Question #f85fc

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Question #f85fc

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Answer 1 · 2016-12-08T18:16:29

$Q (62 hours) = 15.45 {g of}^{42} K$ $Q(62" hours") = 15.45 "g of "^42K$

Explanation:

Using the formula:

$Q (t) = Q (0) e^{α t}$ $Q(t) = Q(0)e^(alphat)$

Using this reference half-life of $^{42} K$ $""^42K$ , set $t = 12.36 hours$ $t = 12.36" hours"$ and make $\frac{Q (t)}{Q (0)} = \frac{1}{2}$ $(Q(t))/(Q(0)) = 1/2$ :

$\frac{1}{2} = e^{α (12.36 hours)}$ $1/2 = e^(alpha(12.36" hours"))$

Use the natural logarithm on both sides:

$ln (\frac{1}{2}) = ln (e^{α (12.36 hours)})$ $ln(1/2) = ln(e^(alpha(12.36" hours")))$

Flip the equation and ln(e) redues to the exponent:

$α (12.36 hours) = ln (\frac{1}{2})$ $alpha(12.36" hours") = ln(1/2)$

$α (12.36 hours) = - ln (2)$ $alpha(12.36" hours") = -ln(2)$

$α = - \frac{ln (2)}{12.36 hours}$ $alpha = -ln(2)/(12.36" hours"$

$Q (62 hours) = (500 {g of}^{42} K) e^{(- \frac{ln (2)}{12.36 hours} (62 hours))}$ $Q(62" hours") = (500 "g of "^42K)e^((-ln(2)/(12.36" hours")(62" hours"))$

$Q (62 hours) = 15.45 {g of}^{42} K$ $Q(62" hours") = 15.45 "g of "^42K$

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