Chapter 4, Problem 9T

(a)

To determine

To find: The model the amount of Krypton-91 after tseconds .

Answer to Problem 9T

  $A\left(t\right)=3{\left(0.5\right)}^{\frac{t}{10}}$

Explanation of Solution

Given: The half-life of krypton-91 $\left(\text{K}\text{r}\right)$ is 10 seconds. At time $t=0$ a heavy canister contains 3 g of this radioactive gas.

Initial amount of krypton is 3 g.

Half-life of krypton is 10 seconds.

It means, in 10 second amount of krypton left 1.5 g.

Let the radioactive decay model, $A\left(t\right)=A_0{\left(0.5\right)}^{\frac{t}{t_{1/2}}}$

Put, $A_0=3,t_{1/2}=10$

  $A\left(t\right)=3{\left(0.5\right)}^{\frac{t}{10}}$

Hence, the model of half-life of krypton is $A\left(t\right)=3{\left(0.5\right)}^{\frac{t}{10}}$

(b)

To determine

To find: The amount of krypton left after 1 minute.

Answer to Problem 9T

46 mg

Explanation of Solution

Given: The model is $A\left(t\right)=3{\left(0.5\right)}^{\frac{t}{10}}$

Put t=60 into model

  $\begin{array}{c}A\left(60\right)=3{\left(0.5\right)}^{\frac{60}{10}}\\ =0.0469\end{array}$

Hence, amount of krypton left after 1 minute is 46 mg.

(c)

To determine

To find: The time when amount of krypton reduce to $1\mu g$

Answer to Problem 9T

3 minutes and 35 seconds

Explanation of Solution

Given: The model is $A\left(t\right)=3{\left(0.5\right)}^{\frac{t}{10}}$

Put $A\left(t\right)={10}^{-6}$

  $\begin{array}{c}3{\left(0.5\right)}^{\frac{t}{10}}={10}^{-6}\\ {\left(0.5\right)}^{\frac{t}{10}}=3.33\times {10}^{-7}\\ \frac{t}{10}\ln 0.5=\ln \left(3.33\times {10}^{-7}\right)\\ t=\frac{10\ln \left(3.33\times {10}^{-7}\right)}{\ln \left(0.5\right)}\\ t\approx 215\end{array}$

Hence, after 3 minutes and 35 seconds