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I have a half-life question that I can't solve. There's very limited information given. Even the half-life formula has not been taught yet.

The mass of a radioactive substance in a certain sample has decreased 32 times in 10 years. Determine the half-life of the substance.

The answer is 2.

From the question I understand that 32 half-lives have gone by in 10 years. How can this be solved using the simplest possible methods? Not using graphs and avoiding logarithms too if possible?

user961627
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3 Answers3

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Just recall that $2^5=32$ and then compute $10/5=2$

Hagen von Eitzen
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  • I see what you mean, but how do you interpret the problem statement? We are told that the mass of a sample decreased $32$ times in $10$ years, but this doesn't tell us how much it decreased... – A.P. Apr 19 '15 at 10:59
  • I don't even see what you mean. What you're saying is right - the teacher said the same thing. But I don't understand this. I can recall 2^5 = 32 etc.. but can you please explain in words how is that related to the question. And what does 2^5 and 10/5 relate to in terms of the problem statement? For example, why divide 10 by 5? I can't make the connection. – user961627 Apr 19 '15 at 11:05
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The solution $2$ assumes the following interpretation of the problem statement:

Suppose you have a radioactive sample of mass $m_0$ which after $10$ years is reduced to mass $m_{10} = \frac{m_0}{32}$. What is the half-life of the substance?

Recall that the half-life of a radioactive substance is the time after which the mass of a sample is halved by radioactive decay. If $10$ is a multiple of the half-life, then $$ m_{10} = \left(\left(\left(m_0 \cdot \frac{1}{2}\right) \cdot \frac{1}{2}\right) \dotsm \right) \cdot \frac{1}{2} = \frac{m_0}{2^n} $$ Since $2^5 = 32$, we know that in $10$ years the mass got halved $5$ times; in other words, once every $\frac{10}{5} = 2$ years.

Note: Still, I find your problem statement ambiguous, at best.

A.P.
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If a substance has "half life" T and initial amount Q then after time t there will have been $\frac{t}{T}$ "half lives" so the initial amount will have been multiplied by $\frac{1}{2}$ $\frac{t}{T}$ times. The amount remaining will be $Q\left(\frac{1}{2}\right)^{\frac{t}{T}}= \frac{Q}{2^{\frac{t}{T}}}$. In this problem "decreased 32 times" means "multiplied by $\frac{1}{32}$". Since $32= 2^5$, that is "multiplied by $\frac{1}{2^5}$. The "$\frac{t}{T}= 5$. We are given that t= 10 so $\frac{10}{T}= 5$, $5T= 10$, $T= 2$. The half life is 2 years.

AP:"We are told that the mass of a sample decreased 32mtimes in 10 years, but this doesn't tell us how much it decreased".

No, because we weren't told how much there was to begin with! All that is important is the amount relative to the original amount.

user247327
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