Question

1 а.write a nuclear equation for the alpha decay of Ann (3.25 points) b. Write a nuclear equation for the beta decay of 3 Sr (3.25 points) e. Write a nuclear equation for the positron emission of 12N (.25 poin) d. Write a nuclear equation to represent positron emission in O-15 (3.25 points) 2. Fluorine-18 undergoes positron emission with a half-life of 1.10 x 102 minutes. If a patient is given a 248 mg dose for a PET scan, how long will it take for the amount of fluorine-18 to drop to 83 mg? (Assume that none of the fluorine is excreted from the body.) (6 points) 3. An archaeologist graduate student found a leg bone of a large animal during the building c new science building. The bone had a carbon-14 decay rate of 14.8 disintegrations per minu per gram of carbon. Living organisms have a decay rate of 15.3 disintegrations per minute How old is the bone? (6 points)

Answer 1

Solution:- (1) (a) In alpha decay, atomic number decreases by two units and mass number decreases by four units.

₉₅Am²⁴¹ --------> ₂He⁴(alpha particle) + ₉₃Np²³⁷

(b) In beta decay, atomic number increases by 1 unit and mass number remains same.

₃₈Sr⁸⁹ -------> _-1e⁰(beta particle) + ₃₉Y⁸⁹

(c) Inpositron decay, atomic number decresaes by 1 and mass number remains same.

₇N¹³ -----> ₆C¹³ + ₁e⁰(positron)

(d)  ₈O¹⁵ ------>  ₇N¹⁵ + ₁e⁰(positron)

(2) Radioactive decay obeys first order kinetics.

ln[A] = - kt + ln[A]₀

Where, [A]₀ initial amount, [A] is final amount, k is decay constant and t is total time.

k could be calculated from the given half life using the formula, k = 0.693/half life

k = 0.693/(1.10 x 10² min) = 0.0063 min^-1

Initial amount is 248 mg and final amount is 83 mg. We are asked to calculate the time it takes to drop the amount of radioactive isotope from 248 mg to 83 mg. Let's plug in the values in the equation..

ln[83] = -0.0063 min^-1(t) + ln(248)

4.419 = -0.0063 min^-1(t) + 5.513

4.419 - 5.513 = -0.0063 min^-1(t)

-1.094 = -0.0063 min^-1(t)

t = 1.094/0.0063min^-1

t = 173.65 min

If we round off this value then it is almost 174 minutes.

(3) Half like for carbon-14 is 5730 years. This problem is similar to #2.

k = 0.693/ 5730 years = 1.21 x 10^-4 year^-1

Initial decay rate, [A]₀ = 15.3 and [A] = 14.8

Let's plug in the values in the equation, ln[A] = - kt + ln[A]₀

ln(14.8) = - 1.21 x 10^-4 year^-1(t) + ln(15.3)

2.695 = -1.21 x 10^-4 year^-1(t) + 2.729

2.695 - 2.729 = -1.21 x 10^-4 year^-1(t)

-0.034 = -1.21 x 10^-4 year^-1(t)

t = 0.034/1.21 x 10^-4 year^-1

t = 281 years

So,the bone is 281 years old.