# The half-life of uranium-238 is 4.51 billion years, while that of uranium-235 is .707 billion years.  Suppose that a sample contains equal amounts of uranium-238 and uranium-235 at the time of its formation.  If the proportion of uranium-238 to uranium-235 in the sample is currently 137.8 to 1, what is the age of the sample? Use the formula for exponential growth or decay: P = Po e^(kt)

where P = life at certain time t          e = constant = 2.712828....

Po = initial life

k = growth or decay factor

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Use the formula for exponential growth or decay: P = Po e^(kt)

where P = life at certain time t          e = constant = 2.712828....

Po = initial life

k = growth or decay factor

t = time

half life means half of the initial life, so P = Po/2

solve for the growth or decay factor, k

half life of uranium-238 at t = 4.51 (in billion)

Po/2 = Po e^(k  * 4.51)    cancel Po on both sides

1/2 = e ^ (4.51k)        it's okay to disregard the billion since the

other given t is also in billion. remember

unit consistency. take the natural log of

both sides

ln(1/2) = ln e^(4.51k)       property: ln e = 1

-0.6931 = 4.51k         divide both sides by 4.51

k of U238 = -0.1536        unitless

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half life of uranium-235 at t = 0.707 (in billion)

Po/2 = Po e^(k  * 0.707)    cancel Po on both sides

1/2 = e ^ (0.707k)  it's okay to disregard the billion since the

other given t is also in billion. remember

unit consistency. take the natural log of

both sides

ln(1/2) = ln e^(0.707k)       property: ln e = 1

-0.6931 = 0.707k         divide both sides by 4.51

k of U235 = -0.9804        unitless

take the ratio

137.8 = Po(u238) e^(-0.1536*t)   age was unknown. let it be t

1         Po(u235) e^(-0.9804*t)      Po(u238) = Po(u235)

Po's cancel out

137.8 = e^(-0.1536t)            property: a^m   = a^(m-n)

1        e^(-0.9804t)                           a^n

137.8 = e^(-0.1536t+0.9804)t

137.8 = e^(0.8268t)            take ln of both sides

4.9258 = 0.8268t         divide both sides by 0.8268

t = 5.9576 billion years.      take note that we disregarded the billion

when we solved for k.

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