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In the early 1960s, radioactive strontium-90 was released during atmospheric testing of nuclear weapons and got into the bones of people alive at the time. If the half-life of strontium-90 is 30 years, what fraction of the strontium-90 absorbed in 1963 remained in people's bones in 2004?
fraction =

Answer :

ikuemuyaolusegun

Answer: fraction remaining at 2004 = 2/5

Step-by-step explanation: The radioactive decay formula is

N = N¹ * e{- lamda * time}

But the fraction remaining after time t which is 2004 - 1963 = 41 years is

N/N¹ = e{- lamda * time}

lamda is the decay constant measured in per second and is equal to

Lamda = In2/{half-life}

So we first convert our half life which is 30 years in the question to seconds

In a day we have = 60*60*24= 86400 secs.

In a year we have = 365*86400= 31536000secs

In 30 years we have=30* 31536000

= 946080000secs

So our half life = 946080000secs

Recall that In2 = 0.693

Therefore

Lamda = 0.693/946080000

=7.325*EXP{-10} per second

So we have gotten our decay constant.

Now let's convert our time t that is 41 yrs to seconds

Following same procedure for converting our half-life, we have

41 years = 41 * 31536000

= 1292976000 seconds

= 1.3*EXP{9} seconds

Now we can now substitute in to our original fraction that is

N/N¹ = e{- lamda * time}

= e{- {7.325*EXP{-10} } *

1.3*EXP{9}}

= e{- 0.952}

N/N¹= fraction remaining at 2004

= exponential of - 0.952

=0.4 = 4/10= 2/5.

NOTE: EXP used above is another way of writing 10^.

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