# 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...

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?

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### 1 Answer

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.