A 70 kg human body typically contains 140 g of potassium. Potassium has a chemical atomic mass of 39.1 u and has three naturally occurring isotopes. One of those isotopes, 40K, is radioactive with a half-life of 1.3 billion years and a natural abundance of 0.012%. Each 40K decay deposits, on average, 1.0 MeV of energy into the body.

Part A What yearly dose in Gy does the typical person receive from the decay of 40K in the body? Express your answer using two significant figures.

Q=Gy

Q =  2.3 * 10⁻⁴ Gy

Explanation:

The mass of the human body, M = 70 kg

Mass of potassium,

Atomic mass of potassium, MM = 39.1 g/mol

The half life of  the isotope 40K,

Number of moles of potassium,

n = 3.58 mols

Let the number of atoms in the potassium be N

Number of mols of potassium,

The 40K isotope has a natural abundance of 0.012% = 0.00012

That means the number of atoms originally in the 40K isotope is,

Number of atoms remaining after decay

Decay constant,

The question is asking for the yearly dose, i.e t = 1 yr

The number of 40K atoms decaying in 1 yr =

The number of 40K atoms decaying in 1 yr = 1 * 10¹¹ atoms

1 40K atom deposits 1.0MeV of energy into the body

1 * 10¹¹ atoms will deposit 1 * 10¹¹MeV of energy into the body = 10¹⁷ eV

1 eV = 1.6 * 10⁻¹⁹ Joules

10¹⁷ eV = (10¹⁷)*(1.6 * 10⁻¹⁹) Joules

10¹⁷ eV = 0.016 J

The amount of dose a typical body receives in Gy = 0.016/70

The amount of dose a typical body receives in Gy = 2.29 * 10⁻⁴

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