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gavmur [86]
3 years ago
14

The radioactive isotope of lead, Pb-209, decays at a rate proportional to the amount present at time t and has a half-life of 3.

3 hours. If 1 gram of this isotope is present initially, how long will it take for 80% of the lead to decay
Chemistry
1 answer:
Marina86 [1]3 years ago
3 0

Answer:

Therefore it will take 7.66 hours for 80% of the lead decay.

Explanation:

The differential equation for decay is

\frac{dA}{dt}= kA

\Rightarrow \frac{dA}{A}=kdt

Integrating both sides

ln A= kt+c₁

\Rightarrow A= e^{kt+c_1}

\Rightarrow A=Ce^{kt}         [e^{c_1}=C]

The initial condition is A(0)= A₀,

\therefore A_0=Ce^{0.k}

\Rightarrow C=A_0

\therefore A=A_0e^{kt}.........(1)

Given that the Pb_{209}  has half life of 3.3 hours.

For half life A=\frac12 A_0 putting this in equation (1)

\frac12A_0=A_0e^{k\times3.3}

\Rightarrow ln(\frac12)= 3.3k     [taking ln both sides, ln \ e^a=a]

\Rightarrow k=\frac{ln \frac12}{3.3}

⇒k= - 0.21

Now A₀= 1 gram, 80%=0.8

and A= (1-0.8)A₀ = (0.2×1) gram = 0.2 gram

Now putting the value of k,A and A₀in the equation (1)

\therefore 0.2=1e^{(-0.21)\times t}

\Rightarrow e^{-0.21t}=0.2

\Rightarrow -0.21t= ln(0. 2)

\Rightarrow t= \frac{ln (0.2)}{-0.21}

⇒ t≈7.66

Therefore it will take 7.66 hours for 80% of the lead decay.

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