Part A:You are using a Geiger counter to measure the activity of a radioactive s

Question

Part A:You are using a Geiger counter to measure the activity of a radioactive substance over the course of several minutes. If the reading of 400. counts has diminished to 100. counts after 61.7 minutes , what is the half-life of this substance?

Part B; An unknown radioactive substance has a half-life of 3.20 hours . If 27.1 g of the substance is currently present, what mass A0 was present 8.00 hours ago?

Part C:Americium-241 is used in some smoke detectors. It is an alpha emitter with a half-life of 432 years. How long will it take in years for 37.0 % of an Am-241 sample to decay?

PART D;A fossil was analyzed and determined to have a carbon-14 level that is 20 % that of living organisms. The half-life of C-14 is 5730 years. How old is the fossil?

Explanation / Answer

Part A

400 counts dropping to 100 counts means two half lives have passed

400 --> 200 one half life
200 --> 100 one half life

so the half life is 61.7/2 =30.85 mins

Part B

Use this equation below to find out how much of a radioactive material is left after a given period of time. You can use the equation in reverse to find out the initial
mass of the radioactive material if you know its current mass.

A = current mass of radioactive material = 27.1 g
A0 = initial mass of radioactive material = to be determined
T½ = half-life of the radioactive material = 3.20 h
t = elapsed time since the original mass was present = 8.00 h

A = A0e^(-0.693t/T½)
A/A0 = e^(-0.693t/T½)
ln (A/A0) = ln [e^(-0.693t/T½)]
ln A - ln A0 = (-0.693t/T½)
ln A = -0.693t/T½ + ln A0

ln 27.1 = -0.693(8.00 h)/3.20 h + ln A0

3.211 = -1.7325 + ln A0

3.211  + 1.7325 = ln A0

ln A0 = 3.211 + 1.7325

ln A0 =4.9435

A0 = e^4.9435

A0 = 140.26  Initial Mass of Radioactive Material

Part C:

t1/2 = 0.693/K

K = 0.693 / 432

K = 1.6042 * 10^-3

K = 2.303/t log(a / (a-x))

1.6042 * 10^-3 = 2.303 / t log (100/(100-37))

t = 71.74 years.

Part D:

Let the amount (No) of carbon sample be 100g

So the amount left (Nt) = 20% of 100g = 20g

Using the formula ,

Nt = No (1/2)t / t 0.5

where t is the amount of time lapsed

and t 0.5 is the half life period

putting all the values we get ,

20 = 100 (1/2)t / 5730

t = 13304.65 years

So the fossil must be 13304.65 years old.

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects