Question

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 amount 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?

Answer 1

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:
t = (ln [y(t) ÷ a]) ÷ -k  (we can say beginning amount = 200 and ending amount = 100
t = (ln [200 ÷ 100]) ÷ -k
t = (ln [2]) ÷ -k
t = 0.69314718056 ÷ --.03
t = 23.1049060187
about 23 years

Answer 2

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.