QuestionHow long (in yr) will it take for 1.00 g of Strontium-90 to decay to 0.200 g, given a half-life of 28.1 yr?

Studdy Solution

STEP 1

Assumptions1. The initial amount of Strontium-90 is1.00 g. The final amount of Strontium-90 is0.200 g3. The half-life of Strontium-90 is28.1 years4. The decay of Strontium-90 follows first-order kinetics, which is a standard assumption for radioactive decay

STEP 2

The formula for the decay of a substance following first-order kinetics is given by $= N0 \times (1/2)^{t/t_{1/2}}$ where- $$ is the final amount of the substance- $0$ is the initial amount of the substance- $t$ is the time that has passed- $t_{1/2}$ is the half-life of the substance

STEP 3

We can rearrange the formula to solve for time $t$ $t = t_{1/2} \times \log2 \left(\frac{0}{}\right)$

STEP 4

Now, plug in the given values for the initial amount $0$ , the final amount $$, and the half-life $t_{1/2}$ to calculate the time $t$.
$t =28.1 \, \mathrm{yr} \times \log2 \left(\frac{1.00 \, \mathrm{g}}{0.200 \, \mathrm{g}}\right)$

STEP 5

Calculate the time $t$ .
$t =28.1 \, \mathrm{yr} \times \log2 (5)$

STEP 6

Evaluate the logarithm and multiply by the half-life to get the final time.
$t =28.1 \, \mathrm{yr} \times2.322 =65.2 \, \mathrm{yr}$ It will take approximately65.2 years for1.00 g of Strontium-90 to be reduced to0.200 g by decay.

Was this helpful?

QuestionHow long (in yr) will it take for 1.00 g of Strontium-90 to decay to 0.200 g, given a half-life of 28.1 yr?

Studdy Solution

STEP 1

Assumptions1. The initial amount of Strontium-90 is1.00 g. The final amount of Strontium-90 is0.200 g3. The half-life of Strontium-90 is28.1 years4. The decay of Strontium-90 follows first-order kinetics, which is a standard assumption for radioactive decay

STEP 2

The formula for the decay of a substance following first-order kinetics is given by $= N0 \times (1/2)^{t/t_{1/2}}$ where- $$ is the final amount of the substance- $0$ is the initial amount of the substance- $t$ is the time that has passed- $t_{1/2}$ is the half-life of the substance

STEP 3

We can rearrange the formula to solve for time $t$ $t = t_{1/2} \times \log2 \left(\frac{0}{}\right)$

STEP 4

Now, plug in the given values for the initial amount $0$ , the final amount $$, and the half-life $t_{1/2}$ to calculate the time $t$.
$t =28.1 \, \mathrm{yr} \times \log2 \left(\frac{1.00 \, \mathrm{g}}{0.200 \, \mathrm{g}}\right)$

STEP 5

Calculate the time $t$ .
$t =28.1 \, \mathrm{yr} \times \log2 (5)$

STEP 6

Evaluate the logarithm and multiply by the half-life to get the final time.
$t =28.1 \, \mathrm{yr} \times2.322 =65.2 \, \mathrm{yr}$ It will take approximately65.2 years for1.00 g of Strontium-90 to be reduced to0.200 g by decay.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionHow long (in yr) will it take for 1.00 g of Strontium-90 to decay to 0.200 g, given a half-life of 28.1 yr?

Studdy Solution

STEP 1

Assumptions1. The initial amount of Strontium-90 is1.00 g. The final amount of Strontium-90 is0.200 g3. The half-life of Strontium-90 is28.1 years4. The decay of Strontium-90 follows first-order kinetics, which is a standard assumption for radioactive decay

STEP 2

STEP 3

We can rearrange the formula to solve for time tttt=t1/2×log⁡2(0)t = t_{1/2} \times \log2 \left(\frac{0}{}\right)t=t1/2​×log2(0​)

STEP 4

Now, plug in the given values for the initial amount 000, the final amount $$, and the half-life $t_{1/2}$ to calculate the time $t$.t=28.1 yr×log⁡2(1.00 g0.200 g)t =28.1 \, \mathrm{yr} \times \log2 \left(\frac{1.00 \, \mathrm{g}}{0.200 \, \mathrm{g}}\right)t=28.1yr×log2(0.200g1.00g​)

STEP 5

Calculate the time ttt.t=28.1 yr×log⁡2(5)t =28.1 \, \mathrm{yr} \times \log2 (5)t=28.1yr×log2(5)

STEP 6

Evaluate the logarithm and multiply by the half-life to get the final time.t=28.1 yr×2.322=65.2 yrt =28.1 \, \mathrm{yr} \times2.322 =65.2 \, \mathrm{yr}t=28.1yr×2.322=65.2yrIt will take approximately65.2 years for1.00 g of Strontium-90 to be reduced to0.200 g by decay.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

We can rearrange the formula to solve for time $t$ $t = t_{1/2} \times \log2 \left(\frac{0}{}\right)$

Now, plug in the given values for the initial amount $0$ , the final amount $$, and the half-life $t_{1/2}$ to calculate the time $t$.
$t =28.1 \, \mathrm{yr} \times \log2 \left(\frac{1.00 \, \mathrm{g}}{0.200 \, \mathrm{g}}\right)$

Calculate the time $t$ .
$t =28.1 \, \mathrm{yr} \times \log2 (5)$

Evaluate the logarithm and multiply by the half-life to get the final time.
$t =28.1 \, \mathrm{yr} \times2.322 =65.2 \, \mathrm{yr}$ It will take approximately65.2 years for1.00 g of Strontium-90 to be reduced to0.200 g by decay.