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
years.
Answer:
The potassium-40 present in 80 kg is 
The effective dose absorbed per year is
per year
Explanation:
From the question we are told that
The mass of potassium in 1 kg of human body is 
The mass of the person is 
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 
Now 1 kg of human body contains
of Potassium
So 80 kg of human body contains k kg of Potassium
=> 

Now from the question potassium-40 is 0.012% of the total potassium so
Amount of potassium-40 present is mathematically represented as


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

Substituting values


Converting to Sieverts
We have



So
for half-life (
) the dose is 
Then for 1 year the dose would be x
=> 
per year