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.
part a
what is the mass of the sample?
express your answer to three significant figures and include the appropriate units.
$m = $
value
units

Explanation:

Step1: Recall decay formula

The decay rate \( R = \lambda N \), where \( \lambda = \frac{\ln 2}{T_{1/2}} \) and \( N = \frac{m}{M}N_A \) ( \( M \) is molar mass, \( N_A \) Avogadro's number). First, convert half - life to seconds. \( T_{1/2}=4.468\times 10^{9}\ yr\times365\ days/yr\times24\ h/day\times3600\ s/h \). Calculate \( T_{1/2} = 4.468\times 10^{9}\times3.154\times 10^{7}\ s\approx1.409\times 10^{17}\ s \). Then \( \lambda=\frac{\ln 2}{T_{1/2}}=\frac{0.693}{1.409\times 10^{17}\ s}\approx4.92\times 10^{-18}\ s^{-1} \).

Step2: Relate decay rate to mass

From \( R=\lambda N=\lambda\frac{m}{M}N_A \), solve for \( m \): \( m=\frac{R M}{\lambda N_A} \). Molar mass of \( ^{238}U \) \( M = 238\ g/mol \), \( R = 475\ decays/s \), \( N_A = 6.022\times 10^{23}\ mol^{-1} \). Substitute values: \( m=\frac{475\ s^{-1}\times238\ g/mol}{4.92\times 10^{-18}\ s^{-1}\times6.022\times 10^{23}\ mol^{-1}} \). First, calculate numerator: \( 475\times238 = 475\times(200 + 38)=95000+17950 = 112950 \). Denominator: \( 4.92\times6.022\times 10^{5}\approx29.63\times 10^{5}=2.963\times 10^{6} \). Then \( m=\frac{112950}{2.963\times 10^{6}}\ g\approx0.0381\ g = 38.1\ mg \) (or more accurately, let's do precise calculation: \( m=\frac{475\times238}{4.92\times 10^{-18}\times6.022\times 10^{23}}=\frac{475\times238}{4.92\times6.022\times 10^{5}} \). \( 4.92\times6.022 = 29.62824 \), \( 475\times238 = 112950 \), \( 112950\div29.62824\approx3812 \), then divide by \( 10^{5} \): \( 3812\div100000 = 0.03812\ g\approx3.81\times 10^{-2}\ g \) or \( 38.1\ mg \)).

Answer:

\( m = \boldsymbol{3.81\times 10^{-2}\ g} \) (or \( 38.1\ mg \))