The initial mass of a radioactive goo is 180 grams. After 195 minutes, the sample has decayed to 2.8125 grams. Find the half-life of the substance in minutes? Find a formula for the amount, G(t), remaining at time t.? Find grams remain after 35 min?

We know that a form of the decay formula is:

#G(t) = G(0)(1/2)^(t/t_(1/2))" [1]"#

Where #G(0)# is the initial amount and #t_(1/2)# is the half-life.

Given #G(0) = 180" g" and G(195" min") = 2.8125" g"#

Substitute the above into equation [1]:

#2.8125" g" = (180" g")(1/2)^((195" min")/t_(1/2))#

Solve for half-life, #t_(1/2)#.

Divide both sides by #180" g"#:

#(2.8125" g")/(180" g") = (1/2)^((195" min")/t_(1/2))#

Take the natural logarithm of both sides:

#ln((2.8125" g")/(180" g")) = ln((1/2)^((195" min")/t_(1/2)))#

Use the property of logarithms that allow one to move the exponent to the outside as a coefficient:

#ln((2.8125" g")/(180" g")) = ((195" min")/t_(1/2))ln(1/2)#

Multiply both sides by #t_(1/2)#

#t_(1/2)ln((2.8125" g")/(180" g")) = (195" min")ln(1/2)#

Divide both sides by #ln((2.8125" g")/(180" g"))#:

#t_(1/2) = (195" min")ln(1/2)/ln((2.8125" g")/(180" g"))#

#t_(1/2) = 32.5" min"#

The formula for #G(t)# is:

#G(t) = G(0)(1/2)^(t/(32.5" min")#

The amount of the 180-gram sample remaining after 35 min:

#G(35" min") = (180" g")(1/2)^((35" min")/(32.5" min"))#

#G(35" min") = 85.327" g"#