The radioisotope phosphorus-32 is used in tracers for measuring phosphorus uptake by plants. What is the half-life of phosphorus-32 if 26.5 days are required for the activity of a sample of phosphorus-32 to fall to 27.7 percent of its original value?

THE HALF LIFE OF THE RADIOACTIVE PHOSPHORUS-32 IS 1.15 DAYS.

Explanation:

Half life is the time required for half of a given sample of a radioactive substance to decay. It is represented as t1/2.

Nt = No e^-yt

Nt = 27.7 %

No = 100%

t = 26.5 days

y = decay constant

t1/2 = half life = unknown

So therefore,we solve for the decay constant and then using the formula Half life = Ln 2/ decay constant, we solve for the half life.

Nt = No e^-yt

27.7 = 100 e ^-26.5 y

0.277 = e^-26.5 y

Take the natural log of both sides

Ln 0.277 = -26.5 y

y = Ln 0.277/ -26.5

y =-1.2837/ -26.5

y =4.84 *10 ^-2

Since half life = Ln 2 / y

Half life = ln 2 / 4.84*10-2

Half life = 1.1456 days.

Hence, the half life for the radioactive phosphorus -32 is 1.15 days.

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