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 (potassium), is radioactive with a half-life of 1.3 billion years and a natural abundance of 0.012 %. Each 40K (potassium) decay deposits, on average, 1.0 MeV of energy into the body. What yearly dose in Gy does the typical person receive from the decay of 40K (potassium) in the body? Express your answer using two significant figures.

Gy = 3.14x10⁻⁴ Gy

Explanation:

To get the dose in Gy we need to use the following expression:

Gy = E / m  (1)

Where:

Gy: dose

E: energy absorbed per atom

m: mass of the human body.

We don't have the energy per atom, but we can calculate that by following the next procedure.

First, let's determine the number of atoms of potassium in our body. For that we need to determine the moles in the 140 g of potassium, with the molecular mass and then, use the avogadro's number:

moles = m/MM

moles = 140 / 39.1 = 3.58 moles

N° atoms = 3.58 * 6.02x10²³ atoms = 2.16x10²⁴ atoms of K.

The abundance of the ⁴⁰K is 0.012% so the atoms of this isotope would be:

N = 2.16x10²⁴ * (0.012/100) = 2.59x10²⁰ atoms of ⁴⁰K.

With this number, and the half life rate, we can determine the number of decay atoms in a year (λ) using the following expression:

λ = ln2 / t(1/2)

λ = ln2 / 1.3x10⁹ = 5.33x10⁻¹⁰ year⁻¹

This number, multiplied by the number of atoms:

R = 5.33x10⁻¹⁰ * 2.59x10²⁰ = 1.38x10¹¹ atoms/year

Now, each atom of K gives an average energy of 1 MeV, so with the atoms we have:

E = 1.38x10¹¹ * 1x10⁶ eV = 1.38x10¹⁷ eV

This value can be expressed in Joules so:

E = 1.38x10¹⁷ eV * (1 J / 6.24x10¹⁸ eV) = 0.022 J

Finally, we can use (1) to get the dose in Gy:

Gy = 0.022 / 70

Hope this helps

answer: pseudoscience