Question

Iron-59 is a radioisotope that is used to evaluate bone marrow function. The half-life of iron-59 is 44.5 days. If you begin with 43.5 mg of this isotope, what mass remains after 55.2 days have passed?

mg

Answer 1

In kt 1 CA ionce ntvation oinitial time t a G(onentvation attev (onstant K is Yate O.693 th k = the halt- life is InC-InCA kt Inc -kt In CAs If e plt Yaph af line straisht Us) t Jet a of -K and slepe with a intercept InCAs

.a 43.5 mjzAo 1 t=559 days 6.693 43.5 44.5day In 93.5 0.85963146079 1 a-85963146079 43-5 2.3622899 CA 43-5 my -1 2.3622 899 SA 18.41433STmy.

Answer 2

In (SA) = kt a time t At initial concentration At concentration after k is rate constant K = 0.693 It is the half-life. InCA - INCA = kt Inca = - kt + InCA If we plot a graph (us) t we get a with a slope of Sk' intercept InCA of Inca straight line and

a = 43.5 mg = Cho - પપs days t=55.9 days 143.5 Incal 0.693 X55.9 day 44.5 day 43.5 In (4305 ) = 0.95963146024 = 0.85963146079 0.85963146079 43-5 = 2.3699899 CA 43-5 mg 2.3629899 'A= 18.4143357 mg.