Question

a sample of \\(_{92}^{238}\text{u}\\) is decaying at a rate of 475 decays/s. the half - life is \\(4.468\times 10^{9}\\) yr. \\(\quad\\) part a\\(\quad\\) what is the mass of the sample?\\(\quad\\) express your answer to three significant figures and include the appropriate units.\\(\quad\\) \\(m = \\) \\(\text{value}\\) \\(\text{units}\\)\\(\quad\\) submit \\(\quad\\) request answer\\(\quad\\) provide feedback

Explanation:

Step1: Recall decay formula and constants

The decay rate \( R \) is related to the number of atoms \( N \) and the decay constant \( \lambda \) by \( R = \lambda N \). The decay constant \( \lambda \) is related to the half - life \( T_{1/2} \) by \( \lambda=\frac{\ln 2}{T_{1/2}} \). Avogadro's number \( N_A = 6.022\times 10^{23}\space mol^{-1}\), and the molar mass of \( ^{238}U \) is \( M = 238\space g/mol=0.238\space kg/mol \).

First, convert the half - life from years to seconds.
\( T_{1/2}=4.468\times 10^{9}\space yr \), and \( 1\space yr = 3.154\times 10^{7}\space s \). So \( T_{1/2}=4.468\times 10^{9}\times3.154\times 10^{7}\space s\approx1.409\times 10^{17}\space s \)

Step2: Calculate the decay constant \( \lambda \)

Using \( \lambda=\frac{\ln 2}{T_{1/2}} \), \( \ln 2\approx0.693 \), so \( \lambda=\frac{0.693}{1.409\times 10^{17}\space s}\approx4.919\times 10^{-18}\space s^{-1} \)

Step3: Find the number of atoms \( N \) from the decay rate

Given \( R = 475\space decays/s \), and \( R=\lambda N \), so \( N=\frac{R}{\lambda} \). Substituting the values of \( R \) and \( \lambda \), we get \( N=\frac{475\space decays/s}{4.919\times 10^{-18}\space s^{-1}}\approx9.656\times 10^{19}\space atoms \)

Step4: Calculate the mass \( m \) of the sample

The number of moles \( n=\frac{N}{N_A} \), and the mass \( m = n\times M=\frac{N\times M}{N_A} \)

Substitute \( N = 9.656\times 10^{19}\space atoms \), \( M = 0.238\space kg/mol \), and \( N_A = 6.022\times 10^{23}\space mol^{-1} \)

\( m=\frac{9.656\times 10^{19}\times0.238}{6.022\times 10^{23}}\space kg \)

First, calculate the numerator: \( 9.656\times 10^{19}\times0.238\approx2.30\times 10^{19} \)

Then, divide by \( 6.022\times 10^{23} \): \( m=\frac{2.30\times 10^{19}}{6.022\times 10^{23}}\space kg\approx3.82\times 10^{-5}\space kg = 38.2\space mg \) (or \( 3.82\times 10^{-2}\space g \))

Answer:

\( 38.2\space mg \) (or \( 3.82\times 10^{-2}\space g \) or \( 3.82\times 10^{-5}\space kg \))