activity of 16 Ci to decay to 2.0 Ci? CI.10 The iceman known as Ötzi was discove
ID: 544916 • Letter: A
Question
activity of 16 Ci to decay to 2.0 Ci? CI.10 The iceman known as Ötzi was discovered in a high mountain pass on the Austrian-Italian border. Samples of his hair and bones had carbon-14 activity that was 50% of that present in new hair or bone. Carbon-14 undergoes beta decay and has a half-life of 5730 yr. (5.2, 5.4) The mummified remains of Ötzi were discovered in 1991 a. How long ago did Ötzi live? b. Write a balanced nuclear equation for the decay of carbon-14 CI.11 K is an electrolyte required by the human body and is found o the isotopes ofExplanation / Answer
Solution:- (CI.10)(a):- Radioactive decay obeys first order kinetics. The equation used is...
ln[A] = -kt + ln[A]0
[A] is final amount, it's activity count here for this problem and [A]0 is the initial activity present in new hair o bone.
k is decay constant and t is the life of the sample, life of Otzi for this problem.
k = 0.693/half life
k = 0.693/5730 yr
k = 1.21 x 10-4 yr-1
If [A]0 is 100 then it's 50% would be 50, that means [A] = 50
Plug in the values in the equation to calculate t.
ln[50] = - 1.21 x 10-4 yr-1(t) + ln[100]
3.912 = -1.21 x 10-4 yr-1(t) + 4.605
3.912 - 4.605 = -1.21 x 10-4 yr-1(t)
-0.693 = -1.21 x 10-4 yr-1(t)
t = 0.693/1.21 x 10-4 yr-1
t = 5727 yr
So, The Otzi was live 5727 years before.
(b) Carbon-14 undergoes beta decay, in beta decay, the atomic number increases by one unit and the mass number remains unchanged. So, the equation would be as given below..
6C14 -----> -1e0 + 7N14
(CI.8):- (a) Volume of bracelet = 25.6 cm3
density of bracelet = 10.2 g/cm3
mass = volume* density
So, mass of bracelet = 25.6 cm3 x 10.2 g/cm3 = 261.12 g
Silver is 92.5% by mass. So, mass of silver in the bracelet could be calculated as..
silver mass% = (mass of silver/mass of bracelet) * 100
92.5 = (mass of silver/261.12)*100
mass of silver = 92.5 x 261.12/100
mass of silver = 241.536 g
(b) number of protons = atomic number
number of neutrons = mass number - number of protons
So, for the first isotope....
number of protons = 47
number of neutrons = 107 - 47 = 60
For second isotope..
number of protons = 47
number of neutrons = 109 - 47 = 49
(c) Average atomic mass is the sum of isotopic mass times abundance.
So, average atomic mass = (106.905 x 0.5184) + (108.905 x 0.4816)
average atomic mass = 107.868 amu
(d) 47Ag112 ------> -1e0 + 48Cd112
(e) Using first order kinetics we would calculate the decay constant and then using half life formula we can calculate the half life.
ln(8.00) = - k(9.3h) + ln(64.0)
2.079 = - k(9.3h) + 4.159
2.079 - 4.159 = - k(9.3h)
-2.08 = - k(9.3h)
k = 2.08/9.3h
k = 0.224 h-1
half life = 0.693/0.224 h-1
half life = 3.09 h
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