Chapter 4, Problem 97RE

(a)

To determine

To Find: The remaining amount from a $10mg$ sample after thousand years.

Answer to Problem 97RE

  $9.97mg$

Explanation of Solution

Given: Uranium - $234$ has a half-life of $2.7\times {10}^5\text{ years}$ .

Formula used: $m(t)=m_0{2}^{\frac{-t}{h}}$

Given: Sample is $10mg$ so $m_0=10$

  $h=2.7\times {10}^5\text{ years}$

Then

  $\begin{array}{l}m(t)=m_0{2}^{\frac{-t}{h}}\\ m(t)=10\cdot 2^{-\frac{t}{2.7\times {10}^5}}\end{array}$

For finding the amount remaining for thousand years

  $t=1000={10}^3$

  $\begin{array}{l}m({10}^3)=(10)2^{\frac{-{10}^3}{2.7\times {10}^5}}\\ m({10}^3)=(10)2^{\frac{-1}{270}}\\ m({10}^3)=\mathrm{9.97432..}\\ m({10}^3)\approx 9.97mg\end{array}$

Hence, remaining amount from a $10mg$ sample after thousand years is $9.97mg$

(b)

To determine

To Find: The Time taken to decompose until its mass is $7mg$

Answer to Problem 97RE

  $1.39\times {10}^5\text{ years}$

Explanation of Solution

Given: Uranium - $234$ has a half-life of $2.7\times {10}^5\text{ years}$ .

Formula used: $m(t)=m_0{2}^{\frac{-t}{h}}$

Given: Sample is $10mg$ so $m_0=10$

  $\begin{array}{l}h=2.7\times {10}^5\text{ years}\\ m(t)=7\end{array}$

Then

  $\begin{array}{l}m(t)=m_0{2}^{\frac{-t}{h}}\\ 7=10\cdot 2^{-\frac{t}{2.7\times {10}^5}}\\ 2^{-\frac{t}{2.7\times {10}^5}}=\frac{7}{10}\\ \ln 2^{-\frac{t}{2.7\times {10}^5}}=\ln \frac{7}{10}\\ -\frac{t}{2.7\times {10}^5}\cdot \ln 2=\ln \frac{7}{10}\\ t=-\frac{2.7\times {10}^5\cdot \ln \frac{7}{10}}{\ln 2}\\ t=138934.756\\ t\approx 1.39\times {10}^5\text{ years}\end{array}$

Hence, Time taken to decompose until its mass is $7mg$ is $1.39\times {10}^5\text{ years}$