An unknown radioactive element decays into non-radioactive substances. In 880 days the radioactivity of a sample decreases by 36 percent.

Question

(a) What is the half-life of the element? 

half-life:     (days) 

(b) How long will it take for a sample of 100 mg to decay to 63 mg? 

time needed:    (days) 

Answer 1

880 days decays 36%

A=Pe^rt is the general formula for continuous change/decay or growth

(1-.36) = e^880r where r= daily rate of change/decay, with r<0 for decay, negative "growth"

.64 = e^880r

take natural logs of both sides

ln.64 = 880r

880r = ln.64

divide both sides by 880

r=(ln.64)/880

1/2= e^rt

plug in the value of r and solve for the remaining variable, t

1/2 =.5= e^(ln.64/880)t

ln(.5) = [(ln.64)/880]t

t = 880ln.5/ln.64

= about 880(1.5531)

=1,366.765 days = half life

it looks about right

as 36% decays in 880 days

and 64% is left

in 2(880)= 1760 days another 36% decayed leaving .64(.64) = 40.96% left

50% is left for a half life and is between 64% and 41%

and 1366 is between 880 and 1760

63/100 = e^(ln.64/880)t

ln.63 = (ln.64)t/880

t = 880ln.63/ln.64

= about 911.053 days for 100 mg to decay to 63 mg

Answer 2

0.64 = e^880k, take ln of both sides

-0.446 = 880k

-.000507 = k

0.5 = e^-0.0005.7t

-0.6531 = -0.000507t, talke ln of both sides

1,288.17 = t

63 = 100(0.5)^1,288.17t

or

63 = 100e^1,288.17t

Your choice.

Answer 3

A = A(t) = Pe^kt, where P = initial amount, A = amount remaining after t days, and k is a constant.

When t = 880, A = (1-.36)P = 0.64P

So, 0.64P = Pe^880k

0.64 = e^880k

880k = ln(0.64)

k = -0.000507

So, A = Pe^-0.000507t

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To determine half-life, let A = 0.5P:

0.5P = Pe^-0.000507t

0.5 = e^-0.000507t

ln(0.5) = -0.000507t

t = half-life = 1367.1542 days

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If P = 100 and A = 63, we have 63 = 100e^-0.000507t

0.63 = e^-0.000507t

ln(0.63) = -0.000507t

So, t = 911.3 days