Unit 2: Lesson 7 Half Life Decay
Presentation
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Chemistry
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10th - 12th Grade
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Practice Problem
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Easy
Standards-aligned
Ryan McCluskey
Used 53+ times
FREE Resource
50 Slides • 7 Questions
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Unit 2: Lesson 7 Nuclear Decay and Half Life
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Multiple Choice
Solve for the missing daughter atom in the following equation:
146C --> 0-1e + ________
145B
146C
147N
42He
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Multiple Choice
Which kind of decay is being represented?
146C --> 0-1e + ________
Positron
Gamma
Beta
Alpha
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Multiple Choice
Solve this equation for alpha decay.
88226Ra --> ___ + 24He
86222Rn
89226Ac
84224Po
88225Ra
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Multiple Choice
Solve this equation for positron decay.
510 B--> ___ + +10e
410Be
610C
710N
613C
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The video slides like the one that follows can be very helpful. Be sure you are taking the time to view.
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Multiple Choice
If the half life of a sample is 5 years how many half lifes have passed in 50 years?
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250
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50
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Fill in the Blanks
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I would be sure to add this formula to my notes.
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Mf = mass final and Mi = Mass initial
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The final mass must always be smaller than the initial.
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Poll
Which method do you find to be easier?
The logical method
The mathematical method
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Use the embedded videos for extra support. The following video demonstrates how to solve these using the logical method.
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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.
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You can use the logical method as long as you have a whole number of half lifes
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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.
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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?
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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
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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
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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
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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
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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
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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
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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.
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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
Unit 2: Lesson 7 Nuclear Decay and Half Life
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