SOLUTION: PLEASE HELP!! The radioactive isotope thorium 234 has a half-life of approximately 578 hours. a) If a sample has an initial mass of 64 mg, a function that models the mass in mg a

Algebra ->  Human-and-algebraic-language -> SOLUTION: PLEASE HELP!! The radioactive isotope thorium 234 has a half-life of approximately 578 hours. a) If a sample has an initial mass of 64 mg, a function that models the mass in mg a      Log On


   

Question 277974: PLEASE HELP!!
The radioactive isotope thorium 234 has a half-life of approximately 578 hours.
a) If a sample has an initial mass of 64 mg, a function that models the mass in mg after t hours is a(t)= _____________ ?
b) The amount remaining after 75 hours will be about __________ mg.
c) The initial mass will decay to 12 mg after ________ hours.
Found 2 solutions by stanbon, Theo:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The redioactive isotope thorium 234 has a half-life of approximately 578 hours.
A(t) = Ao*(1/2)^(t/578)
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a) If a sample has an initial mass of 64 mg, a function that models the mass in mg after t hours is a(t)= 64*(1/2)^(t/578)
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b) The amount remaining after 75 hours will be about __________ mg.
A(75) = 64*(1/2)^(75/578)
A(75) = 64*0.914
A(75) = 58.5 mg
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c) The initial mass will decay to 12 mg after ________ hours.
solve 12 = 64*(1/2)^(t/578)
(1/2)^(t/578) = 0.1875
Take the natural log of both sides to get:
(t/578)ln(1/2) = ln(0.1875)
t/578 = 2.4150
t = 1395.89 hrs.
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Cheers,
Stan H.

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
half life = 578 hours.

exponential decay is assumed.

this means that 1/2 = 1 * (1+g)^t where g = the growth rate and t = time in hours.

.5 = (1+g)^t

since t = 578, then this equation becomes:

.5 = (1+g)^578

take the log of both sides of this equation to get:

log(.5) = log((1+g)^578)

this is equivalent to:

log(.5) = 578 * log(1+g)

divide both sides of this equation by 578 to get:

log(1+g) = log(.5) / 578 which becomes:

log(1+g) = -.000520814

this means that 1+g = .998801502 which means that g = .998801502 - 1 = -.001198498

to see if this is correct, substitute in original equation to see if it is true.

original equation is:

.5 = (1+g)^578

this becomes:

.5 = .998801502^578 = .5 confirming the value of g is good.

now that we know the value of (g), we should be able to answer the other questions.

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a) If a sample has an initial mass of 64 mg, a function that models the mass in mg after t hours is:

a(t) = 64 * (.998801502)^t

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b) The amount remaining after 75 hours will be about:

a(t) = 64 * (.998801502)^75 = 58.49503274

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c) The initial mass will decay to 12 mg after about:

a(t) = 64 * (.998801502)^t becomes:

12 = 64 * (.998801502)^t

divide both sides by 64 to get:

12/64 = (.998801502)^t

take log of both sides of the equation to get:

log(12/64) = log(.998801502^t) which becomes:

log(12/64) = t*log(.998801502)

divide both sides of this equation by log(.998801502) to get:

t = log(12/64)/log(.998801502) to get:

t = 1395.891675 hours.

the sample will decay to 12 mg in approximately 1395.89 hours.

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extra:

if the function is correct, then the 64 mg should decay to 32 mg in 578 hours.

formula is:

32 = 64 * .998801502^t

divide both sides of this equation by 64 to get:

32/64 = .998801502^t

take log of both sides of this equation to get:

log(32/64) = log(.998801502^t) which becomes:

log(32/64) = t*log(.998801502).

divide both sides of this equation by log(.998801502) to get:

t = log(32/64)/log(.998801502) which becomes:

t = 578

any minor discrepancy in the number is due to rounding.