activity of 16 Ci to decay to 2.0 Ci? CI.10 The iceman known as Ötzi was discove

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 of

Explanation / 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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