Question

Chromium-51 is a radioactive isotope used to label red blood cells to identify if they remain intact in the body. It has a half-life of 27.7 days. If 0.25 mg is an approximate dose amount, how much remains after 83.1 days?

Answer 1

$\left[\begin{array}{cccc}0 \ days&27,7days&55,4days&\textbf{83,1days}\\&&&\\0,25mg&0,125mg&0,0625mg&\textbf{0,03125mg}\end{array}\right]$

Answer 2

Answer:

Amount of chromium-51 remaining after 83.1 days = 0.031 mg

Explanation:

Radioactive disintegration follows the exponential law which can be mathematically expressed as:

$N(t)=N(0)e^{-0.693t/t1/2}------(1)$

where N(0) = initial amount of the radioactive substance

N(t): radioactive substance left after time t

t1/2 = half life of the radioisotope

In the case of chromium-51:

N(0) = 0.25 mg

t = 83.1 days

t1/2 = 27.7 days

Substituting these values in equation (1) gives:

$N(t)=0.25e^{-0.693*83.1/27.7} = 0.031 mg$