Answer:
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.
