Phosphorus-32 has a half life of 14 days. Starting with 8.00 g of 32-P, how many grams will remain after 70 days?
Answer
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Hint: We know that half life of a radioactive substance is the time that is needed by the isotope to react half of its initial mass. Here, half life phosphorus-32 is 14, that means, in 14 days the original mass of phosphorus-32 will be reduced to half.
Complete step by step answer:
Let’s discuss the half life in detail. The original mass of isotope will be halved for each passing of a half life.
After 1st half life= 50% of initial mass left
After 2nd half life= 25% of initial mass left
After 3rd half life= 12.5% of initial mass left
After 4th half life= 6.25% of initial mass left
After 5th half life= 3.125% of initial mass left
Here, in case of Phosphorus-32 first we have to calculate the number of half lives. The formula is,
Number of half lives=$\dfrac{{{\text{Total}}\,\,{\text{time}}}}
{{{\text{Time}}\,\;{\text{for}}\,\,{\text{one}}\,\,{\text{half}}\,\,{\text{life}}}}$
The total time is 70 days and time for one half life is 70. So, the number of half lives is,
Number of half lives=$\dfrac{{70}}
{{14}} = 5$
Now, we have to calculate the mass of P-32 left after 5 half lives.
We know, after the 5th half life 3.125% of initial mass left.
$ \Rightarrow $ Mass of P-32 left${\text{ = 3}}.{\text{125}}\% \,{\text{of 8}}.00{\text{ g}} = \dfrac{{3.125}}
{{100}} \times 8 = 0.25\,{\text{g}}$
Note: It is to be remembered that radioactivity is a phenomenon in which nuclei emit particles due to the unstable nucleus. There are three types of radioactivity namely alpha, beta and gamma. Henri Becquerel was the first scientist who discovered this phenomenon and after that Marie Curie named it as ‘radioactivity’.
Complete step by step answer:
Let’s discuss the half life in detail. The original mass of isotope will be halved for each passing of a half life.
After 1st half life= 50% of initial mass left
After 2nd half life= 25% of initial mass left
After 3rd half life= 12.5% of initial mass left
After 4th half life= 6.25% of initial mass left
After 5th half life= 3.125% of initial mass left
Here, in case of Phosphorus-32 first we have to calculate the number of half lives. The formula is,
Number of half lives=$\dfrac{{{\text{Total}}\,\,{\text{time}}}}
{{{\text{Time}}\,\;{\text{for}}\,\,{\text{one}}\,\,{\text{half}}\,\,{\text{life}}}}$
The total time is 70 days and time for one half life is 70. So, the number of half lives is,
Number of half lives=$\dfrac{{70}}
{{14}} = 5$
Now, we have to calculate the mass of P-32 left after 5 half lives.
We know, after the 5th half life 3.125% of initial mass left.
$ \Rightarrow $ Mass of P-32 left${\text{ = 3}}.{\text{125}}\% \,{\text{of 8}}.00{\text{ g}} = \dfrac{{3.125}}
{{100}} \times 8 = 0.25\,{\text{g}}$
Note: It is to be remembered that radioactivity is a phenomenon in which nuclei emit particles due to the unstable nucleus. There are three types of radioactivity namely alpha, beta and gamma. Henri Becquerel was the first scientist who discovered this phenomenon and after that Marie Curie named it as ‘radioactivity’.
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