Question

Potassium has a density of 0.86 g/cm^3, an atomic weight of 39.0983 g/mol, and contains 0.012% of Potassium-40 which is radioactive by beta-decay with a half-life of 1.25x10^9 years.

a. What is the total and average binding energy of potassium-40?

b. If the average mass of potassium within the human body is about 140 g, estimate the average activity (in bq) of [40,19] K in the body.

Answer 1

a)

The atomic mass of 40K is 39.0983 amu. Calculate the nuclear binding energy of this

nucleus and the corresponding nuclear binding energy per nucleon

Mass number = 40

No. of protons = 19, then

Neutrons = Mass number – proton number = 19protons and 40-19= 21 neutrons are present

Mass of proton = 1.00727647

Mass of neutrón = 1.008665

Total mass = 19* 1.00727647 + 21*1.008665= 40.32021793 amu

The mass defect:

39.0983 -40.32021793= -1.22191793 amu

Mass = 1.22191793/(6.022*10^23) = 2.029089*10^-24 g or -2.029089*10^-27 kg

Energy released (Einstein E = mC^2)

E = m*C^2 = -2.029089*10^-27 * (3*10^8)^2 = -1.826180*10^-10 J

Nuclear binding energy = E = -1.826180*10^-10 J

Total energy --> 1.826180*10^-10 J

Average per nucleon:

Nuclear binding energy per nucleon

Nucleon = neutron + protons = 40

E per nucleon = E / nucleon =( -1.726*10^-10 J)/( 40) = 4.56545*10^-11 J/nucleon

b)

mass of K total in body = 140 g of K

mass of K40 --> 0.012/100 * 140 = 0.0168g of 40 K

mol = mass/MW = 0.0168 / 39.0983 = 0.0004296862 mol of 40 K

atoms of K40 = mol*NA = (0.0004296862)(6.022*10^23) = 2.5875*10^20 atoms of K40

activity = k*amount

k = from half life

HL = ln(2)/K

k = ln(2) / (HL) = (ln(2)) / (1.25*10^9) = 5.54517*10^-10 1/y

change to 1/s

5.54517 1/y * 1 y / 365 d * 1 d /24 h * 1 h / 60 min * 1 min & 60 s

k = 1.758*10^-7 1/s

activity = k*amount

Activit6y = (1.758*10^-7) * (2.5875*10^20) = 4.54*10^13 counts per second = 4.54*10^13 Bq