Respuesta :
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
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?
[tex]\dfrac{\lambda}{\lambda _1} = 2^{\dfrac{a}{b}}[/tex],
where λ₁ = remaining amount of sample, λ = original amount of sample, a = total time length for disintegration, b = half life.
How to find the remaining time for sample to be with 0.25 activity level?
Putting values
[tex]\lambda = 2 \: \rm curie\\\lambda _1 = 0.25 \: \rm curie\\b = 10[/tex]
[tex]\dfrac{2}{0.25} = 2^{\dfrac{a}{10}}\\\\\dfrac{a}{10} = 3\\\\a = 30[/tex]
Thus, after 30 years, the activity level of this sample will be 0.25
Thus, option C : 30 years is correct.
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