Question

The decay constant of a radioactive isotope is \(1.21\times 10^{-4}\ \text{year}^{-1}\). The half-life of the isotope is $\underline{\hspace{3cm}}$ years. [round off to nearest integer]

Hint:

Remember the trio: \(N=N_0 e^{-\lambda t}\), \(t_{1/2}=\dfrac{\ln 2}{\lambda}\), and mean life \(\tau=\dfrac{1}{\lambda}\). If \(\lambda\) is per year, $t_{1/2}$ automatically comes out in years—no extra conversion needed.

Answer 1

Step 1: Start from the radioactive decay law.
\[ N(t)=N_0\,e^{-\lambda t}, \] where $\lambda$ is the decay constant. By definition of the half-life $t_{1/2}$, \[ N(t_{1/2})=\frac{N_0}{2}. \]

Step 2: Derive the half-life formula.
\[ \frac{N_0}{2}=N_0 e^{-\lambda t_{1/2}} \ \Rightarrow\ \frac{1}{2}=e^{-\lambda t_{1/2}} \ \Rightarrow\ \ln\!\left(\frac{1}{2}\right)=-\lambda t_{1/2} \ \Rightarrow\ t_{1/2}=\frac{\ln 2}{\lambda}. \]

Step 3: Insert the numerical value with correct units.
\[ \lambda = 1.21\times 10^{-4}\ \text{year}^{-1},\qquad \ln 2 = 0.693147\ldots \] \[ t_{1/2}=\frac{0.693147}{1.21\times 10^{-4}} = 5728.4876\ \text{years}. \]

Step 4: Round to the requested precision and verify.
Nearest integer $\Rightarrow\ 5728$ years. Check by back-substitution: \[ e^{-\lambda t_{1/2}}=e^{-1.21\times 10^{-4}\times 5728.49} =e^{-0.69315}\approx 0.5000, \] which matches the definition of half-life. Alternative (log$_{10}$) route (optional).
Using $\ln 2 = 2.302585\times\log_{10}2$ with $\log_{10}2=0.30103$ also gives \[ t_{1/2}=\frac{2.302585\times 0.30103}{1.21\times 10^{-4}}\approx 5.728\times 10^{3}\ \text{years}. \]

Final Answer:\quad \[ \boxed{5728\ \text{years}} \]