Respuesta :
Answer:
Step-by-step explanation:
Hello!
The complete table attached.
The following model allows you to predict the decade rate of a substance in a given period of time, i.e. the decomposition rate of a radioactive isotope is proportional to the initial amount of it given in a determined time:
y= C [tex]e^{kt}[/tex]
Where:
y represents the amount of substance remaining after a determined period of time (t)
C is the initial amount of substance
k is the decaing constant
t is the amount of time (years)
In order to know the decay rate of a given radioactive substance you need to know it's half-life. Rembember, tha half-life of a radioactive isotope is the time it takes to reduce its mass to half its size, for example if you were yo have 2gr of a radioactive isotope, its half-life will be the time it takes for those to grams to reduce to 1 gram.
1)
For the first element you have the the following information:
²²⁶Ra (Radium)
Half-life 1599 years
Initial quantity 8 grams
Since we don't have the constant of decay (k) I'm going to calculate it using a initial quantity of one gram. We know that after 1599 years the initial gram of Ra will be reduced to 0.5 grams, using this information and the model:
y= C [tex]e^{kt}[/tex]
0.5= 1 [tex]e^{k(1599)}[/tex]
0.5= [tex]e^{k(1599)}[/tex]
ln 0.5= k(1599)
[tex]\frac{1}{1599}[/tex] ln 0.05 = k
k= -0.0004335
If the initial amount is C= 8 grams then after t=1599 you will have 4 grams:
y= C [tex]e^{kt}[/tex]
y= 8 [tex]e^{(-0.0004355*1599)}[/tex]
y= 4 grams
Now that we have the value of k for Radium we can calculate the remaining amount at t=1000 and t= 10000
t=1000
y= C [tex]e^{kt}[/tex]
[tex]y_{t=1000}[/tex]= 8 [tex]e^{(-0.0004355*1000)}[/tex]
[tex]y_{t=1000}[/tex]= 5.186 grams
t= 10000
y= C [tex]e^{kt}[/tex]
[tex]y_{t=10000}[/tex]= 8 [tex]e^{(-0.0004355*10000)}[/tex]
[tex]y_{t=10000}[/tex]= 0.103 gram
As you can see after 1000 years more of the initial quantity is left but after 10000 it is almost gone.
2)
¹⁴C (Carbon)
Half-life 5715
Initial quantity 5 grams
As before, the constant k is unknown so the first step is to calculate it using the data of the hald life with C= 1 gram
y= C [tex]e^{kt}[/tex]
1/2= [tex]e^{k5715}[/tex]
ln 1/2= k5715
[tex]\frac{1}{5715}[/tex] ln1/2= k
k= -0.0001213
Now we can calculate the remaining mass of carbon after t= 1000 and t= 10000
t=1000
y= C [tex]e^{kt}[/tex]
[tex]y_{t=1000}[/tex]= 5 [tex]e^{(-0.0001213*1000)}[/tex]
[tex]y_{t=1000}[/tex]= 4.429 grams
t= 10000
y= C [tex]e^{kt}[/tex]
[tex]y_{t=10000}[/tex]= 5 [tex]e^{(-0.0001213*10000)}[/tex]
[tex]y_{t=10000}[/tex]= 1.487 grams
3)
This excersice is for the same element as 2)
¹⁴C (Carbon)
Half-life 5715
[tex]y_{t=10000}[/tex]= 6 grams
But instead of the initial quantity, we have the data of the remaining mass after t= 10000 years. Since the half-life for this isotope is the same as before, we already know the value of the constant and can calculate the initial quantity C
[tex]y_{t=10000}[/tex]= C [tex]e^{kt}[/tex]
6= C [tex]e^{(-0.0001213*10000)}[/tex]
C= [tex]\frac{6}{e^(-0.0001213*10000)}[/tex]
C= 20.18 grams
Now we can calculate the remaining mass at t=1000
[tex]y_{t=1000}[/tex]= 20.18 [tex]e^{(-0.0001213*1000)}[/tex]
[tex]y_{t=1000}[/tex]= 17.87 grams
4)
For this exercise we have the same element as in 1) so we already know the value of k and can calculate the initial quantity and the remaining mass at t= 10000
²²⁶Ra (Radium)
Half-life 1599 years
From 1) k= -0.0004335
[tex]y_{t=1000}[/tex]= 0.7 gram
[tex]y_{t=1000}[/tex]= C [tex]e^{kt}[/tex]
0.7= C [tex]e^{(-0.0004335*1000)}[/tex]
C= [tex]\frac{0.7}{e^(-0.0004335*1000)}[/tex]
C= 1.0798 grams ≅ 1.08 grams
Now we can calculate the remaining mass at t=10000
[tex]y_{t=10000}[/tex]= 1.08 [tex]e^{(-0.0001213*10000)}[/tex]
[tex]y_{t=10000}[/tex]= 0.32 gram
5)
The element is
²³⁹Pu (Plutonium)
Half-life 24100 years
Amount after 1000 [tex]y_{t=1000}[/tex]= 2.4 grams
First step is to find out the decay constant (k) for ²³⁹Pu, as before I'll use an initial quantity of C= 1 gram and the half life of the element:
y= C [tex]e^{kt}[/tex]
1/2= [tex]e^{k24100}[/tex]
ln 1/2= k*24100
k= [tex]\frac{1}{24100}[/tex] * ln 1/2
k= -0.00002876
Now we calculate the initial quantity using the given information
[tex]y_{t=1000}[/tex]= C [tex]e^{kt}[/tex]
2.4= C [tex]e^{( -0.00002876*1000)}[/tex]
C= [tex]\frac{2.4}{e^( -0.00002876*1000)}[/tex]
C=2.47 grams
And the remaining mass at t= 10000 is:
[tex]y_{t=10000}[/tex]= C [tex]e^{kt}[/tex]
[tex]y_{t=10000}[/tex]= 2.47 * [tex]e^{( -0.00002876*10000)}[/tex]
[tex]y_{t=10000}[/tex]= 1.85 grams
6)
²³⁹Pu (Plutonium)
Half-life 24100 years
Amount after 10000 [tex]y_{t=10000}[/tex]= 7.1 grams
From 5) k= -0.00002876
The initial quantity is:
[tex]y_{t=1000}[/tex]= C [tex]e^{kt}[/tex]
7.1= C [tex]e^{( -0.00002876*10000)}[/tex]
C= [tex]\frac{7.1}{e^( -0.00002876*10000)}[/tex]
C= 9.47 grams
And the remaining masss for t=1000 is:
[tex]y_{t=1000}[/tex]= C [tex]e^{kt}[/tex]
[tex]y_{t=1000}[/tex]= 9.47 * [tex]e^{( -0.00002876*1000)}[/tex]
[tex]y_{t=1000}[/tex]= 9.20 grams
I hope it helps!
