10 half life problems and answers examples

Answer

1. A radioisotope decays from 150 mg to 120.2 mg in 5 days. Calculate the half-life of this isotope.

Solution

          From $N_t$ = ${\frac{1}{2}}^{\frac{t}{t_{1/2}}}$$N_o$                                                                                                 

we rearrange this equation to take the form

              $t_{1/2}$ = $\frac{t \log 2}{\log {\displaystyle \frac{N_o}{N_t}}}$

                     =$\frac{5\log 2}{\log {\displaystyle \frac{150}{120.2}}}$

                     = 15.6482 days

2. Iodine-131 has a half-life of 8 days. What is the mass of this nuclide that remaines when 40 g of it decays in 48 hours?

Solution

We use the relalation $N_t$ = ${\frac{1}{2}}^{\frac{t}{t_{1/2}}}$$N_o$

                                     = $0.5^{\frac{48}{8}}$X 40

                                     = 0.6250 g

3. A sample of Pd-100 decayed to a mass of 30 mg in 16 days. Given that the half-life of Pd is 4 days, calculate the initial mass of the sample

Solution

We are required to find $N_o$, when we have $N_{t }$ and t. rearranging the equation used in example 2, we obtain;

              $N_o$ = $2^{\frac{t}{t_{1/2 }}}$$N_t$

                   = $2^{\frac{16}{4}}$X $30$

                   = 480 g

4. Given that 50.0 g of C-14 decayed in 4 half-lives, how much of this isotope is left? C-14 Half-Life = 5730 Years.

Solution

We utilize the equation that relate amount remaining, initial mass and number of half-lives,n.

              $N_t$ = ${\frac{1}{2}}^n$X $N_o$

                   = ${\left(\frac{1}{2}\right)}^4$X 50

                   = 3.125 g

5. What is the half-life of an isotope that is 80 % remained after 16 days?

Solution

      % remaining= $\frac{80}{100}$

 Therefore $N_t$= 80, $N_o$ = 100, Now using the half-life equation in example 1, we have

                  $t_{1/2}$ = $\frac{16 \log 2}{\log {\displaystyle \frac{100}{80}}}$

                         = 49.7 days 

6. A certain radioisotope has a half-life of 9 days. What percentage of an initial mass of this isotope remains after 25 days?

This question requires that we calculate $N_t$, when we have t and $t_{1/2}$,  

                  $N_t$ = $0.5^{\frac{t}{t_{1/2}}}$ $N_o$,

             percentage = $\frac{N_t}{N_o} x 100$

                              = $0.5^{\frac{t}{t_{1/2}}}$ x 100

                              = 14.58 %

7.A radioactive source has a half-life of 80 s. How long will it take for 5/6 of the source to decay?

Solutions

Fraction remaining = $\frac{1}{6}$, in this case $N_t$= 1, $N_o$ = 6, we are required to determine t, when we have $t_{1/2}$, now from $N_t$= $N_o$ x $0.5^{\frac{t}{t_{1/2}}}$,we obtain the relation for calculating t as;

                        $t$ = $\frac{t_{1/2 }\log {\displaystyle \frac{N_o}{N_t}}}{\log 2}$

                          = $\frac{80 \log {\displaystyle \frac{1}{6}}}{\log 2}$

                          = 206.797 seconds

8. Achaelogist dated Holy shroud using radiocarbon (${}^{14}\mathrm{C}$), he found out that the amount of ${}^{14}\mathrm{C}$ found in Holy shroud is 92 % of that in the living tossue (half-life of carbon-14=5730 years)

a) Calculate the decay constant for ${}^{14}\mathrm{C}$ assuming that it decreases solely due to nuclear decay 

  Solution

         λ = $\frac{0.693}{t_{1/2}}$

          = 0.0001 $yea{r}^{-1}$

b) Calculate the age of Shroud of Turin

           $N_t$ = $N_o$ $e^{-\lambda t}$

             $t$ = $\frac{1}{\lambda }\ln \frac{N_o}{N_t}$

                = $\frac{5730}{0.693}$$\ln \frac{100}{72}$

                = 690 years

9. The activity of a radioactive source decreased from 6000  per minute to 500 counts per minute in 10 hours. Calculate the half-life of the radioactive source.

Solution

We use the equation $A_t$ = ${\left(\frac{1}{2}\right)}^{\frac{t}{t_{1/2}}}$ $A_o$, where $A_t$ is the activity in time t, $A_o$ is the original activity

                                   $500$= ${\left(\frac{1}{2}\right)}^{\frac{10}{t_{1/2}}}$$\times$ $6000$

                                   $t_{1/2}$ = $\frac{10 \log 2}{\log 12}$

                                          = $2.789 hours$

We can also use the relation $A_t$ = ${\left(\frac{1}{2}\right)}^n$$A_o$, where n is the number of half-lives

                                           $A_t$ = $\frac{A_o}{2^n}$

                                            $n$ = $\frac{t}{t_{1/2}}$= $\frac{l\mathrm{og} \frac{A_o}{A_t}}{\log 2}$ 

                                          $t_{1/2}$ = $\frac{ 10 \log {\displaystyle 2}}{\log {\displaystyle \frac{6000}{500}} }$

                                                  = 2.789 hours

10. Calculate the percentage of the original radioisotope that remained after three half-lives elapsed

  $\% remained$ =

                     = $0.5^3 \times 100$

                     = $12.5 \%$

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