Mathematics: Circle the answer choice below that does not equal the following: - 48/16 1) 48/-16 2) -3 3) 3 4) -(48/16)

Circle the answer choice below that does not equal

Ответ:

matius0711

26.09.2021

Mathematics

116 years

Step-by-step explanation:

For this type of problems we have to use the formula that model the radioactive decay, that it's:

$N = N_{0} e^{-\lambda t}$ ,

where $N_{0}$ correspond to the initial amount of mass of the element, $\lambda$ correspond to the decay constant of the element, correspond the time measured in years and correspond to the amount of mass left after years.

In this case the original amount of mass is unknown but we don't need it to resolve the question, so let's make use of the half-life data that we've been given:

If the original amount of mass is $N_{0}$ then after = 29 years the amount of mass left is the half of $N_{0}$ .

$N = N_{0}\frac{1}{2} = N_{0}e^{-29\lambda }$ , we cancel equal terms on both sides and obtain that $e^{-29\lambda} = \frac{1}{2}$ , now we take natural logarithm on both sides (due to exponential function and logarithmic function are inverses) and then we obtain $\lambda$ by dividing for -29, so $\lambda = \frac{-log(\frac{1}{2})}{29}$

Now we know the formula that model the radioactive decay for Strontium-90:

$N = N_{0}e^{\frac{log(\frac{1}{2})}{29}t}$

So we've been asked for the time in years were the mass left will be the 1/16 of it's original mass,

By that time in years we have that $N = N_{0}\frac{1}{16} = N_{0}e^{\frac{log(\frac{1}{2})}{29}t}$ , we cancel equal terms and take natural logorithm on both sides $log(\frac{1}{16}) = \frac{log(\frac{1}{2})}{29}t}$ , finally we obtain after dividing that $t = \frac{29 log(\frac{1}{16})}{log(\frac{1}{2})} = 116$ years

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Circle the answer choice below that does not equal the following: - 48/16
1) 48/-16
2) -3
3) 3
4) -(48/16)

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