billion years. This means it takes this long for just half of any given amount of Uranium-238 to decay. Its decay progress is crucial not just in scientific calculations like dating the age of Earth, but also in understanding how uranium is slowly converted into other elements, namely lead-206, through a series of other elements in between.

Here's why the decay of Uranium-238 is important:
  • It has been used to date the Earth, allowing scientists to estimate that our planet is approximately 4.5 billion years old.
  • The process of Uranium-238 decaying over time provides essential data for geological and archaeological studies.
This decay series contributes significantly to the natural radioactivity found on Earth today, and while the rate at which Uranium-238 decays is very slow due to its long half-life, it remains a significant factor in many scientific analyses concerning earth sciences.
Exponential Decay Formula
Radioactive decay follows an exponential decay pattern, meaning the rate of decay is proportional to the amount present at any given time. The formula to express this is:
  • \[ N = N_0 \times e^{-\lambda t} \]
In this formula:
  • \( N \) is the remaining quantity.
  • \( N_0 \) is the initial quantity.
  • \( \lambda \) is the decay constant, which is related to the half-life.
  • \( t \) is the elapsed time.
The decay constant \( \lambda \) can be derived from the half-life with the relation \( \lambda = \frac{\ln(2)}{t_{1/2}} \). This formulation explains how radioactive decay happens in a smooth, continuous process over time. The exponential decay formula is foundational in fields like physics and chemistry, helping scientists accurately predict the behavior of radioactive substances over time.

Using this formula gives a clear understanding of how quickly or slowly a substance decays, providing critical information for a variety of scientific and practical applications.

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Most popular questions from this chapter

One of the hazards of nuclear explosion is the generation of \(\mathrm{Sr}^{90}\) and its subsequent incorporation in bones. This nuclide has a half life of \(28.1\) years. Suppose one microgram was absorbed by a new born child, how much \(\mathrm{Sr}^{90}\) will remain in his bones after 20 years? (a) \(61 \mu \mathrm{g}\) (b) \(61 \mathrm{~g}\) (c) \(0.61 \mu \mathrm{g}\) (d) none

Which of the following statements is/are incorrect, when the nuclide, \({ }_{19} \mathrm{~K}^{40}\), is subject to \(\mathrm{K}\)-electron capture? (a) \({ }^{40} \mathrm{Ar}\) is produced with emission of \(\beta\)-particle. (b) \({ }^{40} \mathrm{Ar}\) is produced with emission of \(\alpha\)-particles. (c) \({ }^{40} \mathrm{Ar}\) is produced without emission of radiation (d) \({ }^{40} \mathrm{Ar}\) is produced with emission of X-radiation.

The nucleus resulting from \({ }_{92} \mathrm{U}^{238}\) after successive emission of two \(\alpha\) and four \(\beta\) particle is (a) \({ }_{90} \mathrm{Th}^{230}\) (b) \({ }_{92} \mathrm{U}^{230}\) (c) \({ }_{88} \mathrm{Ra}^{230}\) (d) \({ }_{94} \mathrm{Pu}^{230}\)

If a substance with half-life of 3 days is taken to another place in 12 days. What is the amount of substance left now? (a) \(1 / 8\) (b) \(1 / 32\) (c) \(1 / 4\) (d) \(1 / 16\)

When a radioactive element emits successively one \(\alpha\). particle and two \(\beta\) particles, the mass number of the daughter element (a) is reduced by 4 units (b) remains the same (c) is reduced by 2 units (d) is increased by 2 units

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