The radioisotope Phosphorus-32 is used in leukemia therapy. The half-life of this isotope os 14.26 days. Which equation determines the percent of an initial amount of the isotope remaining after t days?

A= 14.26 (1/2)^100/t
A= 14.26(1/2)^t/100
A= 100(1/2)^14.26/t
A= 100(1/2)^t/14.26

Respuesta :

the remaining value= the initial value × (1/2)^(t/T), t is the number of years, T is half life.
So D is the right answer. 

The amount of isotope remaining after t days is [tex]A=100(\frac{1}{2} )^\frac{t}{14.26}[/tex]

Half life is the amount of time that it takes a substance to decay to half of its original value. It is given by:

[tex]N(t)=N_o(\frac{1}{2} )^\frac{t}{t_\frac{1}{2} }[/tex]

Where N(t) is the amount of substance remaining, N₀ is the initial amount, t is the time and t₁₂ is the the half life.

Given that:

  • Half life = 14.26 days, initial amount = 100

Hence:

[tex]A=100(\frac{1}{2} )^\frac{t}{14.26}[/tex]

The amount of isotope remaining after t days is [tex]A=100(\frac{1}{2} )^\frac{t}{14.26}[/tex]

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