Questionl'équation de la désintégration du polonium $218, A(t)=A_{0}\left(\frac{1}{2}\right)^{\frac{t}{3,1}}$ , où $A$ est la quantité non désintégrée après $t$ min et $A_{0}$ , la quantité initiale. a) Après 90 s , quelle sera la quantité non désintégrée d'un échantillon initial de 50 mg ? b) En combien de temps cet échantillon dimi-nuera-t-il à $10 \%$ de son poids initial de 50 mg ? c) Ta réponse en b) changerait-elle si la masse de l'échantillon initial était modifiée. Explique ta réponse.

Studdy Solution

STEP 1

1. The decay of polonium-218 follows an exponential decay model.
2. The half-life of polonium-218 is given as 3.1 minutes.
3. We are given an initial amount $A_0 = 50$ mg.
4. We need to convert seconds to minutes where necessary.

STEP 2

1. Calculate the remaining amount after 90 seconds.
2. Determine the time for the sample to decrease to 10% of its initial weight.
3. Analyze the effect of changing the initial mass on the time to reach 10% of the initial mass.

STEP 3

Convert 90 seconds to minutes since the half-life is given in minutes:
$90 \, \text{seconds} = \frac{90}{60} \, \text{minutes} = 1.5 \, \text{minutes}$

STEP 4

Use the decay formula to find the remaining amount after 1.5 minutes:
$A(t) = A_0 \left(\frac{1}{2}\right)^{\frac{t}{3.1}}$
Substitute $A_0 = 50$ mg and $t = 1.5$ minutes:
$A(1.5) = 50 \left(\frac{1}{2}\right)^{\frac{1.5}{3.1}}$
Calculate the exponent:
$\frac{1.5}{3.1} \approx 0.4839$
Calculate the remaining amount:
$A(1.5) = 50 \times \left(\frac{1}{2}\right)^{0.4839}$
$A(1.5) \approx 50 \times 0.717$
$A(1.5) \approx 35.85 \, \text{mg}$

STEP 5

To find the time when the sample is 10% of its initial weight, set $A(t) = 0.1 \times A_0 = 5$ mg:
$5 = 50 \left(\frac{1}{2}\right)^{\frac{t}{3.1}}$
Divide both sides by 50:
$\frac{1}{10} = \left(\frac{1}{2}\right)^{\frac{t}{3.1}}$
Take the logarithm of both sides:
$\log_{10}\left(\frac{1}{10}\right) = \log_{10}\left(\left(\frac{1}{2}\right)^{\frac{t}{3.1}}\right)$
$-1 = \frac{t}{3.1} \log_{10}\left(\frac{1}{2}\right)$
Solve for $t$ :
$t = \frac{-1}{\log_{10}\left(\frac{1}{2}\right)} \times 3.1$
Calculate $\log_{10}\left(\frac{1}{2}\right) \approx -0.3010$ :
$t = \frac{-1}{-0.3010} \times 3.1$
$t \approx 3.32 \times 3.1$
$t \approx 10.29 \, \text{minutes}$

STEP 6

Analyze if changing the initial mass affects the time to reach 10% of the initial mass:
The decay formula $A(t) = A_0 \left(\frac{1}{2}\right)^{\frac{t}{3.1}}$ shows that the time to reach a certain percentage of the initial mass depends on the ratio $\frac{A(t)}{A_0}$ , not on $A_0$ itself. Thus, the time to reach 10% of the initial mass is independent of the initial mass.
The solutions are: a) After 90 seconds, the remaining amount is approximately 35.85 mg. b) It takes approximately 10.29 minutes for the sample to decrease to 10% of its initial weight. c) The time to reach 10% of the initial mass does not change with different initial masses.

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