Respuesta :

1.37% of cesium–131 will remain after 60 days

Explanation:

From the question given above, the following data were obtained:

Half-life (t½) = 9.7 days

Time (t) = 60 days

Percentage remaining after 60 days =?

Next, we shall determine the number of half-lives that has elapsed. This can be obtained as follow:

Half-life (t½) = 9.7 days

Time (t) = 60 days

Number of half-lives (n) =?

n = t / t½

n = 60 / 9.7

Finally, we shall determine the percentage remaining. This can be obtained as follow:

Let the original amount be N₀

Let the amount remaining be N

Number of half-lives (n) = 60 / 9.7

N = N₀ / 2ⁿ

Divide both side by N₀

N/N₀ = 1/2ⁿ

N/N₀ = 1 / 2⁽⁶⁰÷⁹•⁷⁾

N/N₀ = 0.0137

Multiply by 100 to express in percentage

N/N₀ = 0.0137 × 100

N/N₀ = 1.37%

Therefore, the percentage remaining after 60 days is 1.37%

NOTE; N/N₀ is the fraction remaining.

Learn more: https://brainly.com/question/14883322

ACCESS MORE
EDU ACCESS