Question

Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and thus our bodies) as three isotopes: Potassium-39, Potassium-40, and Potassium-41. Their current abundances are 93.26%, 0.012% and 6.728%. A typical human body contains about 3.0 grams of Potassium per kilogram of body mass. 1. How much Potassium-40 is present in a person with a mass of 80 kg? 2. If, on average, the decay of Potassium-40 results in 1.10 MeV of energy absorbed, determine the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body. Assume an RBE of 1.2. The half-life of Potassium-40 is 1.28 x 10°years.

Answer 1

Complete Question

Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and thus our bodies) as three isotopes: Potassium-39, Potassium-40, and Potassium-41. Their current abundances are 93.26%, 0.012% and 6.728%. A typical human body contains about 3.0 grams of Potassium per kilogram of body mass. 1. How much Potassium-40 is present in a person with a mass of 80 kg? 2. If, on average, the decay of Potassium-40 results in 1.10 MeV of energy absorbed, determine the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body. Assume an RBE of 1.2. The half-life of Potassium-40 is
`1.28 * 10^9`years.

Answer:

The potassium-40 present in 80 kg is
`Z = 0.0288 *10^(-3)\ kg`

The effective dose absorbed per year is
`x = 2.06 *10^(-24)` per year

Step-by-step explanation:

From the question we are told that

The mass of potassium in 1 kg of human body is
`m = 3g= (3)/(1000) = 3*10^(-3) \ kg`

The mass of the person is
`M = 80 \ kg`

The abundance of Potassium-39 is 93.26%

The abundance of Potassium-40 is 0.012%

The abundance of Potassium-41 is 6.78 %

The energy absorbed is
`E = 1.10MeV = 1.10 *10^(6) * 1.602 *10^(-19) = 1.7622*10^(-13) J`

Now 1 kg of human body contains
`3.0*10^(-3)\ kg` of Potassium

So 80 kg of human body contains k kg of Potassium

=>
`k = ( 80 * 3*10^(-3))/(1)`

`k = 0.240\ kg`

Now from the question potassium-40 is 0.012% of the total potassium so

Amount of potassium-40 present is mathematically represented as

`Z = (0.012)/(100) * 0.240`

`Z = 0.0288 *10^(-3)\ kg`

The effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body is mathematically evaluated as

`D = (E)/(M)`

Substituting values

`D = (1.7622*10^(-13))/(80)`

`D = 2.2*10^(-15) J/kg`

Converting to Sieverts

We have

`D_s = REB * D`

`D_s = 1.2 * 2.2 *10^(-15)`

`D_s = 2.64 *10^(-15)`

So

for half-life (
`1.28 *10^9 \ years`) the dose is
`2.64 *10^(-15)`

Then for 1 year the dose would be x

=>
`x = (2.64 *10^(-15))/(1.28 * 10^9)`

`x = 2.06 *10^(-24)` per year