Question

1. the radioactive decay of a substance is governed by the equation

$n(t)=n_{0}e^{-lambda t}$
where $n(t)$ is the number of particles as a function of time, $n_{0}$ is the initial
number of particles, and $lambda$ is the decay constant.
the half - life, $t_{1/2}$, is the time when $n(t_{1/2})=n_{0}/2$. the sodium isotope
$^{22}na$ has a half - life of 2.6 years. what is the decay constant, $lambda$, for $^{22}na$?

Explanation:

Step1: Substitute into formula

Given \(N(t)=N_0e^{-\lambda t}\), when \(t = t_{1/2}\), \(N(t_{1/2})=\frac{N_0}{2}\). So \(\frac{N_0}{2}=N_0e^{-\lambda t_{1/2}}\).

Step2: Simplify the equation

Divide both sides by \(N_0\) (since \(N_0
eq0\)), we get \(\frac{1}{2}=e^{-\lambda t_{1/2}}\).

Step3: Take natural logarithm

Take the natural logarithm of both sides: \(\ln(\frac{1}{2})=-\lambda t_{1/2}\).

Step4: Solve for \(\lambda\)

Since \(\ln(\frac{1}{2})=-\ln 2\), then \(-\ln 2=-\lambda t_{1/2}\). Given \(t_{1/2} = 2.6\) years, so \(\lambda=\frac{\ln 2}{t_{1/2}}\).
Substitute \(t_{1/2} = 2.6\) into the formula: \(\lambda=\frac{\ln 2}{2.6}\approx\frac{0.693}{2.6}\approx0.2665\) per year.

Answer:

\(\lambda\approx0.2665\) per year.