1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
lana66690 [7]
2 years ago
12

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

Mathematics
1 answer:
Julli [10]2 years ago
4 0

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 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)

\frac{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

\frac{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= \frac{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= \frac{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= \frac{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= \frac{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-life 24100 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= \frac{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!

You might be interested in
sarita use 54 ounces of apples to make an apple pie how many pounds and ounces of apples did sarita use
LUCKY_DIMON [66]
One pound is 16oz. So you take 54 and minus it by 16 until you get something less that 16. Then you take how many times you subtracted it, and that is how many pounds. Then you take the "leftover" number and they you have your answer. 
so Sarita used 3 pounds and 8 oz. 
3 0
3 years ago
Read 2 more answers
In the figure, r//s
lys-0071 [83]
M < 2 = m < 6...corresponding angles

m < 5 = m < 8....vertical angles are congruent
5 0
3 years ago
Read 2 more answers
Emma is 9 and one fourths Year’s old. How many months old is Emma?
krok68 [10]

She is 4 years an d3 months old

5 0
3 years ago
Read 2 more answers
Factorise 21ax+10by-35ay-6bx<br>with working pls​
wolverine [178]

Answer:

<h2>( 7a - 2b ) ( 3x - 5y )</h2>

<u>Step-by-step explanation:</u>

21ax + 10by - 35ay - 6bx

21ax - 35ay - 6bx + 10by

7a ( 3x - 5y ) - 2b ( 3x - 5y )

( 7a - 2b ) ( 3x - 5y )

3 0
2 years ago
What equation has infinite many solutions?
Tresset [83]
3 because
2n+3(n-4)+5=5n-7
2n+3n-12+5=5n-7
5n-7=5n-7
5n-5n=7-7
0
6 0
2 years ago
Other questions:
  • Simplify x to the 1/3 power (x to the 1/2 power + 2x to the 2 power )
    15·1 answer
  • The fraction 4/5 is equivalent to 0.8 and 80%. How do I write 0.8 and 80% as unreduced fractions?
    7·1 answer
  • I need help with 1 2 3
    12·1 answer
  • Determine the number of triangles ABC possible with the given parts.<br> b = 24, c = 29, B = 43°
    15·1 answer
  • What is the probability of picking a blue coin out of a bag that contains 5 blue and 10 red coins
    9·1 answer
  • PLEASE HELP :D ( these are multi step equations )
    7·1 answer
  • What fraction of all the 10-digit numbers with distinct digits have the property that the sum of every pair of neighboring digit
    14·1 answer
  • Trig how do you do I do not know
    15·1 answer
  • Riley invested $8,500 in an account paying an interest rate of 3.8% compounded daily. Assuming no deposits or withdrawals are ma
    10·1 answer
  • Solve the following pair of equations by elimination method: x+y=4, 2x+y=3​
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!