Question

Bromine-82 has a half-life of 36 hours. A sample containing Br-82 was found to have an activity of 1.4x10^5 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

Q163849

Bromine-82 has a half-life of 36 hours. A sample containing Br-82 was found to have an activity of 1.4x10^5 disintegrations/min. How many grams of Br-82 were present in the sample? Assume that there were no other radioactive nuclides in the sample.

Solution:

The rate of radioactive disintegration is directly proportional to the number of radioactive elements present at that time.

$A ∝ N.$

where A is the number of disintegration per unit time

N is the number of radioactive nuclei present at the given moment.

$A = λ N; \space \space \space \space \space λ \space is \space the \space decay \space constant.$ ..............Equation 1

From the given half-life we can find the decay constant by using the formula

$t_{1/2} = \frac{ln2}{ λ } = \frac{0.693}{ λ}$

which can be rearranged as

$λ = \frac{0.693}{t_{1/2}}$

Step 1: To find the decay constant in units of min^-1.

$t_{1/2} = 36\space hours * \frac{60\space minutes }{1\space hour } = 2160 \space minutes$

So the decay constant will be equal to

$λ = \frac{0.693}{2160\space minutes } = 0.0003208\space min^{-1} = 3.208 * 10^{-4} min^{-1}$

Step 2: To find the number of nuclei present at that time.

In question, we are given

Rate of disintegration, A = 1.4x10⁵ disintegrations/min.

and we have calculated, λ = 3.208 * 10^-4 min^-1 .

plug this in equation 1 we have

$A = λ N;$

1.4x10⁵ disintegrations/min = 3.208 * 10^-4 min^-1 * N

divide both the side by 3.208 * 10^-4 min^-1 we have.

$\frac{1.4*10^{5} disintegrations/min}{3.208 * 10^{-4} min^{-1}} = \frac{ 3.208 * 10^{-4} min^{-1} * N }{3.208 * 10^{-4} min^{-1} }$

$N = \frac{1.4*10^{5} disintegrations/min}{3.208 * 10^{-4} min^{-1}} = 4.364 * 10^{8} atoms.$

So there are 4.364 * 10⁸ atoms of bromine-82 present at that moment in the given sample.

Step 3: Convert the number of atoms of Bromine to grams.

We will need the atomic mass of Bromine-82. We are not provided the atomic mass in the given problem.

You can assume that atomic mass equal to 82 g/mole.

I have taken the atomic mass of Br-82 from a source and it is equal to 81.9168g/mol.

$moles \space of \space Br(82) = 4.364 * 10^{8} Br(82) \space atoms * \frac{1mol }{6.022*10^{23} \space atoms \space of \space Br(82)}$

= 7.247 * 10^-16 mole of Br.

$grams \space of \space Br(82) = 7.247 * 10^{-16} mole\space of\space Br * \frac{81.9168\space grams \space of Br }{1\space mol\space Br}$

= 5.934 * 10^-14 grams of Bromine-82.

in 2 significant the answer will be 5.9 * 10^-14 grams of Bromine-82

So there were 5.9 *10^-14 grams of bromine-82 in the given sample.

Please let me know if you have not understood any part of the problem.

Thank you. :)