The piece of wood from a recently cut tree shows 20 decays per minute. A wooden piece of the same size placed in a museum (obtained from a tree cut many years back) shows 2 decays per minute. If the half-life of C14 is 5730 years, then what is the age of the wooden piece placed in the museum?

Let A₀ be the initial activity of C14 in the recently cut tree, and A be the activity of C14 in the wooden piece from the museum. We are given that A₀ = 20 decays per minute and A = 2 decays per minute. The half-life of C14 is t₁/₂ = 5730 years. The formula for radioactive decay is:

A = A₀e^(-λt)

where λ is the decay constant, and t is the time elapsed. The decay constant is related to the half-life by:

λ = ln(2) / t₁/₂

Substituting the given values, we have:

λ = ln(2) / 5730 ≈ 1.21 x 10⁻⁴ year⁻¹

Now, we can solve for t using the equation:

2 = 20e^(-λt)

Dividing by 20:

0.1 = e^(-λt)

Taking the natural logarithm of both sides:

ln(0.1) = -λt

t = -ln(0.1) / λ

Substituting the value of λ:

t = -ln(0.1) / (1.21 x 10⁻⁴) ≈ 19039 years

Therefore, the age of the wooden piece placed in the museum is approximately 19039 years.