Question

The disintegration rate of a certain radioactive sample at any instant is 4250 disintegrations per minute. 10 minutes later, the rate becomes 2250 disintegrations per minute. The approximate decay constant is
(Take log₁₀1.88 = 0.274)

• A. 0.02 min^–1
• B. 2.7 min^–1
• C. 0.063 min^–1
• D. 6.3 min^–1

Answer 1

The Correct Option is C

\(A_0=4250\)
\(A=2250\)
\(A=A_0e^{−λt}\)
\(2250=4250e−λt\)
\(\frac {2250}{4250}=e^{−λt}\)
\(λ(10)=ln (\frac {4250}{2250})\)
\(λ(10)=0.636\)
\(λ=0.063\)

So, the correct option is (C): \(0.063\ min^{-1}\)