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

Question

Accepted Answer

Answer: C. 19039 years Fresh wood: 20 decays/min (current activity if alive / just cut). Old museum sample: 2 decays/min. Half-life of C‑14 = 5730 years. Find the age of wooden piece in museum. Step 1: Ratio of activities \frac{A}{A_0} = \frac{2}{20} = 0.1 So, the museum sample’s activity is 10% of that in fresh wood. Step 2: Radioactive decay law A = A_0 e^{-\lambda t} \frac{A}{A_0} = e^{-\lambda t} So, 0.1 = e^{-\lambda t} t = \frac{\ln(0.1)}{-\lambda} Step 3: Decay constant \lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{5730 \,	ext{yr}} \approx 1.2097 	imes 10^{-4} \,	ext{yr}^{-1} Step 4: Solve for t t = \frac{\ln(0.1)}{-1.2097 	imes 10^{-4}} \ln(0.1) = -2.3026 t = \frac{2.3026}{1.2097 	imes 10^{-4}} t \approx 19039 \,	ext{years} ✅ Final Answer: Option C: 19039 years

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

Answer

Explanation

More JEE Physics questions