Question

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

Answer 1

Answer:

Q =  2.3 * 10⁻⁴ Gy

Explanation:

The mass of the human body, M = 70 kg

Mass of potassium, $m_{p} = 140 g$

Atomic mass of potassium, MM = 39.1 g/mol

The half life of  the isotope 40K, $t_{1/2} = 1.3 * 10^{9}$

Number of moles of potassium, $n = \frac{m_{p} }{MM}$

$n = \frac{140}{39.1}$

n = 3.58 mols

Let the number of atoms in the potassium be N

Number of mols of potassium, $n = \frac{N}{6.02 * 10^{23} }$

$N = 6.02 * 10^{23} * n\\N = 6.02 * 10^{23} * 3.58\\N = 2.156 * 10^{24} atoms$

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

That means the number of atoms originally in the 40K isotope is, $N_{0} = 0.00012 * 2.156 * 10^{24}$

$N_{0} = 2.5875165 * 10^{20} atoms$

Number of atoms remaining after decay $N_{f} = N - N_{0}$

Decay constant, $\lambda = \frac{ln 2}{t_{1/2} }$

$\lambda = \frac{ln 2}{1.3 * 10^9} \\\lambda = 533.19 * 10^{-12} yr^{-1}$

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

$N_{f} = N_{0} e^{- \lambda t} \\N_{f} = (2.5875165 * 10^{20} ) e^{- 533.19 * 10^{-12} *1}\\N_{f} = 2.587516499 * 10^{20}$

The number of 40K atoms decaying in 1 yr = $N_{0} - N_{f} = (2.5875165 *10^{24} ) - (2.587516499 * 10^{20})$

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