Science:Math Exam Resources/Courses/MATH110/April 2012/Question 06 (b)

MATH110 April 2012

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Question 06 (b)

In a fission nuclear weapon, a sphere of uranium-235 is compressed to a fraction of its size, forcing at least one atomic nucleus to split and release neutrons. These neutrons force other nuclei to release their neutrons, and so on. The result is that the number of neutrons grows exponentially. Let P(t) be the number of neutrons t seconds after detonation. The number of neutrons satisfies the differential equation $\displaystyle {\frac {dP}{dt}}=10^{8}P(t).$ (b) According to your model in part (a), how long does it take a given number of neutrons to double?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Say you started with 5 neutrons (or 500, or 5 million) - how would you incorporate this information into your formula from part (a)?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We will call our given number of neutrons at the start (when t = 0), $P_{0}$. Note that when we calculate the number of neutrons at the start, we get $P(0)=1$. So in order for our function $P(t)$ to give us the starting number of neutrons, $P_{0}$, we need to multiply $e^{10^{8}t}$ by $P_{0}$ to get our final formula, $P(t)=P_{0}e^{10^{8}t}$. To calculate how long it takes for the neutrons to double, we set our function $P(t)$ equal to double the starting amount, or $2P_{0}$. This gives us $2P_{0}=P_{0}e^{10^{8}t}$ After canceling the $P_{0}$ from both sides, taking a natural logarithm of both sides, and simplifying using log rules we get $\ln(2)=\ln(e^{10^{8}t})$ $\displaystyle \ln(2)=10^{8}t$ ${\frac {\ln(2)}{10^{8}}}=t$

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