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horsena [70]
3 years ago
14

A certain isotope decays so that the amount A remaining after t years is given by: A = A0 · e ^-0.03t where A0 is the original a

mount of the isotope. To the nearest year, the half-life of the isotope (the amount of time it takes to decay to half the original amount) is how many years?
Mathematics
2 answers:
Katena32 [7]3 years ago
7 0
<span>Ending Amt = Bgng Amt * e ^-0.03t
In this equation, the "-0.03" is the decay factor or "k"
We can now solve for half-life by this equation:
</span>t = <span>(<span>ln [y(t) ÷ a]<span>)<span> ÷ -k  (we can say beginning amount = 200 and ending amount = 100
</span></span></span></span>t = <span>(<span>ln [200 ÷ 100]<span>)<span> ÷ -k
</span></span></span></span>t = <span>(<span>ln [2]<span>)<span> ÷ -k
</span></span></span></span>t = 0.69314718056<span> ÷ --.03
t =</span><span><span><span> 23.1049060187 </span>
about 23 years
</span></span>
earnstyle [38]3 years ago
5 0

Answer:

The number of years is approximately 23 years.          

Step-by-step explanation:

Given : A certain isotope decays so that the amount A remaining after t years is given by : A = A_0\cdot e^{-0.03t} where A_0 is the original amount of the isotope.

To find : To the nearest year, the half-life of the isotope (the amount of time it takes to decay to half the original amount) is how many years?

Solution :

The decay model is given by A = A_0\cdot e^{-0.03t}

We have given that,  

The amount of time it takes to decay to half the original amount.

i.e. A=\frac{A_0}{2}

Substitute the values in the formula,

 A=A_0e^{-0.03t}

 \frac{A_0}{2}=A_0e^{-(0.03)t}

 \frac{1}{2}=e^{-(0.03)t}

Taking natural log both side,

 \ln\frac{1}{2}=\ln e^{-(0.03)t}

 -\ln2=-(0.03)t\ln e

 t=\frac{-\ln2}{-0.03}

 t=23.10

Therefore, The number of years is approximately 23 years.

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