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The half-life of radium-226 is 1590 years. (a) A sample of radium-226 has a mass of 100 mg. Find a formula for the mass of the sample that remains after t years. (b) Find the mass after 1000 years correct to the nearest milligram. (c) When will the mass be reduced to 40 mg? SOLUTION (a) Let m(t) be the mass of radium-226 (in milligrams) that remains after t years. Then dm/dt = km and m(0) = 100, so this theorem gives m(t) = m(0)ekt = ekt. In order to determine the value of k, we use the fact that m(1590) = 1 2 . Thus e1590k = so e1590k = and 1590k = ln 1 2 = − ln(2) k = . Therefore m(t) = . We could use the fact that eln(2) = 2 to write the expression for m(t) in the alternative form m(t) = . (b) The mass after 1000 years is as follows. (Round your final answer to the nearest milligram.) m(1000) = ≈ mg

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Answer:

See explanation

Explanation:

a) Formula for the mass of the sample that remains after t years = N= Noe^-kt

Where;

N = mass at time t years

No = mass at time t= 0

k = decay constant

t = time taken

So,

N = 100e^-kt

b) First,

t1/2 = -ln(1/2)/k

t1/2 = 0.693/k

t1/2 = half life of radium-226 =1590 years

1590 = 0.693/k

k = 0.693/1590

k = 4.36 * 10^-4

So,

N= 100e^-(4.36 * 10^-4 * 1000)

N= 65 mg

c) From

N = 100e^-kt

40 = 100e^-(4.36 * 10^-4t)

40/100 = e^-(4.36 * 10^-4t)

0.4 = e^-(4.36 * 10^-4t)

ln(0.4) = ln(e^-(4.36 * 10^-4t))

-0.9163 = -4.36 * 10^-4t

t = 0.9163/4.36 * 10^-4

t = 2102 years

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