MATHEMATICS HIGH SCHOOL

A sample of radioactive waste has a half-life of 10 years and an activity level of 2 curies. After how many years will the activity level of this sample be 0.25 curie?a. 10 yearsb. 20 yearsc. 30 yearsd. 40 yearse. 80 years

Answers

Answer 1

Answer:

We can use formula for half life of radioactive substance to get the needed time interval's length.

Option: C: 30 years.

After 30 years, the activity level of this sample will be 0.25

Given that:

Length of half life = b = 10 years
Activity level = 2 curies

Let after y years, the activity level of this sample be 0.25.

What is the relationship between half life and amount remaining?

$(\lambda)/(\lambda _1) = 2^{(a)/(b)}$ ,

where λ₁ = remainingamount of sample, λ = original amount of sample, a = total time length for disintegration, b = halflife.

How to find the remaining time for sample to be with 0.25 activity level?

Putting values

$\lambda = 2 \: \rm curie\n\lambda _1 = 0.25 \: \rm curie\nb = 10$

$(2)/(0.25) = 2^{(a)/(10)}\n\n(a)/(10) = 3\n\na = 30$

Thus, after 30 years, the activity level of this sample will be 0.25

Thus, option C : 30 years is correct.

Learn more about half life of radioactive substances here:

brainly.com/question/13234158

Answer 2

Answer:

Answer:

c. 30 years

Step-by-step explanation:

The expression for half life of a radioactive substance is given as

λ/λ₁ = 2⁽ᵃ/ᵇ⁾.............. Equation 1

Where λ₁ = remaining sample, λ = original sample, a = Total disintegration time, b = half life.

Given: λ = 2 curies, λ₁ = 0.25 curies, b = 10 years.

Substituting these values into equation 1

2/0.25 = 2⁽ᵃ/¹⁰⁾

8 = 2⁽ᵃ/¹⁰⁾

2³ = 2⁽ᵃ/¹⁰⁾

Since the base in both side of the equation is the same, we can equate the base.

3 = a/10

a/10 = 3

a = 10×3

a = 30 years.

Thus it will take 30 years for the activity level of the sample to be 0.25 curie

The right option is c. 30 years

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Answers

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Answers

Answer:

Step-by-step explanation:

Sean purchased a printer and laptop

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Answers

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Answers

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Answers

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(on e2020)

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