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Chapter 11: Q52E (page 520)

The half-life $T_{1 / 2}$ is not the average lifetime $蟿$ of a radioactive nucleus. We find the average lifetime by multiplying $t$ by the probability per unit time $P (t)$ that the nucleus will 鈥渓ive鈥� that long, then integrating over all time.
(a) Show that $P (t)$ should be given by ${位别}^{鈭� 位迟}$
(b) Show that $蟿 = T_{1 / 2} / In 2$ .

Short Answer

Expert verified

(a) The coefficient shall be $位$ . And probability will be $P (t) = {位别}^{鈭� 位迟}$ .

(b) The average life period is $蟿 = \frac{T_{1 / 2}}{In 2}$ .

Step by step solution

01

Given data

Given data:

$P (t) 鈭� N (t)$

Formula used:

${鈭�}_{0}^{鈭�} P (t) dt = {鈭�}_{0}^{鈭�} {ke}^{- 位迟} dt$

02

Probability

a)

We know that $P (t)$ should be proportional to $N (t)$ for the nucleus to live long, which is proportional to $e^{鈭� 位迟}$ assuming the coefficient of proportionality is $k$ , the total probability for the nucleus to live shall be 1.

$\begin{matrix} {鈭�}_{0}^{鈭�} P (t) dt = {鈭�}_{0}^{鈭�} {ke}^{鈭� 位迟} dt \\ 1 = {\frac{{ke}^{鈭� 位迟}}{鈭� 位} |}_{0}^{鈭�} \\ 1 = \frac{k}{位} \\ k = 位 \end{matrix}$

Thus, the coefficient shall be $位$ . And probability will be $P (t) = {位别}^{鈭� 位迟}$ .

03

Determine τ=T12In2

Now we can calculate the average lifetime:

$\begin{matrix} 蟿 = {鈭�}_{0}^{鈭�} {位迟e}^{鈭� 位迟} dt \\ = {鈭� {te}^{鈭� 位迟} |}_{0}^{鈭�} 鈭� (鈭� {鈭�}_{0}^{鈭�} e^{鈭� 位迟} dt) \\ = {鈭�}_{0}^{鈭�} e^{鈭� 位迟} dt \\ = \frac{1}{位} \end{matrix}$

Using the relation between the time period and disintegration constant as:

$位 = \frac{In 2}{T_{1 / 2}}$

The average life period can be expressed as:

$蟿 = \frac{T_{1 / 2}}{In 2}$

Thus, the average life period is $蟿 = \frac{T_{1 / 2}}{In 2}$ .

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Most popular questions from this chapter

What does the semiemperical binding energy formula predict forthe mass of a carbon- 12 atom? Comment on your answer.

Calculate $Q$ for neutron decay.

Calculate the binding energy per nucleon of carbon- $12$ .

For the lightest of nuclei, binding energy per nucleon is not a very reliable gauge of stability. There s no nucleon binding at all for a single proton or neutron yet one is stable (so far as we know) and the other is not(a) Helium-3 and hydrogen-3 (tritium) differ only in the switch of a nucleon. Which has the higher binding energy per nucleon? (b) Helium-3 is stable, while tritium, in fact, decays into helium-3.Does this somehow violate laws?

Question:How is it that a high binding energy is a low energy?

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