Research and prepare a report on a chosen element from the periodic table of elements having a half-life greater than 1 year.

The Periodic Table of Elements arranges all chemicals in an informative array. This listing of elements make up all of the materials and compounds that are used in daily life. Each element in the Periodic Table has a unique history, properties, and uses. One unique aspect of each element is its half-life. Defined as the amount of time for the quantity of an element to reduce by ½, the half life is an important feature of all elements.

In the core project for MATH 1314, research and prepare a report on a chosen element from the periodic table of elements having a half-life greater than 1 year. Prepare a 300 word (minimum) report including where your element is found in nature as well as how your element is utililized in your chosen field of study. Include citations.

I (Dr. Alpert) will be using this purple type for example …. Please delete my example and insert your element (unless you want to use my element as yours – that is OK with me). Your response should be in black print, please.
The citations will be the link where you obtained your information.
The element I am using for my example is Titanium-44 that has a half-life of 63 years. (socratic.org)

Using the rules of Logarithms, calculate the k value in the half-life of the element using the function: N(t)=N_0 e^(-kt) or the formula: N_0 e^(-kt)= N_0/2.

If I begin with 100grams of Titanium-44, in 63 years there will be just 50 grams. I will use this formula:
Amount at time “t” = Original amount (1/2)(t/half-life).
A(t) = 100 (1/2)(t/63)
Using this half-life concept, I can easily generate at least 10 points in the form (time, amount)
(0, 100); (63, 50); (126, 25); (189, 12.5); (252, 6.25) (315, 3.125); (378, 1.5625); (441, 0.78125);
(504, .390625); (567, 0.1953125); (630, 0.09765625)
Notice the X-values are multiples of 63 years and the y-values are the amount of Titanium-44 left after X years.
This is a very nice form of the formula. The original amount of 100 grams is obvious, and the half-life of 63 years is obvious in the exponent.
Amount at time “t” = Original amount (1/2)(t/half-life).
A(t) = 100 (1/2)(t/63)

However, this project asked me to put the formula in a format using the number e, and to show my computation. So here it is:
100 e-k*63 = 100/2 (definition of half-life)
100 e-k*63 = 50 (simplify 100/2)
e-k*63 = 0.5 (divide each side of the equation by the original amount of 100)
ln e-k*63 = ln (0.5) (take the natural logarithm of each side)
(-k*63) lne = ln(0.5) (law of logs, page 495 #3)
(-k*63) = ln(0.5) (ln e is just 1 because the base of the log and the argument undo each other)
-k = ln (0.5) / 63 (divide each side by 63 in order to isolate the variable)
K=-0.0110023362
Therefore, the function in the requested format is:
A(t) = 100 e(-0.0110023362 t)

Putting this formula in Y1, and setting the table of my calculator to start at 0 and change by 63, the same ten points are generated:
(0, 100); (63, 50); (126, 25); (189, 12.5); (252, 6.25) (315, 3.125); (378, 1.5625); (441, 0.78125);
(504, .390625); (567, 0.1953125); (630, 0.09765625)
Notice the X-values are multiples of 63 years and the y-values are the amount of Titanium-44 left after X years.

Using the function derived in question 2, calculate and graph 10 points based on the half-life of your element. Be sure to appropriately label each of the axes of your graph, and appropriately label your graph. In paragraph form (200 word minimum), address the following:
What trend do you see?
Give at least two observations from your graph (possibilities include the Y-intercept and its meaning, is the graph always increasing? Is the graph always decreasing? Is there a change in direction?)
Your graph will have a Y-Intercept as a point. Use critical thinking strategies to determine if this is applicable in a real-world scenario.
Critical Thinking: Discuss any negative OR positive impacts of your element. Based on this information, should the usage of this element be altered in any way.