Question

a. Find an expression in terms of $\tau$for the half-life $t1/2$of a sample of excited atoms. The half-life is the time at which half of the excited atoms have undergone a quantum jump and emitted a photon.

b. What is the half-life of the $3p$state of sodium?

Short answer

a. $t1/2=\frac{0.693}{\lambda }\approx 0.693\tau$

b. The Half life of the$3p$state of sodium is$1.0x{10}^{9}s$.

Step-by-step solution

Part a Step 1 : Given information

We have to find an expression in terms of $\tau$for the half life $t1/2$of a sample of exited atoms.

Part a Step 2 : Simplification

We know from first order reaction that,

$\frac{dN}{dt}=-\lambda N$

Hence,

$$\begin{gathered} \int \frac{dN}{N}=-\lambda \int dt \\ {\log }_e\left(N\right)=-\lambda t+C \end{gathered}$$

If $N_t$is number of nuclei at time =$t$,than $N=N_0$when $t=0.$

Implies that , ${\log }_e\left(N_0\right)=C$

Substituting the value of C we get,

${\log }_e\left(N\right)=-\lambda t+{\log }_e\left(N_0\right)$

${\log }_e\left(N\right)-{\log }_e\left(N_0\right)=-\lambda t$

${\log }_e\left(\frac{N}{N_0}\right)=-\lambda t$

$\frac{N}{N_0}=e^{-\lambda t}$$\Rightarrow N=N_0{e}^{-\lambda t}$

the half life $t_{1/2}$is when $t=t_{1/2}$and $N=\frac{N_0}{2}$

$$\begin{gathered} \frac{N_0}{2}=N_0{e}^{-\lambda t} \\ 2=e^{-\lambda t_{1/2}} \end{gathered}$$

taking $\log$on both sides we get,

$$\begin{gathered} {\log }_e\left(2\right)=\lambda t_{1/2} \\ 0.6931=\lambda t_{1/2} \\ {t}_{1/2}=\frac{0.6931}{\lambda } \end{gathered}$$

$t_{1/2}=\frac{\ln \left(2\right)}{\lambda }=\frac{0.6931}{\lambda }\approx 0.693\tau$($\tau$=Mean lifetime)

Part b. Step 1 : Given information

We have to find the half life of$3p$state of sodium.

Part b. Step 2 : Simplification

We will find the lifetime of sodium atoms at the $3p$state using the relationship between the spectral linewidth and uncertainty principle:

$∆E\times ∆t=\frac{h}{4\pi }$

The information is full width half maximum line width of sodium atomic absorption spectrum is $\lambda =9.2\times {10}^{-14}m.$

The atomic transition from ground $3s$to excited states has a wavelength $\lambda =589.0nm.$

we will use the relation : $∆E=\frac{hc∆\lambda }{{\lambda }^2}$

plugging in the values we get,

$∆E=\frac{h.(3\times {10}^8).(9.2\times {10}^{-14})}{{(589.0\times {10}^{-9})}^2}$

Now using the uncertainty relationship , we have that

$∆t=\frac{h}{4\pi ∆E}$$=\frac{{(589.0\times {10}^{-9})}^2}{4\pi (3\times {10}^8).(9.2\times {10}^{-14})}$

$∆t=1.000\times {10}^{9}s$

The Half life of$3p$state of sodium is$1.0\times {10}^{9}s$.