Question

Bromine-82 has a half-life of 36 hours. A sample containing Br-82 was found to have an activity of

1.4 x 105 disintegrations/min. How many grams of Br-82 were present in the sample? Assume that

there were no other radioactive nuclides in the sample.

Answer 1

As we know,

Decay constant, $k =\dfrac{0.693}{t_{\frac{1}{2}}}=\dfrac{0.693}{36\text{hour}}=0.01925hr^{−1}$

$N_o=1.4\times 10^5\text{disintegration/min}=\dfrac{1.4}{60}\times 10^5{\text{disintegration/hour}}$

Also,

$k=\dfrac{2.303}{t}log\dfrac{N_0}{N_t},$

$log\dfrac{N_0}{N_t}=\dfrac{kt}{2.303}=\dfrac{0.01925\times 36}{2.303}=0.3009$

or $\dfrac{N_0}{N_t}=Antilog0.3009=1.999=2$

$N_o=N_t\times{2}=\dfrac{1.4\times 10^5\times 2}{60}=4.66\times10^3\text{disintegration/hour}$

Grams of Br-82 present in the sample is $N_t-No=(4.66-2.33)\times 10^3=2.33\times 10^3$

=$\dfrac{2.33}{4.66}\times 100=50$ %

Grams of $Br-82=100-\dfrac{50}{100}(82)=82-41=41g$