Unit 2: Lesson 7 Half Life Decay

Unit 2: Lesson 7 Nuclear Decay and Half Life

Multiple Choice

Solve for the missing daughter atom in the following equation:

¹⁴₆C --> ⁰_-1e + ________

• ¹⁴₅B

• ¹⁴₆C

• ¹⁴₇N

• ⁴₂He

Multiple Choice

Which kind of decay is being represented?

¹⁴₆C --> ⁰_-1e + ________

• Positron

• Gamma

• Beta

• Alpha

Multiple Choice

Solve this equation for alpha decay.

₈₈²²⁶Ra --> ___ + ₂⁴He

• ₈₆²²²Rn

• ₈₉²²⁶Ac

• ₈₄²²⁴Po

• ₈₈²²⁵Ra

Multiple Choice

Solve this equation for positron decay.

₅¹⁰ B--> ___ + +₁⁰e

• ₄¹⁰Be

• ₆¹⁰C

• ₇¹⁰N

• ₆¹³C

​The video slides like the one that follows can be very helpful. Be sure you are taking the time to view.

Multiple Choice

If the half life of a sample is 5 years how many half lifes have passed in 50 years?

• 10

• 250

• 5

• 50

Fill in the Blanks

I would be sure to add this formula to my notes.

Mf = mass final and Mi = Mass initial

The final mass must always be smaller than the initial.

Poll

Which method do you find to be easier?

• The logical method

• The mathematical method

​Use the embedded videos for extra support. The following video demonstrates how to solve these using the logical method.

​Use the embedded videos for extra support. The following video demonstrates how to solve these using the mathematical method. Notice they don't solve number of half lives first but include it in their formula.

You can use the logical method as long as you have a whole number of half lifes

​Solving for n

​When solving for the n or number of half lifes in our equation we can also use logic or math. In this class I will always keep the number of half lifes as a whole number when doing this so that the logical method will work. But it is a good idea to practice solving for n mathematically as well.

​Solving for n

After 24.0 days, 8.00 milligrams of an original 128.0 milligram sample remain. What is the half-life of the sample?

​Solving for n

After 24.0 days, 8.00 milligrams of an original 128.0 milligram sample remain. What is the half-life of the sample?

​

​M_(f) = M_(i) x 1/2ⁿ

​Solving for n

After 24.0 days, 8.00 milligrams of an original 128.0 milligram sample remain. What is the number of half-lifes the sample has been through?

​

​M_(f) = M_(i) x 1/2ⁿ

​8.00 = 128.0 x 1/2ⁿ

​Solving for n

After 24.0 days, 8.00 milligrams of an original 128.0 milligram sample remain. What is the number of half-lifes the sample has been through?

​

​M_(f) = M_(i) x 1/2ⁿ

​8.00 = 128.0 x 1/2ⁿ

8.00/128.0 = 0.0625 = 1/2ⁿ

​Solving for n

After 24.0 days, 8.00 milligrams of an original 128.0 milligram sample remain. What is the number of half-lifes the sample has been through?

​

​M_(f) = M_(i) x 1/2ⁿ

​8.00 = 128.0 x 1/2ⁿ

8.00/128.0 = 0.0625 = 1/2ⁿ

​log(0.0625) = log(1/2) n Multiplying by the log drops n from the exponent

​Solving for n

After 24.0 days, 8.00 milligrams of an original 128.0 milligram sample remain. What is the number of half-lifes the sample has been through?

​

​M_(f) = M_(i) x 1/2ⁿ

​8.00 = 128.0 x 1/2ⁿ

8.00/128.0 = 0.0625 = 1/2ⁿ

​log(0.0625) = log(1/2) n

​log(0.0625)/log(0.5) = n

​Solving for n

After 24.0 days, 8.00 milligrams of an original 128.0 milligram sample remain. What is the number of half-lifes the sample has been through?

​

​M_(f) = M_(i) x 1/2ⁿ

​8.00 = 128.0 x 1/2ⁿ

8.00/128.0 = 0.0625 = 1/2ⁿ

​log(0.0625) = log(1/2) n

​log(0.0625)/log(0.5) = n

​4 half lifes = n

​Solving for n

After 24.0 days, 8.00 milligrams of an original 128.0 milligram sample remain. What is the half-life of the sample?

​

​I could also do this logically

​128/ 2 = 64

​64/2 = 32

​32/2 = 16

​16/2 = 8

​

​I had to cut 128 in half 4 times to get to 8. This means 4 half lifes occured. This only works with whole number half lifes.

​Solving for n

After 24.0 days, 8.00 milligrams of an original 128.0 milligram sample remain. What is the length of each half life?​

Now that we know 4 half lifes have occured we can determine that each half life must be 6 days since 24 days have passed for this decay to occur.

​

​n = total time / 1 half life

​4 = 24/x

​

​x = 6 days