Question

Use the following half-life graph to answer the following question:

Answer 1

Answer:

The correct answer is 12.5.

Step-by-step explanation:

When looking at the x axis, you locate the 2. After 2 minutes, the point connects at (2,12.5). Therefore, the correct answer is 12.5.

Answer 2

Answer:

At t = 2 minutes, remaining quantity of the radioactive element is 12.5 mg.

Step-by-step explanation:

To get the answer of this question we will solve this further with the help of the equation $A_{t}=A_{0}e^{-kt}$

where k = decay constant

t = time for decay

$A_{0}$ = Initial quantity taken

From the graph attached we can say that 50 mg of a radioactive element remained half in 1 minute.

So the equation becomes

$50=25e^{-k(1)}$

Now we take natural log on both the sides of the equation

ln50 = ln[25.$e^{-k}$

3.912 = ln25 + $ln(e^{-k})$

3.912 = 3.219 + (-k)lne

3.912 - 3.219 = -k [since lne = 1]

0.693 = -k

k = -0.693

Now we will calculate the remaining quantity of the element after 2 minutes

$A_{t}=50.e^{-(0.693)(2)}$

              = $50.e^{-1.386}$

              = $\frac{50}{e^{1.386}}$

              = $\frac{50}{3.9988}$

              = 12.50 mg

Now we confirm this value from the graph.

At t = 2 minutes, remaining quantity of the radioactive element is 12.5 mg.