Question #f25a9

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Question #f25a9

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Answer 1 · 2017-09-06T14:01:17

$0.693 \cdot \frac{N}{n}$ $0.693 * N/n$

Explanation:

As you know, a radioactive decay is essentially a first-order reaction

$A \to products$ $"A " -> " products"$

which means that you can express the rate of the reaction, i.e. the rate of disintegration, in differential form like this

$- \frac{d [A]}{d t} = k \cdot [A]$ $- (d["A"])/(dt) = k * ["A"]$

Here

$k$ $k$ is the rate constant

$[A]$ $["A"]$ is the concentration of the radioactive element

In your case, you know that this radioactive element emits $n$ $n$ alpha particles per second. This is equivalent to saying that the rate of change of its concentration per unit of time is equal to

$\frac{d [A]}{d t} = n$ $(d["A"])/(dt) = n$

Moreover, you know that your sample contains $N$ $N$ atoms of this radioactive element, so

$[A] = N$ $["A"] = N$

Plug this into the first equation to get

$- n = k \cdot N$ $-n = k* N$

It's important to realize that the minus sign is there to show that the concentration of $A$ $"A"$ is decreasing as the reaction proceeds, which means that you don't really need it for your purposes.

$n = k \cdot N$ $n = k * N$

This will get you

$k - \frac{n}{N}$ $k - n/N$

Now, the half-life of a first-order reaction is given by

$t_{1/2} = \frac{ln (2)}{k} \approx \frac{0.692}{k}$ $t_"1/2" = ln(2)/k ~~ 0.692/k$

This means that you have

$t_{1/2} = \frac{0.693}{\frac{n}{N}} = 0.693 \cdot \frac{N}{n}$ $t_"1/2" = 0.693/(n/N) = 0.693 * N/n$

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Question #f25a9

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