High School

A freshly prepared sample of a certain radioactive isotope has an activity of [tex]10.0 \, \text{mCi}[/tex]. After [tex]4.00 \, \text{h}[/tex], its activity is [tex]8.00 \, \text{mCi}[/tex].

Find the decay constant, [tex]\lambda[/tex].

Answer :

The decay constant is approximately -0.05575 h^(-1).

To find the decay constant, we can use the formula for radioactive decay:

A = A₀ * e^(-λt)

Where:
A is the final activity (8.00mCi)
A₀ is the initial activity (10.0mCi)
λ is the decay constant (which we need to find)
t is the time (4.00 h)

To solve for λ, we can rearrange the equation as follows:

A/A₀ = e^(-λt)

Taking the natural logarithm of both sides:

ln(A/A₀) = -λt

Now we can solve for λ:

λ = -ln(A/A₀) / t

Substituting the given values:

λ = -ln(8.00mCi/10.0mCi) / 4.00 h

Simplifying:

λ = -ln(0.8) / 4.00 h

Calculating:

λ ≈ -0.223 / 4.00 h

λ ≈ -0.05575 h^(-1)

Therefore, the decay constant is approximately -0.05575 h^(-1).

To know more about radioactive decay refer to

https://brainly.com/question/1770619

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