$\left(\log _{10} 1.88=0.274\right.)$ લો.
At \(t=10\) disintegration rate \(=2250\,dpm\)
\(A=A_{0} e^{-\lambda t}\)
\(2250=4250\,e ^{-\lambda .(10)}\)
\(\Rightarrow \lambda(10)=\ln \left(\frac{4250}{2250}\right)\)
\(\Rightarrow \lambda=0.063 min ^{-1}\)