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The unstable isotope \(^4$$^0\)K is used for dating rock samples. Its half-life is 1.28 \(\times\) 10\(^9\) y. (a) How many decays occur per second in a sample containing 1.63 \(\times\) 10\(^-$$^6\) g of \(^4$$^0\)K? (b) What is the activity of the sample in curies?

Short Answer

Expert verified
4.37 Bq and 1.18 × 10^-10 Ci

Step by step solution

01

Calculate the number of atoms

First, calculate the number of potassium-40 (\(^4$$^0\)K) atoms in the sample. Use the formula \( N = \frac{m}{M} \times N_A \), where \( m \) is the mass of the sample, \( M \) is the molar mass, and \( N_A \) is Avogadro's number. The molar mass of \(^4$$^0\)K is approximately 40 g/mol, and \( N_A = 6.022 \times 10^{23} \) atoms/mol. First convert the mass into grams: \( 1.63 \times 10^{-6} \) g.
02

Use the half-life to find decay constant

To find the decay constant \( \lambda \), use the formula \( \lambda = \frac{\ln(2)}{t_{1/2}} \) where \( t_{1/2} \) is the half-life of the isotope. Substituting the half-life of \(^4$$^0\)K, \( 1.28 \times 10^9 \) years, convert this time into seconds (1 year = 365.25 days = 24 hours/day = 3600 seconds/hour).
03

Determine activity in decays per second

The activity \( A \) is calculated using \( A = \lambda N \), where \( N \) is the number of atoms as found in Step 1, and \( \lambda \) is the decay constant from Step 2. Substitute these values to get decays per second (becquerels).
04

Convert activity to curies

Convert the activity from becquerels to curies, using the conversion factor \( 1 \text{ Ci} = 3.7 \times 10^{10} \text{ Bq} \). Divide the result from Step 3 by \( 3.7 \times 10^{10} \).

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Key Concepts

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

Half-life Calculations
The concept of half-life is fundamental in understanding radioactive decay. It refers to the time required for half the quantity of a radioactive substance to undergo decay.
To calculate the decay activities in a sample, we utilize the half-life to derive important decay-related constants, such as the decay constant \( \lambda \).
To find \( \lambda \), use the formula:
  • \( \lambda = \frac{\ln(2)}{t_{1/2}} \)
  • \( t_{1/2} \) is the half-life of the substance
For a substance like Potassium-40, given its half-life is \( 1.28 \times 10^9 \) years, we first convert this time period into seconds to maintain consistency in our calculations.This involves multiplying by the number of seconds in a year (\( 3.15576 \times 10^7 \) seconds/year). Calculating \( \lambda \) allows us to determine the rate at which a radioactive isotope will decay.
By knowing this rate, we can then compute other important quantities, such as the sample's activity, which is expressed in decays per second or bequerels.
Potassium-40 Dating
Potassium-40 dating is a method used to date geological materials. This technique exploits the radioactive decay of Potassium-40 (\(^4$$^0\)K) to Argon-40 (\(^4$$^0\)Ar), a process helpful for dating ancient rocks.
Potassium-40 has a substantial half-life of \( 1.28 \times 10^9 \) years, which makes it suitable for dating samples that are millions to billions of years old.
This method of dating relies on measuring the ratio of \(^4$$^0\)K to \(^4$$^0\)Ar in a sample, as argon is a gas and escapes from molten material but gets trapped after the rock solidifies.
Once the age is determined, this can provide key insights into the geological and meteorological history of the Earth or other celestial bodies.
Geologists measure the amount of both elements in the rock, correcting for any initial argon, to calculate the time since the rock solidified.
Understanding and applying this method requires careful measurements and some assumptions about the initial state, but it remains a cornerstone technique in geological timekeeping.
Decay Constant
The decay constant \( \lambda \) is a crucial value that indicates the rate of decay of a radioactive element.
It defines the probability per unit time that a nucleus will decay. The larger the value of \( \lambda \), the faster the decay process occurs.
The decay constant is connected to the half-life of the element through the relation \( \lambda = \frac{\ln(2)}{t_{1/2}} \), where \( \ln(2) \) is the natural logarithm of 2, approximately 0.693.
In practical terms, the decay constant helps calculate the activity of a radioactive sample.
For example, in the context of Potassium-40, once you have \( \lambda \), you can determine how many decays occur per second using \( A = \lambda N \), where \( A \) is the activity and \( N \) is the number of radioactive atoms.
The ability to calculate activity is invaluable in various applications, including radiometric dating, medical diagnostics, and understanding nuclear reactions.

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Most popular questions from this chapter

Which reaction produces \(^{131}\)Te in the nuclear reactor? (a) \(^{130}Te + n \rightarrow ^{131}Te\); (b) \(^{130}I + n \rightarrow ^{131}Te\); (c) \(^{132}Te + n \rightarrow ^{131}Te\); (d) \(^{132}I + n \rightarrow ^{131}Te\).

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