# Plutonium isotope Pu-239 has a half-life of 24,100 years. Suppose that 10 grams of Pu-239 were...

## Question:

Plutonium isotope Pu-239 has a half-life of 24,100 years. Suppose that 10 grams of Pu-239 were released in the Chernobyl nuclear accident. How long will it take for the 10 grams to decay to 1 gram?

## Exponential Decay Formula:

Exponential decay formula basically used in radioactive decay, but it has many applications. The following formula is used for radioactive decay application.

$$y(t) = ae^{kt} $$

{eq}y(t) = {/eq} mass of radioactive element after {eq}t {/eq} years.

{eq}a = {/eq} initial amount of mass of radioactive element.

{eq}k = {/eq} decay constant.

## Answer and Explanation:

We are given:

The half-life of {eq}\text{ pu-239 } {/eq} is {eq}24100 {/eq} years.

Intitial mass of the {eq}\text{ pu-239 is } 10 {/eq} grams.

After half-life, the mass of the {eq}\text{ pu-239 } {/eq} will be {eq}\frac{10}{2} \, or \, 5 {/eq} grams.

Now, from these information and comparing exponential decay equation, we can write:

{eq}a = 10 {/eq}

{eq}y(t) = 5 {/eq}

And:

{eq}t = 24100 {/eq}

Inserting these values into exponential decay equation and we have:

{eq}5 = 10e^{24100k} \\ \Rightarrow \frac{5}{10} = e^{24100k} \\ \Rightarrow \frac{1}{2} = e^{24100k} \\ \Rightarrow 24100k = \ln (\frac{1}{2}) \hspace{1 cm} \left[ \because e^a = b \therefore a = \ln (b) \right] \\ \Rightarrow k = \frac{\ln (\frac{1}{2})}{24100} \\ \Rightarrow k \approx -0.000028761 {/eq}

Now:

When {eq}10 {/eq} grams of {eq}\text{ pu-239 } {/eq} decay to {eq}1 {/eq} gram.

Then:

{eq}a = 10 {/eq}

{eq}y(t) = 1 {/eq}

And:

{eq}k = -0.000028761 {/eq}

Again, using these values and exponential decay equation, we have:

{eq}1 = 10e^{-0.000028761t} \\ \Rightarrow \frac{1}{10} = e^{-0.000028761t} \\ \Rightarrow -0.000028761t = \ln (\frac{1}{10}) \hspace{1 cm} \left[ \because e^a = b \therefore a = \ln (b) \right] \\ \Rightarrow t = - \frac{\ln (\frac{1}{10})}{0.000028761} \\ \Rightarrow t \approx 80059 {/eq}

Hence:

Approximately {eq}80059 {/eq} years is required to decay from {eq}10 {/eq} grams to {eq}1 {/eq} gram.

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