Question

Modeling Radioactive Decay In Exercise, complete the table for each radioactive isotope.

Amount Amount
after after
Half-life Initial 1000 10,000
Isotope (In years) quantity years years
239Pu 24,100 7.1 grams

Answer 1

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
`e^(kt)`

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
`e^(kt)`

0.5= 1
`e^(k(1599))`

0.5=
`e^(k(1599))`

ln 0.5= k(1599)

`(1)/(1599)` 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
`e^(kt)`

y= 8
`e^((-0.0004355*1599))`

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
`e^(kt)`

`y_(t=1000)`= 8
`e^((-0.0004355*1000))`

`y_(t=1000)`= 5.186 grams

t= 10000

y= C
`e^(kt)`

`y_(t=10000)`= 8
`e^((-0.0004355*10000))`

`y_(t=10000)`= 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
`e^(kt)`

1/2=
`e^(k5715)`

ln 1/2= k5715

`(1)/(5715)` 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
`e^(kt)`

`y_(t=1000)`= 5
`e^((-0.0001213*1000))`

`y_(t=1000)`= 4.429 grams

t= 10000

y= C
`e^(kt)`

`y_(t=10000)`= 5
`e^((-0.0001213*10000))`

`y_(t=10000)`= 1.487 grams

3)

This excersice is for the same element as 2)

¹⁴C (Carbon)

Half-life 5715

`y_(t=10000)`= 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

`y_(t=10000)`= C
`e^(kt)`

6= C
`e^((-0.0001213*10000))`

C=
`(6)/(e^(-0.0001213*10000))`

C= 20.18 grams

Now we can calculate the remaining mass at t=1000

`y_(t=1000)`= 20.18
`e^((-0.0001213*1000))`

`y_(t=1000)`= 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

`y_(t=1000)`= 0.7 gram

`y_(t=1000)`= C
`e^(kt)`

0.7= C
`e^((-0.0004335*1000))`

C=
`(0.7)/(e^(-0.0004335*1000))`

C= 1.0798 grams ≅ 1.08 grams

Now we can calculate the remaining mass at t=10000

`y_(t=10000)`= 1.08
`e^((-0.0001213*10000))`

`y_(t=10000)`= 0.32 gram

5)

The element is

²³⁹Pu (Plutonium)

Half-life 24100 years

Amount after 1000
`y_(t=1000)`= 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
`e^(kt)`

1/2=
`e^(k24100)`

ln 1/2= k*24100

k=
`(1)/(24100)` * ln 1/2

k= -0.00002876

Now we calculate the initial quantity using the given information

`y_(t=1000)`= C
`e^(kt)`

2.4= C
`e^(( -0.00002876*1000))`

C=
`(2.4)/(e^( -0.00002876*1000))`

C=2.47 grams

And the remaining mass at t= 10000 is:

`y_(t=10000)`= C
`e^(kt)`

`y_(t=10000)`= 2.47 *
`e^(( -0.00002876*10000))`

`y_(t=10000)`= 1.85 grams

6)

²³⁹Pu (Plutonium)

Half-life24100 years

Amount after 10000
`y_(t=10000)`= 7.1 grams

From 5) k= -0.00002876

The initial quantity is:

`y_(t=1000)`= C
`e^(kt)`

7.1= C
`e^(( -0.00002876*10000))`

C=
`(7.1)/(e^(-0.00002876*10000))`

C= 9.47 grams

And the remaining masss for t=1000 is:

`y_(t=1000)`= C
`e^(kt)`

`y_(t=1000)`= 9.47 *
`e^((-0.00002876*1000))`

`y_(t=1000)`= 9.20 grams

I hope it helps!