Question

The half-life of bromine-74 is 25min.How much of a 6.0 mg sample is active after 75min?

Answer 1

Hint:To solve the question, we should know about the half-life of an element and the equation relating the half-life of an element, time and concentration.
Half-life is calculated using equation,$Half-life=\dfrac{time}{no.\,of\,periods}$

Complete step-by-step answer:From the lower classes we are studying about the periodic table. And we have different elements into various blocks. And generally say that the elements in the f block are radioactive elements specially elements after atomic number 90. The actinides series has the radioactive elements. The radioactive elements are those which are very unstable and the atom tends to attain stability by emission of radiation by undergoing disintegration.
The radioactive elements have unstable neutrons, if the nucleus is unstable then the atom will collapse, so in order to attain stability they have decreased their energy. And these molecules give off atomic radiations, and they attain the configuration of a stable atom. Because of this emission of radiation, the atom gets converted into another atom which has stable configuration, or sometimes it may get converted to an unstable radioactive element which again undergoes disintegration till it attains a stable state.
Half-life of a radioactive element is the time required for the element to undergo disintegration and reduce its initial concentration to half of the concentration.
Here the half-life of the bromine-74 is 25minute. So we could first calculate the number of periods.
The equation is as follows: $Half-life=\dfrac{time}{no.\,of\,periods}$
$No.\,of\,period=\dfrac{Time\,given}{{{t}_{{1}/{2}\;}}}$, ${{t}_{{}^{1}/{}_{2}}}$ is the half-life period
$No.\,of\,period=\dfrac{75}{25}=3$
The number of periods is three and in each period half the amount of the initial concentration of the elements disintegrates.
So in the first period we could say, 6mg disintegrates to $\dfrac{6}{2}$
And then the $\dfrac{6}{2}$disintegrates to $\dfrac{6}{4}$ and then to $\dfrac{6}{8}$
$6mg\to \dfrac{6}{2}\to \dfrac{6}{4}\to \dfrac{6}{8}$
Hence the final value will be 0.75mg. After 75minutes only 0.75mg of the sample is active.

Note:To do the problems related to nuclear chemistry, radioactivity we should be very thorough with the concepts of chemicals kinetic like to find the rate of a reaction, its equation etc.
And if the rate constant is given then we could find the half-life period of an element as follows:
${{t}_{{}^{1}/{}_{2}}}=\dfrac{0.693}{k}$, where k is the rate constant.