Half-Life Simulation Lab Count out about 200 M&M®s and place them up in the side
ID: 1021811 • Letter: H
Question
Half-Life Simulation Lab
Count out about 200 M&M®s and place them up in the side that has "M" in the tray. The tray represents a rock sample and the M&M®s that faces the M side represents the entire radioactivity the sample contains initially.
Shake the tray to flip over some of the M&M®s. One shake represents the passage of 5,000 years.
Remove all of the M&M®s that flipped over and count them. These represent the radioactivity that underwent nuclear decay and transmutated into a different element.
Repeat Steps 2 and 3 until all of the M&M®s are gone
1- Explain why your "radioactive decay" curve is not perfectly smooth; in other words, besides the fact that this is not really radioactive and you are shaking a tray instead of the passage of real time, why are real radioactive decay curves very predictable
3.What do mathematicians call the shape of this curve?
4.If the half-life of your sample is 5,000 years, calculate the value for k, the rate constant. Be sure to show your calculations.
5.Assume that your "radioactivity" was C-14 (carbon-14, carbon with mass number 14) and that it undergoes beta decay. Write the balanced nuclear equation for this reaction.
6.Carbon dating can be used to get a measurement of the age of a carbon-containing sample. (Scientists actually use mass spectroscopy and measure ratios of isotopes, which is a bit more complicated than can be replicated here.) Using the curve you created in Step 1, determine the age of a sample with this amount of radioactivity left:
a.75%
b.50%
c.25%
d.0%
The number of radioactive particle VS. Time 200 O 150 100 50 0 4 10 12 14 16 18 Time (Multiple by 5000 years)Explanation / Answer
(4) All radio active deacy follow first order kinetics only.
Hence,
k = 0.693 / t1/2
k = 0.693/5000
k = 1.386*10-4 per year
(5) 146C -----------> 147N + 0-1e
(6)
first order rate constant equation,
k = (2.303/t)Log[a/(a-x)]
(a)
t = (2.303/1.386*10-4) Log(100/75) = 2.076*103 years
(b)
t = (2.303/1.386*10-4) Log(100/50) = 5.00*103 years
(c)
t = (2.303/1.386*10-4) Log(100/25) = 1.00*104 years
(d)
t = (2.303/1.386*10-4) Log(100/0) = infinite time is required
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