The isotope has a half-life of 5,715 years.After 20,000 years,a sample of the isotope is reduced 0.7 grams.What was the initial size of the sample (in grams)? How large was the sample after the first 2,000 years? Round your answers to four decimal places.
A)7.9172 g,6.2119 g
B)10.2924 g,8.0755 g
C)4.7503 g,3.7271 g
D)6.3338 g,4.9695 g
E)3.9586 g,3.1059 g

Question

The isotope  has a half-life of 5,715 years.After 20,000 years,a sample of the isotope is reduced 0.7 grams.What was the initial size of the sample (in grams)? How large was the sample after the first 2,000 years? Round your answers to four decimal places. ​
A)7.9172 g,6.2119 g 
B)10.2924 g,8.0755 g 
C)4.7503 g,3.7271 g 
D)6.3338 g,4.9695 g 
E)3.9586 g,3.1059 g

Quizplus · Accepted Answer

We can use the formula for exponential decay: $A=A_0e^{-kt}$, where $A$ is the final amount, $A_0$ is the initial amount, $k$ is the decay constant (which depends on the half-life), and $t$ is the time elapsed. We can solve for $A_0$ and $A$ using the given information.

After 20,000 years, the final amount is 0.7 grams, so we have:

$0.7=A_0e^{-k(20000)}$

After 5,715 years (which is one half-life), the amount is reduced by half, so we can use this to find $k$:

$\frac{1}{2}=e^{-k(5715)}$

Taking the natural logarithm of both sides, we get:

$-\ln 2 = -k(5715)$

$k = \frac{\ln 2}{5715}$

Substituting this into the first equation and solving for $A_0$, we get:

$A_0 = \frac{0.7}{e^{-k(20000)}} = \frac{0.7}{e^{-\frac{\ln 2}{5715}(20000)}} \approx 7.9172 	ext{ grams}$

To find the size of the sample after 2,000 years, we can use the same formula with $t=2000$:

$A = A_0e^{-kt} = 7.9172e^{-\frac{\ln 2}{5715}(2000)} \approx 6.2119 	ext{ grams}$

Therefore, the answer is A.

The Isotope Has a Half-Life of 5,715 Years